EK Chem Ch7 Acid, Bases,Titration Flashcards
Titration
Titration is the process of finding / to determine the concentration of an unknown solution (the analyte) by reacting it with a solution of known concentration (the titrant). A solution of known concenteartion is used to determine the concentration of an unknown solution (titrand or analyte)
The analyte is generally placed in an Erlenmeyer flask, while the titrant is placed in a burette so that the volume of solution added can be monitored.
The titrant is added to the analyte until the endpoint is reached. Calculations are then performed to find the unknown concentration of the analyte. Titrations are typically performed for acid/base reactions but are not limited to them.
equivalence point titration
At equivalence points during the titration, the number of acid or base groups added to the solution is equivalent to the number of base/acid groups in the original unknown solution. We can calculate our unknown concentration or volume using the formula NaVa = NbVb, where N and V are the normality (mol/L) and volume of the acidic and basic solutions, respectively. It is important to convert from molarity (M) to normality (N) for polyprotic acids and polyvalent bases.
flat regions of titration curves
=represent buffering solutions (a roughly equal mix of an acid/base and its conjugate), while the steep, near-vertical sections of the curve contain equivalence point(s), which indicate that enough of the titrant has been added to completely remove one equivalent (acid or base group) from each of the original molecules in the unknown solution. Species with multiple acid or base groups (e.g. H3PO4 or Ca(OH)2) will have multiple equivalence points during the titration.
end point of titration
The final key point of any titration is the endpoint. To be successful, there must be some method for observing the endpoint of the reaction. The type of titration reaction that is being used will determine the method used for observing the endpoint. For example, in an acid-base titration, a specific pH value will be the endpoint (monitored by color-changing indicators), while for precipitation reactions, the endpoint is realized by the appearance of a precipitate. Regardless of the details of the reaction involved, the goal of titrations is always to use known volumes/concentrations to determine unknown volumes/concentrations.
Based on Figure 2, what is the approximate pKa of the carboxylic acid group in glycine?
answer is 2.4, In the Henderson-Hasselbalch equation, pH = pKa + log[B]/[A], where B is the conjugate base of the weak acid (represented by A). If [B] = [A], the pH will equal the pKa. This occurs for the first plateau in the titration curve for glycine shown in Figure 2. Estimating the pH at the middle of the first plateau gives a value closest to 2.4, as shown below.
Ka
Acidic species can be strong (for the exam, assume that strong acids completely dissociate in water) or weak. When an acid dissociates, it releases a proton to make the surrounding solution acidic. However, weak acids only partially dissociate and at equilibrium coexist in a deprotonated state (A-) and a protonated state (HA), according to the equation HA ⇌ H+ + A-. The concentration ratio of products and reactants is constant given fixed conditions and is called the acid dissociation constant (Ka). Ka is defined by the equation below. Ka = [H+][A-]/[HA]
ACID IONIZATION CONSTANT or ACID DISSOCIATION CONSTANT
Ka equation
Ka = [H+][A-]/[HA] The square brackets indicate the concentration of the respective aqueous species. Ka expresses how easily an acid releases a proton (i.e. its strength). In addition, this equation shows how the dissociation state of weak acids vary according to the [H+] level in the solution. A commonly tested family of acids on the MCAT are carboxylic acids (those containing –COOH), such as lactic acid and amino acids, which normally have a Ka of approximately 10-3 to 10-6. As this shows, expressing acidity in terms of the Ka constant alone involves inconvenient numbers that are not very intuitive.
pka
Therefore, pKa was introduced as an index to express the acidity of weak acids, where pKa = -logKa. For example, the Ka values for lactic acid (HC3H5O3) and nitrous acid (HNO2) are 8.3 × 10-4 and 4.1 × 10-4, respectively. The pKa values for these acids are 3.1 and 3.4, respectively, which are simpler expressions that are easier to understand and compare. The smaller the pKa value, the stronger the acid (since as X increases, pX decreases). Therefore, the pKa values above tell us that lactic acid is a stronger acid than nitrous acid.
pka for diprotic
For diprotic amino acids (i.e. those amino acids for which the side chain is neither acidic nor basic), the pI can straightforwardly be captured as the average of the two pKa values.
pka for triprotic
For triprotic amino acids (i.e. those that have an acidic or basic side chain), the pI can be obtained by averaging the two acidic pKa values for acidic amino acids or the two basic pKa values for basic amino acids. Another way of saying this is that for acidic amino acids, the pI is the average of the two lowest pKa values; for basic amino acids, the pI is the average of the two highest pKa values. Be careful to avoid averaging all three values given; the pI is always the average of two, not three, pKa values.
buffer 1
Acid-base buffers confer resistance to a change in the pH of a solution when hydrogen ions or hydroxide ions are added or removed to solution. An acid-base buffer typically consists of a weak acid and its conjugate base. ***The most important buffer to know for the MCAT is the bicarbonate buffer system, which is shown below. H2O (aq) + CO2 (g) ⇌ H2CO3 (aq) ⇌ H+ (aq) + HCO3− (aq) Buffers resist pH changes best when the pH values are at or near the pKa value for the acid/base used, because that is when the conjugate acid and base have equal concentrations. Optimal buffering occurs when the pH is within approximately 1 pH unit from the pKa value of the system. The pH of a buffer solution can be calculated using the Henderson-Hasselbalch equation: pH = pKa + log [conjugate base] / [acid].
bicarbonate buffer system
***The most important buffer to know for the MCAT is the bicarbonate buffer system, which is shown below. H2O (aq) + CO2 (g) ⇌ H2CO3 (aq) ⇌ H+ (aq) + HCO3− (aq) Carbonic acid (H2CO3) has the conjugate base of HCO3−. Buffers work because the concentrations of the weak acid and its salt are large compared to the number of protons or hydroxide ions added or removed. When protons are added to the solution from an external source, some of the bicarbonate in the buffer is converted to carbonic acid, using up the protons added; when hydroxide ions are added to the solution, protons are dissociated from some of the carbonic acid in the buffer, converting it to bicarbonate and replacing the protons lost.
bicarbonate buffer system 2
On Test Day, you may see the bicarbonate buffer concept tested from a chemical, biochemical, or biological perspective. For example, hyperventilation (rapid shallow breathing) results in excess CO2 being expelled from the blood, causing the pH to rise. In response, the buffer needs to release more H+ to lower the pH back to physiological norms. An additional fact to be aware of is that other mechanisms in the body are also used to regulate pH, since carbonic acid works best at a pH below physiological conditions, because its pKa1 (pKa1 = 6.3, pKa2 = 10.3) is much lower than the normal pH of blood (7.4).
What is the approximate pH of a saturated aqueous solution of hydrochloric acid whose molarity is 10.6 M?
= -1 Hydrochloric acid is a strong acid and completely dissociates in aqueous solution. In this solution, the hydronium ion concentration is 10.6 M, which can be approximated as 10 M to make the math easier. The pH is the -log of the hydronium ion concentration: -log[10] = -log[101] = -1. While the typical pH range is normally thought of as ranging from 0 to 14, if the concentration of hydronium ion is greater than 1 M, negative pH values are possible. It is also possible to have pH values greater than 14, i.e. if the hydroxide concentration is greater than 1 M.
How acidic or basic a solution is can be expressed in terms of pH or pOH……..
are defined as follows: pH = −log [H+] and pOH = −log [OH−]. For example, a solution with an H+ concentration of 10 ^−4 M will have a pH of 4, and a solution with an OH− concentration of 10^−9 M will have a pOH of 9. pH and pOH values can be estimated given a certain concentration using the following shortcut: p(N × 10−M) = (M−1).(10−N), such that a solution with an H+ concentration of 4 × 10−8 will have a pH = (8−1).(10−4) = 7.6.
Kw
pH and pOH are related to each other through the equation pH + pOH = pKw, where Kw is the autoionization constant of water (Kw = [H3O+][OH−] = 1 × 10−14 at 25°C). Thus, at 25°C, pH + pOH = 14. Thus, a solution with a low pH will automatically have a high pOH, and vice versa. pH is most commonly used, but it is important to be able to interconvert pH and pOH values if needed.
Acidic solutions have a high …………
Acidic solutions have a high H+ concentration and a low pH.
Basic solutions have a low…..
Basic solutions have a low H+ concentration and a high pH.
Bronsted Lowry Acid
proton donor
Bronsted lowry base
proton acceptor
lewis acid
electron pair acceptor
lewis base
electron pair donor
Take B, flip it get D, nice way to remember electron donor******
lewis acid/base
no H plus moving, definition more broad!!
strong acid
100% ionization, donate protons very easily** say process occurs 100% so 100% ionization, equilbirum so far to the right one arrow going right* everything turns into our products!
ex HCl
H2O + HCl –> H3O+ and Cl-
Ka= [H3O+] X [Cl-]/ [HCl] leave out water** is a pure liquid concentration doesnt change so LEAVE IT OUT of equilbirum expression**, super strong number in numerator and small number in denomatinor, so gives you a very big/high number of Ka, Ka much much greater than 1 here* so that is how we recognize a strong acid, an acid ionization constant much much greater than 1**
titration 2
definition
procedure for determining concentration of a solution**
equilbirum expression
Concentration of products/ concentration of reactants
stronger acid….
weaker conjucate base, so Cl- is a weak conjugate base!
H2O+ HCl–> Cl- + H3O+, Cl- weak conjugate base
can think about competing base strength Bronsted Lowrey base water here accepting proton, Cl- anion tries to pick up proton from hydronium (H3O+) for reverse reaction BUT since HCl so good at donating ions Cl- not good at accepting
water much stronger base than chloride anion*
acetic acid
CH3COOH bronsted lowrey acid
H2O BLB
CH3COOH + H2O–> (eq arrows) CH3COO- + H3O+
gives us acetate anion (CH3COO-) plus H3O+**
WEAK ACID don’t donate protons very well, so acetic acid will stay mostly protonated so when think about this reaction coming to an equilbrium will have a high concentration of reactants here**
Ka= [CH3COO-] [H3O+]/ [CH3COOH]
so very small number on top/ large bottom
small number divide by large number gives you Ka value much less than 1* so this value will be much less than 1, and that is how we recognize a weak acid look at Ka value**
autoionization constant for water
Kw, autoionization constnat or ion product ocnstant so not Ka*
H2O + H2O-> H3O+ + OH-
Kw= products/reactants, but dont worry about reactants because pure water so = [H3O+] [OH-]
= 1.0 X 10^-7 M (concentration of hydronium numbers, H3O+) concentration of hydroxide at 25 degrees celcius of pure water is 1.0 X 10^-7
Kw= 1.0 X 10 ^-14, since much less than 1 equilbirum lies far to the left, why have such low concentration of ions***
concentrations of hydronium equal to concentration of hydroxide [H3O+]= [OH-] water neutral*** so solution neutral
H3O+ > OH-
acidic solution
[H3O+]=[OH-]
neutral
[H3O+] < [OH-]
basic
Lemon juice ex. H3O+= 2.2 X 10^-3 M, OH-?
2.2 X 10^-3 M X = 1.0 X 10^-14
because [H3O+][OH-]= 1.0 X 10^-14
x=OH-= 4.5 X 10^-12 M, concentration of H3O+ is greater than concentration of OH-, so this is an acidic solution, ex lemon juice is acidic
ph of water
-log [H3O+]
pH H2O, -log (1.0 X 10^-7)
pH= 7.00 NEUTRAL
or if pH = 3.82, 3.82= -log [H3O+]
to solve that 10^-3.82 = 1.5 X 10^-4 concentration of hydronium ions, if get two significant figures here, need two significant figures for our answer
how to get concentration of H3O+
10^-ph***
calculate the pH of an aq ammonia solution with a [-OH] of 2.1 X 10^-3 M
want us to find pH, which equals - log[H3O+]
BUT they gave us the concentration of -OH, so use the equation [H3O+][-OH] = 1.0 X 10^-14**
plug in, X (2.1 X 10^-3) = 1.0 x 10^-14
so X, the H3O+ concentration is 4.8 X 10^-12. , so use that number to calculate pH
pH= -log (4.8 X 10^-12)= 11.32, so 2 significant figures in log so 2 sig figs to right of decimal point, pH>> 7 its basic!
what is pOH equal to?
pH= -log (H+) or -log(H3O+)
pOH= -log[-OH]
pOH of water
calculate the pH of an aq ammonia solution with a [-OH] of 2.1 X 10^-3 M cnt. 2
pOH= -log (1.0 X 10^-7) same as did for pH above, so pOH= 7.00
remember pH + pOH = 14
ex if pOH= -log(2.1 X 10^-3)
pOH= 2.68, so pOH is 2.68 once again 2 sig figs in 2.1 log so 2 sig figs after decimal place, sovle for pH so the pH plus POH= 14, so 14-2.68 gives us 11.32=pH same pH got when used it a different way so doesnt matter when calculate ammonia solution same anser either way*
pKa
= - logKa
HF
hydrofluoric acid
Ka=3.5 X 10^-4
WEAK ACID, less than 1* greater than acetic acid (Ka 1.8 X 10^-5) and methanol 2.9 X 10^-16
largest value for Ka, but smaller value pKA lower value for pKa more acidic your acid***
methanol
weak base equilbrium
Kb
acid will be BLAcid ammonia will be BLB Nh3, protonate ammonia form ammonium ion, NH4+ and-OH
so if ammonia functioned as bl base, so ammonium conjugate acid to Nh3, water was blacid so conjugate base is -OH
formula
B+ H2O–> -OH + BH
Kb= base dissociation constant, base ionization constant
products over reactanst! [BH][-OH]/ B * reactants leaving out water!
Kb 2
higher value for Kb stronger base, more of your products you will make here***
weak bases
Nh3 ammonia
C6H5NH2 aniline, ammonia KB 1.8 X 10^-5
aniline is 4.3 X 10^-10, so ammonia stroner base than aniline even though both weak bases
pKb= -log Kb= -log (1.8 X 10^-5)
pKb= 4.74
calculate pH of 0.500 M solution of NH3 aq
NH3 + H2O– [NH4+] + -OH
NH3 NH4+ -OH
I 0.500 0 0
C -x +x +x
E 0.500- x x x
- 8 X 10^-5 = (x)(x)/0.5000 -x x<<< 0.500
- 8 X 10^-5 = x^2/0.500
x= 0.0030 M, x refers to OH concentration** hydroxide anions
find pOH= -log (OH- ions concentration) = 2.52, pH+ 2.52 = 14.00
pH= 11.48
Kw
equilbirum constant for net reaction, add rxn together to get net reaction multiple equilbirum constants to get net equ constant
Ka X Kb = Kw
Kb for methylamine is 3.7 X 10^-4, calculate Ka value for the methlyammonium ion
Ch3NH2 is the methylamine
Ch3NH3” methylammonium ion
Ka X Kb = Kw
Ka X (3.7 X 10^-4)= 1.0 X 10^-14
Ka= 2.7 x 10^-11
Kb for methylamine is 3.7 X 10^-4, calculate Ka value for the methlyammonium ion
part 2 (further step)
Ka X Kb = Kw
logKa X Kb = log Kw
-logKa + -logKb = -logKw
pKa + pKb = 14.00
if pKa is =-log (2.7 X 10^-11) (from part 1)
pka = 10.57 (2 sig figs becuase 2.7 2 sig figs) pka of methylammonium ion, asks for pKb for methlamine
to find pkb use equation–> 10.57 + pKb= 14.00, so pKb = 3.43
so this is the relatonship btw Ka and Kb**
strong acid/base properties of salts
acid reacts with base to give you water and salt, neutralization reaction
HCl + NaOH–> H2O + NaCl “salt”
Strong acid= HCl
Strong base = NaOH
with a very strng acid get a very weak conjugate base, so Cl- anion cannot take protons from water very well, so pH unaffected and our pH = 7, when have salt formed from strong acid adn strong base* these salts form neutral solutions, so your pH should be equal to 7!
Ch3COOH+ NaOH—>
salt properties
weak acid and strong base–> H2O and CH3COO- and Na+ ions of sodium and acetate can bond
in aqueous solution of sodium acetate, know pH of water is 7 so if think about what we have in solution sodium cations will not react with water and affect pH, but acetate anions will —-> CH3COO- (Bronsted-Lowry Base) + H2O (Bronsted lowry acid)–> -
if take H+ away from H2O get OH-, so what has happened in oru solution?? well because acetate anion present inc OH ions present in solution, so pH is no longer 7 with increased pH the pH should be greater than 7, so aqeous solution of sodium acetate has pH greater than 7; so if have salt formed from weak acid adn strong base get basic solutions with pH greater than 7***
strong acid weak base
salt properties
HCl + NH3–> NH4+ and Cl- (ammonium chloride salt)
HCl= strong acid; NH3= weak base
NH4+ reacts with water and effect pH, pH of water 7 so Cl- anions will not effect pH; so ammonium ion is an acid, focuses as BLA donates a proton to water, H2O funcions as bronsted lowry base and accept a proton
NH4+ + H2O–>get NH3 and H3O+
so in our solution we have inc H3O+ concentration, so dec the pH from 7 so have an acidic solution so the pH is now less than 7** so in aqueous solution of ammonium chloride less than 7, SO SALT FORMED FROM strong acid adn weak base gives you an acidic solution!
common ion effect
ex CH3COOH + H2O–> H3O+ + (conjugate base, acetate) CH3COO-
acetic acid donates proton to H2O to form H2O+
if inc Ch3COO-
if add CH3COO- Na+ inc concentration of acetate anion, according to Le Chatlier’s inc concentratio of one of your products inc equilibrium to left, so some of acetate anion react with hydronium ion which dec hydronium ion concentration and therefore inc pH of resulting solution, so acetate anion is common ion! common on effect two sources of acetate anion, one CH3COOh other is sodium acetate Ch3COO- Na+ you added in, so two sources and expect pH to be higher than just acetic acid alone***
calculate the pH of a solution that is 1.00 M Ch3COOH (Ka = 1.8 X 10^-5) and 1.00 M Ch3COONa
common ion effect problem
CH3COOH + H2O –> H3O+ + CH3COO-
anothre source for acetate anions with CH3COONa acetate anion comes from two sources one ionization of acetic acid, other is sodium acetate add in to create solution
at equilbirum would have: look in picture, can write equilbirum expression using Ka becuase acetic acid donating proton to water
Ka= [H3O+]X [CH3COO-]/ CH3COOH
notice in picture gaining acetate anion concentration so add in 1.00 to products** becuase dealing with CH3COONa
*1.00 + x and 1.00-x close to 1 since x<<1.00
*remember x represents concentration of hydronium ions* so for pH =-log(1.8X 10^-5)= 4.74
acetic acid only 1.00 M Ch3COOH pH= 2.38, different pHes pH higher for two sources again for acetate ion, with addition of sodium acetate and acetic acid
acetic acid common ion problem
part 2
Calculate the pH of a solution that is 0.15 M NH3, Kb= 1.8 X 10^-5 and 0.35 NH4NO3
ammonia= weak base, takes proton from water**
equation: NH3 + H2o–> NH4+ + -OH
so ammonium nitrate is another source of NH4+** that also gets an inc in concentration of 0.35 in ICE diagram
ammonium two sources so it is a common ion*
so after ICE step 2: Kb= [NH4+][-OH]/ NH3
Kb= (0.35 + x) (x) / (0.15 -x) if x<<<1.00, x is an extremely small number so can get rid of it
1.8 X 10^-5=(0.35) x/0.15
x (x is the -OH concentration)= 7.7 X 10^-6 M
step 3, to calculate pH, pOH= -log (7.7 X 10^-6), pOH= 5.11
pH+ pOH= 14.00
pH+ 5.11= 14.00
pH= 8.89
buffer solutions 2
resist changes in pH***
buffer solution 3
inc concentration of H+ in solution, decreases concentration of A-
effectively we have removed protons from solution; how a buffer can resist changes in pH if add acid!
if add base= this time if we add a strong base we will inc the concentration of -OH ions in solution, we have an acid that can react with our base, some weak acid present here so OH ion picks up a proton from HA so OH- and H+ gives us H2O will lose some of our weak acid* if OH- takes a proton away from HA left with A- so inc amount of A- we have in solutions** so once again we have effectively buffered again against the change in pH becuase Oh added reacts with acid present effectively removed hydroxide ions from solution
in a buffer substantial amounts of HA and A- present
how to derive Hendleson Hasselbach equation
at equilbirum this reaction= HA + H2O –> H3O+ + A-
Ka= [H3O+][A-]/HA, can separate these two into log terms–>
**if log AB= log A + log B, then -log AB= -log A - log B
- log Ka= -log [H3O+] - log [A-/HA]
- log Ka= pka and -log[H3O+]= pH
soo pka= pH - log [A-]/[HA]
HH equation= pH= pKa + log [A-]/[HA]*** very useful when doing buffer calculations**
what is the pH of a buffer solution that is 0.24 M NH3 and 0.20 M NH4Cl?
use HH equation
Ka= 5.6 X 10^-10
base is concentration of A- is NH3 concentration in buffer solution is 0.24 M, over concentratoon of acid NH4+ concentration is 0.20 M
pKa= -log (5.6 X 10^-10)= 9.25, so pH= 9.25 + log [0.24/0.20]
so pH= 9.33**
step 2- NOW if 0.005 mol of NaOH is added to 0.50 L of the buffer solution (0.20 M NH3 and 0.20 M NH4Cl) what is the resulting pH?
0.005 mol/ 0.50 L= concentration of NaOH= 0.01 M (NaOH is strong base so its also our concentration of -OH ions in solution)
base will react with acid, so OH- acts with NH4+–> reacts and goes to completion, NH4+ donates H to OH- becomes H2O and NH3 (ammonia)
NH4+ + OH- –> H2O + NH3
think about our concentrations–>
- 20 (NH4+) + 0.01 (OH-) –> H2) + 0.24 (NH3)
- 0.01 -0.01 + 0.01 (since Nh4+ turns into NH3 if lose 0.01 gain 0.01 ________________________________
= 0.19 M 0 0.25 M
pH= 9.25 + log (0.25 M/ 0.19M)
so pH= 9.25 + .119 =9.37
so added base pH went up a little bit but a very small amount, so shows us how buffer solution resists drastic changes in pH
when add some acid in buffer solution
If 0.03 mol of HCL is added to 0.50 L of the buffer solution (0.24 M NH3 and 0.20 M NH4Cl) what is the resulting pH?
same buffer solution as last problem, NH3 and NH4Cl
0.03 mol/ 0.50 L= 0.06 M, so that is really concentration of H3O+ since strong acid think H+ and Cl- really
acid add reacts with base present in buffer solution, so in this case base will react!
NH3 + H3O+–> NH4+ (picks up proton) + H2O (donates proton) 0.24 0.06 M –> 0.20
(-0.06) (-0.06) (all of it will react)—> +0.06 (because whatever lose of NH3 gain for NH4+ since NH3 turns into NH4+)
__________________________________
0.18 M 0 0.26 M
pH= 9.25 + log (0.18)/ 0.26, so pH= 9.25 -0.16 = pH= 9.09
remember NH4+ is the acid here*****
- so for original buffer solution had original pH of 9.33, so pH went down a little bit but our buffer solution is able to resist drastic changes in pH*
titration 2
experiment process outlined
- can find concentartion of HCl doing titration, put acid in the flask
- next need to add a few drops of an acid base indicator ex phenolpthalein turns pink in presence of base but colorless in acid* if in presence of acid clear solution, then standard solution known concentration of NaOH that is 0.100 M
- allow NaOH to drip into flask with our indicator and acid, acid will react get neutralization reaction–> HCl + NaOH–> H2O + NaCl
- so if add certain volume of base, see our solution turn light pink and stays light pink means all of the acid has been neutralized by the base, and we have a tiny amount of base present causing acid base indicator to remain pink; tiny excess of bsae means weve neutralized all the acid present, when indicator changes color is called end point of a titration
- change in volume on top tube is the amount of base we used for titratioN! say it is 48.6 mL then took 48.6 mL of our base to completely neutralize acid we had present so can now calculate concentration of HCl
next part continued……
end point of titration 2
when the acid/base indicator changes color is called end point of a titration
phenolpthalein
- an acid base indicator example used in titration
- turns pink in presence of base but colorless in acid* if in presence of acid clear solution
molarity=
mol/L
titration 2 card continued
….to calculate concentration of acid present
goal= how many moles of base used to neutralize acid present?
if start with NaOH= 0.100 M (molarity = mol/L)
we can take our volume here 48.6 mL and convert it to L, JUST MOVE DECIMAL PLACE 3 places to the left 0.0486 L***
equals mol/ 0.0486 L
x = moles of NaOH necessary to neutralize acid present, so 0.100= x/ 0.0486
= 0.00486 mol of NaOH used in our equation, then look at balanced equaton have a 1:1 molar ratio in equation==> (1) HCl+ (1) NaOH–> H2O + NaCl
equivalence point= just amount of standard solution has been added to completely react with solution being titratd, at equivalence point all acid has been neutralized so completely reacted and because we have a 1 to 1 mole ratio, use that many 0.00486 mol of NaOH must be how many moles of HCl in solution
0.00486 mol HCl present in flask before started titration, know that becuase of 1:1 mol ratio
convert mL to L
take volume 48.6 mL and convert it to L BY JUST MOVING DECIMAL PLACE 3 places to the left 0.0486 L***
mol/L= mol/ 0.0486 L
equivalence point 2
= just amount of standard solution has been added to completely react with solution being titratd
at equivalence point all acid has been neutralized so completely reacted and because we have a 1 to 1 mole ratio, use that many 0.00486 mol of NaOH must be how many moles of HCl in solution
final step calculate original concentration of HCl started with=
now original volume of HCl was 20 mL, so need to convert that into L, move decimal place 3 places get = 0.0200 L
[HCl] = 0.00486 mol/ 0.0200 L= 0.243 M
*short cut to find concentration (M) of HCl*
short cut way= Molarity X volume of acid = Molarity X Volume of base used
MV = MV
if molarity of acid –> X 0.20 mL = 0.100 M (concentration of base) X 48.6 mL (volume of base used to completely neutralize acid) so mL would cancel and solve for x
M= 0.100 M X 48.6 mL/ 0.20 mL= 0.243 M
this acts well when dealing with strong acid and strong base in a 1:1 molar ratio**
find concentration of acidic solution (the solute)
In a titration, 27.4 mL of a 0.0154 M solution of Ba(OH)2 is needed to neutralize 20.0 mL of HCl. What was the concentration of the acid solution?
barium hydroxide concentration = 0.0154 M, molarity = mol/L
Step 1.
0.0154 = X/ 0.0274 L
x= 4.22 X 10^-4 or 0.000422 mol of barium hydroxide Ba(OH)2
Step 2.
now Ba(OH)2 + HCl–> H2O + BaCl2- ions are Ba 2+ and Cl-1
Ba 2+ corses with Cl-1 so BaCl2
balance equation= Ba(OH)2 + 2 HCl –> 2 H2O + BaCl2
mole ratio for Ba(OH)2 to HCl, so for every 1 mol of Ba(OH)2 we have 2 mols of HCl so we know we need 0.000422 mol of Ba(OH)2 needed in our concentraton, so we had twice as many of HCl
so one way: 0.000422 mol X 2= HCl M
or ratio
Ba(OH)2/ HCl = 1/2 = 0.000422/ x
get x = 0.000844 mol of HCl we have at our equivalence point
Step 3.
now just calculate concentration of acid solution
[HCl] = 0.000844 moles / 0.0200 L (the 20.0 mL started with)
HCl= 0.0422 M
if try to do MV= MV with not 1:1 molar ratio
Ba(OH)2 and HCl previous problem
MV base = MV acid
(0.0154 M)(27.4 mL)= x (20.0 mL)
x= 0.021098 M
can see that is NOT THE CORRECT ANSWER AS ABOVE< here get concentration half of concentration got the longer way, of 0.0422 M
so have to multiply answer by 2** to get the correct answer X = 0.0211 M X 2= 0.0422 M, can figure out by looking at balanced equation but tricky for a lot of students but only use short cut way when mol ratio 1;1 not for when mol ratio is not 1:1
titration of a strong acid with a strong base
titration start with 20.00 mL of 0.500 M HCl with adding strong bas 0.500 M NaOH
as add base pH of course increases*
HCl strong acid, ionizes 100%
so equation is HCl + H2O –> H3O+ + Cl- (100% ionization, so 0.500 M of HCl is the same concentration of H3O+ ions have in solution)
0.500M–> pH= -log [H3O+] = pH = -log (0.500 M) = 0.301
titration of a strong acid with a strong base 2
what is pH if add 10.00 mL of 0.500 M NaOH?
we know base adding, OH- ions will neutralize hydronium ions alreayd present, so calculate first how many mols of H3O+ ions are present
conver 20.00 mL to L, 0.0200 L
H3O+= 0.500 M, since Molarity = moles /L
0.500 M= mol/ 0.0200 L = mol= 0.0100 H3O+
step 2. add NaOh strong base, concentration same as OH- ions since strong base, so 0.500 M of NaOH is the same for OH- concentration, convert 10 mL to liters
[-OH]= 0.500 M= mol/ 0.010 L
solve for moles! 0.500 X 0.010 = 0.005 moles of hydroxide ions we have
step 3. so found moles of both acid and base, neutralization reaction that occurs H3O+ + OH –> H2O + H2O
know started with 0.0100 M of H3O+ and added 0.005 M of OH-
all OH- will react and neutralize same amount of hydronium ions, so going to loose all of our OH ions because base will completely react with acid!
H3O+ + OH—>
- 0100 0.00500
- 0.00500 - 0.00500
___________________
=0.00500 mol and 0
left w/ zero moles of our base!
so 0.0100- 0.005= 0.005 so that is how many moles of H3O+ are left over; we have neutralized half of H3O+ present, so another half left over! 1/2 of acid neutralized so 1/2 of acid left!
new concentration of H3O+ in solution = mol/L
= 0.0050 mol/ .030 L = [H3O+] = 0.17 M
****new volume added 10 to 20.00mL= 0.03 L
continuation of titration of strong acid with strong base 3.
new concentration of H3O+ in solution = mol/L
= 0.0050 mol/ .030 L = [H3O+] = 0.17 M
****new volume added 10 to 20.00mL= 0.03 L
pH= -log (H3O+), of course use concentration of H3O+ ions!
pH= -log (0.17 M) = pH= 0.77
so after add 10.00 mL of base, this is the pH on titration curve
Part C of problem titration strong acid with strong base
(notecard 4 for this problem)
What is pH after the addition of 20 mL of a 0.500 M NaOH?
how many mol of NaOH adding to original solution? 0.500 M NaOH is also concentration of OH in solution since talking about a strong base!
[-OH]= 0.500 M, M= mol/L……….= 0.500 M = mol/ 0.0200
so 0.500 M X 0.0200 L = 0.0100 mol OH- ions adding to original solution; which is exactly amount of mol of acid found prviously 0.0100 mol H3O+ or H+
H3O+ + OH- –> 2 H2O
- 0100 M (acid) 0.0100 M (base)
- so this time we have enough base to completely neutralize our acid, everything is reacting in a 1:1 ratio here, so all of our base will react and completely neutralize our acid! so when that happens the pH should be just the pH of water, which we know is equal to 7
- another ex of this idea- add NaOH + HCl react–> H2O + NaCl, acid and base neutralize each other so just pH of water will be left because sodium and chloride anions dont react enough with wter to change pH, so after adding 20.00 mL of base equal to 7!
this is the equivalence point! added enough mol of base to completely neutralize acid present!
part d what is hte pH after the addition of 20.20 mL of 0.500 M NaOH?
step 1. like before find moles of OH- added, so concentration of OH- = 0.500 M
- 500 M = mol / 0.02020 L (converted from 20.20 mL)
- 5 X 0.02020 = .0101 how many moles of OH- ions we have
step 2.
remeber OH- reacted with H3O+; this time more base than acid! all acid will be completely neturalized and will have none left, and most of the base will react! only started with 0.01 mol of H3O+ ions so this time more base than we have acid
H3O+ OH- –> 2H2O
- 0100 0.0101
- 0.0100 -0.0100
= 0 0.0001 mol, very small amount of base left over and all acid neutralized
step 3. total volume
started titration with 20 mL of acid solution, now added 20.20 mL more= 40.20 mL
step 4. concentration of OH- ions =mol /L = 0.0001/ .04020 L
= .002 M of OH- ions
step 5. find pH, now can do this by finding pOH= -log [OH-] = -log (.002)………so =pOH= 2.7
so then pH + pOH = 14.00, plug in! pH= 11.3!!
Titration of a weak acid with a strong base 1
titration of 50 mL of 0.200 M CH3COOH (so acidic solution) with 0.0500 M NaOH
a) What is the pH after the addition of 0.0 mL of NaOH?
do only need to think abotu weak acid equilibrium problem here! CH3COOH donates proton to H2O
CH3COOH + H2O—> H3O+ +CH3COO- (acetate anion)
I 0.200 M 0 0
C -x +x +x
E 0.200 -x x x
Ka= (x^2)/ (0.200-x)
do the usual….. 1.8 X 10^-5 = x^2/ 0.200
x (H3O+ concentration) = 0.0019 at equilibruim, find pH so plug in this concentration into pH calculation = -log [H3O+]
pH = -log (0.0019 ) = 2.72
SO BEFORE we have added any base, 0.0 mL of NaOH pH should be 2.72!! can see on titration curve
titration of a weak acid with a strong base 2
titration of 50 mL of 0.200 M CH3COOH (so acidic solution) with 0.0500 M NaOH
b) What is the pH after the addition of 100 m of 0.0500 M of NaOH?
if strong base NaOH is 0.0500 M, so -OH ion concentration is 0.0500 M
step 1.
- 0500 M = mol/ 0.1000 L (100.0 mL to L, move decimal over 2 places)
- 0500 X 0.1000 = 0.00500 mol -OH
step 2.
started with 50 mL of 0.2 M of CH3COOH
[CH3COOH] = 0.200 M = mol/ 0.0500 L
0.2 X 0.05 = 0.0100 mol of CH3COOH
step 3.
OH- ions will neutralize acid present, so get neutralization reaction, OH takes proton from CH3COOH (Acetic acid)
CH3COOH + OH- —> H2O + CH3COO-
- 0100 mol 0.005 mol 0
- 0.005 -0.005 mol +0.005
_______________________________
- 005 mol acid 0 mol + 0.005 mol
- OH- ions neutralize CH3COOH and so all of OH- recats, use up all of it; so end up with small amount 0.0050 M of acid left over, how much acid not neutralized! if lose that much CH3COOH gain that much acetate anion, finish with 0.005 mol of acetate anion
step 4. after neutralization, find volume to find concentration of CH3COOH and CH3COO-, have moles for both so then obvi need to find volume
-adding 100 mL started with 50 mL, so new volume 150 mL or 150 L
[CH3COOH] = 0.0050 mol/ 0.1500 L = 0.033 M concentration
[CH3COO-] same exact numbers, 0.0050 mol/ 0.15 L = 0.033 M
so equal concentrations of weak acid and conjguate base, so buffer solution as drip OH ions into original acidic solution slowly forming buffer solution, here equal concentrations! goal to find pH so since buffer solution now can use HH
pH= pka + log [A-]/[HA], CH3COOH Ka = 1.8 X 10^-5
pKa= 4.75, the -log (Ka)
so pH= 4.74 + log (0.033/0.033) aka 1, so log (1)= 0
…… so pH = 4.74* half equivalence point; neutralized half of the acid, and half of the acid remains, pH equal to pKa at this point!
titration of a weak acid with a strong base 3
Problem cnt. titration of 50 mL of 0.200 M CH3COOH (so acidic solution) with 0.0500 M NaOH
PART C) What is the pH after the addition of 200 mL of 0.0500 M of NaOH?
step 1. so [-OH]= 0.0500 M = mol/ 0.200
0.05 X 0.2 L = 0.0100 mol of -OH
[CH3COOH]= 0.200 M = mol / 0.0500 mL, 0.2 x 0.05 = 0.0100 mol of CH3COOH; so notice same number of mol of acid as we do of base, base will neutralize the acid! so started with 0.0100 mol of acetic acid and same for base! mole ratio is 1:1, base completely neutralizes all of acid
step 2. neutralization reaction = Oh takes acidic proton
CH3COOH + -OH –> H2O + CH3COO-
- 0100 0.0100 (all base) reacts 0
- 0.0100 -0.0100 +0.0100
_____________________________________
=0 0 0.0100 mol of CH3COO-
so we loose all of acid and base, so this is our equivalence point ofr this titration!!! if we are losing acetic acid converting into acetate so go frm 0 mol of CH3COO-, the acetic acid we lose turns into acetate
step 3.
[CH3COO-] = 0.0100 mol/ 0.2500 L = 0.0400 M acetate ions in solution at equivalence point, and reacts with water (know equation just add water gets to equilbirum so acetate anion acts as base and takes a proton from water, turning it into CH3COOH + OH- (from proton took from H2O))
step 4.
initial concentration of acetate is 0.04 M so just do the same thing as above; acetate here acting as base! so write Kb**
CH3COO- + water –> CH3COOH + OH-
I 0.04 M 0 0
C - x +x + x
E 0.0400 -x x x
Kb= (x)(x)/ 0.0400-x
Ka for acetic acid is 1.8 X 10-5
so KaKb= 1.0 x 10-14, and Kb = 5.6 x10^-10
5.6 x 10^-10= x^2/ 0.0400….. x= concentration of OH ions, 4.7 X 10^-6
step 5.
pOH= -log[-OH], -log(4.7x10^-6) = 5.33
pH+ pOH= 14 so pH + 5.33 = 14
pH= 8.67 at equivalence point, but titration of weak acid with strong base** so basic salt solution at equivalence point, pH in basic range*
REMEMBER
pKA= -log KA
so if Ka= 10^-4, -log (10^-4)
pKA= 4
Khan academy Acid/base questions
Which of the following best describes a chemical species that is measured to have Kb = 3.2 x 10^{-18} ?
Whether a chemical is an acid or base depends on whether the Ka or Kb is greater!
Because KaKb = Kw = 10^-14, for this chemical the Ka= 3.125 X 10^3 if use Kb to solve for Ka. This suggests it is an acid.
Because Ka≫1, the species is known as a strong acid.***
Khan academy acid/base Q
Suppose a large organic molecule X is classified as a Lewis acid, while another large molecule Y is classified as a Bronsted-Lowry acid.
Which of the following most accurately describes a similarity in their behaviors in solution?
=Both molecules will tend to acquire a net negative charge
*A Bronsted-Lowry acid tends to donate hydrogen ions, which have positive charges in solution. DONATES H+
This leaves the rest of the molecule with a net negative charge.
*A Lewis acid tends to accept electrons in solution, leaving the molecule with a net negative charge
Thus both X and Y will tend to acquire a net negative charge in solution
Khan academy Acid/base Q
Which of the following describes the pH of an equilibrated, stoichiometric mixture of ammonia, NH3 and hydrochloric acid HCl?
- Ammonia is a weak base
- Hydrochloric acid is a strong acid
- The hydrochloric acid will donate a proton to the ammonia to form ammonium, NH4+
- Ammonium reacts with water to form hydronium, NH4+ + H2O–> NH3 + H3O+
- A mixture of a strong acid with a weak base acquires an overall acidic pH < 7***
A mixture of a strong acid with a weak base acquires a what pH overall?
an overall acidic pH < 7***
remember 2 for math for this chapter
1. pKa = -log (Ka), if Ka= 1 X 10-1 …. pKa= 1**** becuse -log (1 x 10-1) = 1
2. 10/ .1 = 100, so log (100) = 2
3. log (10^4) = 4, so if pH= -log(H3O+)= -4
pH 2
- pH is the negative logarithm of the molar concentration of hydrogen ions, H+ or H3O+
- Equivalently, the pH is the logarithm of the molar concentration of hydronium
chemistry of buffers and buffers in our blood
- Remember, pH of a solution is dependent on the concentration of hydronium ions or simply put as H+!!!
- since we define a buffer as= A buffer is an aqueous solution that resists changes in pH upon the addition of an acid or a base…. adding water to a buffer or allowing water to evaporate from the buffer does not change the pH of a buffer significantly. *****
- Buffers basically constituent a pair of a weak acid and its conjugate base, or a pair of a weak base and its conjugate acid*
How do you prepare a buffer?
A buffer is essentially prepared in two ways
- mixing a large volume of a weak acid with its conjugate base (eg. acetic acid – acetate ion, CH3COOH - CH3COO-
- mixing a large volume of weak base with its conjugate acid (eg. ammonia – ammonium ion, NH3 - NH4+
Phosphate buffer is a very commonly used buffer in research labs. It falls under the second category. It’s made up of a weak base HPO4 ^2- and its conjugate acid H2PO4-. The pH of a phosphate buffer is usually maintained at a physiological pH of 7.4
Phosphate buffer
Phosphate buffer is a very commonly used buffer in research labs. It falls under the second category. It’s made up of a weak base HPO4 ^2- and its conjugate acid H2PO4-. The pH of a phosphate buffer is usually maintained at a physiological pH of 7.4
How does a buffer work?
Buffer, as we have defined, is a mixture of a conjugate acid-base pair that can resist changes in pH when small volumes of strong acids or bases are added.
- When a strong base is added, the acid present in the buffer neutralizes the hydroxide ions (OH-)
- When a strong acid is added, the base present in the buffer neutralizes the hydronium ions (H3O+)
ex. When a strong acid is added, (buffer contains acetic acid and conjugate base acetate ion) the acetate ions neutralize the hydronium ions producing acetic acid (which is already a a component of the buffer)
on the other hand, when a strong base is added, acetic acid consumes the hydroxide ions producing acetate ions (which is also already a constituent of the buffer).
Suppose we have a buffer that contains a weak base NH3, and its conjugate acid, ammonium ion NH4+ ….
When a strong acid is added, ammonia neutralizes the hydronium ions producing ammonium ions (which is already a component of the buffer).
On the other hand, when a strong base is added, ammonium ions consume the hydroxide ions producing ammonia (which is also already a constituent of the buffer).
So, in order to be an effective buffer….
- The number of moles of the weak acid and its conjugate base must be significantly large compared to the number of moles of strong acid or base that may be added.
- The best buffering will occur when the ratio of [HA] to [A-] is almost 1:1. In that case pH= pKa! Buffers are considered to be effective when the ratio of [HA] to [A-] ranges anywhere between 10:1 and 1:10.
Buffering system of blood
Maintaining a constant blood pH is critical for the proper functioning of our body.
The buffer that maintains the pH of human blood involves a carbonic acid **** H2CO3 - bicarbonate ion HCO3- system!
When any acidic substance enters the bloodstream, the bicarbonate ions neutralize the hydronium ions forming carbonic acid and water. Carbonic acid is already a component of the buffering system of blood. Thus hydronium ions are removed, preventing the pH of blood from becoming acidic.
On the other hand, when a basic substance enters the bloodstream, carbonic acid reacts with the hydroxide ions producing bicarbonate ions and water. Bicarbonate ions are already a component of the buffer. In this manner, the hydroxide ions are removed from blood, preventing the pH of blood from becoming basic.
buffering system of blood 2
In the process of neutralizing hydronium ions or hydroxide ions, the relative concentrations of carbonic acid and bicarbonate ions start fluctuate in the blood stream.
**But this slight change in the concentrations of the two components of the buffering system doesn’t have any adverse effect; the critical thing is that this buffering mechanism prevents the blood from becoming acidic or basic, which can be detrimental.
The pH of blood is maintained at ~7.4 by the carbonic acid- bicarbonate ion buffering system****
Why is it so critical to maintain the pH of our blood?
Believe it or not, if our blood pH goes to anything below 6.8 or above 7.8, cells of the body can stop functioning and the person can die. This is how important the optimum pH of blood is!
Enzymes are very specific in nature, and function optimally at the right temperature and the right pH
If the pH of blood goes out of range, the enzymes stop working and sometimes enzymes can even get permanently denatured, thus disabling their catalytic activity. This in turn affects a lot of biological processes in the human body, leading to various diseases.
titration of a weak acid with a strong base 4
Problem cnt. titration of 50 mL of 0.200 M CH3COOH (so acidic solution) with 0.0500 M NaOH
PART D) What is the pH after the addition of 300 mL of 0.0500 M of NaOH?
Step 1.
[OH-]= 0.0500 M= mol/ 0.300 L
= 0.0150 mol of -OH, part C say needed 0.01 M of OH- ions to completely neutralize acid we had present
so 0.0150 -0.0100 mol= 0.0050 mol of OH- left over after neutralization, after all acid has been neutralized, goal is to find the pH
step 2. find pOH if get concentration of OH after neutralization has occured** 300 mL + 50 mL= 350 mL, convert to L
[OH-]=0.0050/.350 L = 0.014 M
step 3. pOH= -log (0.014)= pOH= 1.85
pH+ pOH= 14, so pH+ 1.85 = 14; pH= 12.15 after add 300 mL of our base!
strong acid ex.
sulfuric acid, H2SO4
nitric acid, HNO3
hydrochloric acid, HCL
weak acid ex.
acetic acid we have talked about at length… CH3COOH
hydrofluoric acid, HF
oxalic acid (COOH2)
strong base ex.
NaOH
potassium hydroxide, KOH
lithium hydroxide, LiOH
weak bases ex.
ammonium hydroxide NH4OH
ammonia NH3
Rule of thumb is:
=Weak acids have strong conjugate bases, while weak bases have strong conjugate acids.
for ex, if HA is a weak acid, then its conjugate base A- will be a strong base. Similarly, if A- is a weak base, then its conjugate acid HA will be a strong acid.
analyte, titrand
vs. titrant
= known volume, unknown concentration
vs.
known concentration and known volume
titrant and analyte 2
Titrant and analyte is a pair of acid and base.
Acid-base titrations are monitored by the change of pH as titration progresses.
Equivalence point 3
= point in titration at which the amount of titrant added is just enough to completely neutralize the analyte solution.
At the equivalence point in an acid-base titration, moles of base = moles of acid and the solution only contains salt and water.
Indicator 2
=For the purposes of this tutorial, it’s good enough to know that an indicator is a weak acid or base that is added to the analyte solution, and it changes color when the equivalence point is reached i.e. the point at which the amount of titrant added is just enough to completely neutralize the analyte solution.
The point at which the indicator changes color is called the endpoint. So the addition of an indicator to the analyte solution helps us to visually spot the equivalence point in an acid-base titration.
methyl red indicator changes
red to yellow
ph ~4-6
you want to choose an indicator close to your equilvanece point1!
bromothymol blue
changes color yellow to blue
ph of 6-7.6 or 6-8
when blue color persists reached end of titration and great way to approximate equilvaence point; you have matched end point of titration with equivlence point
phenolphthalien
strong acid strong base, v steep titration curve pH of about 4 to a pH of about 10
can use other two acid base indicators methyl red becuase methyl red changes from red to yellow range about 4-6 and gives you a good approximation of equilvanece point or could use phenolphthalien changes to magenta in range ph 8.2-10
steep curve can really use any indicator
how to choose an indicator
chose one that changes color around equilvaence point! like strong acid weak base, use methyl red because changes color ph 4-6!
weak acid strong base use phenolphthalein because ph>7 and that has a ph of 8-10**
ex titration of a strong acid with a weak base
Suppose our analyte is hydrochloric acid HCl (strong acid) and the titrant is ammonia NH3 (weak base). If we start plotting the pH of the analyte against the volume of NH3 that we are adding from the burette dropwise
h3O+ slowly starst getting consumed by NH3, analkyte is still acidic due ot predominance of H3O+ ions.
Point 3: This is the equivalence point (halfway up the steep curve). At this point, moles of NH3 = moles of HCl in the analyte. The H3O+ ions are completely neutralized by NH3. But again do you spot a difference here??? In the case of a weak base versus a strong acid, the pH is not neutral at the equivalence point. The solution is in fact acidic (pH ~ 5.5) at the equivalence point. Let’s rationalize this.
At the equivalence point, the solution only has ammonium ions NH4+ and chloride ions Cl-. but again if you recall the ammonium ion NH4+ is the conjugate acid of the weak base NH3. So Nh4+ is a relatively strong acid (weak base NH3 has a strong conjugate acid) and thus NH4+ will react with H2O to produce hydronium ions making the solution acidic.
After the equivalence point, NH3 addition continues and is in excess, so the pH increases. NH3 is a weak base so the pH is above 7, but is lower than what we saw with a strong base NaOH (case 1).
A titration curve can be used to determine:
…..
1) The equivalence point of an acid-base reaction (the point at which the amounts of acid and of base are just sufficient to cause complete neutralization).
2) The pH of the solution at equivalence point is dependent on the strength of the acid and strength of the base used in the titration.
- For strong acid-strong base titration, pH = 7 at equivalence point
- For weak acid-strong base titration, pH > 7 at equivalence point
- For strong acid-weak base titration, pH < 7 at equivalence point
titration curve for weak acid weak base
ex titration curve of weak acid strong base
50 mL of 0.5 M barium hydroxide are required to fully titrate a 100 mL solution of sulfuric acid. What is the initial concentration of the acid?
The formula for barium hydroxide and sulfuric acid are Ba(OH)2 and H2SO4
The balanced reaction is Ba(OH)2 + H2 SO4 –> 2 H2 O + Ba SO4
**Because the stoichiometric ratio of the reactants is 1:1, we can use the shortcut formula Mb Vb = Ma Va, MbVb=MaVa
Ma = Vb/Va Mb, 50/100 X .5 M= .25 M
Which of the following describes the equivalence point on a graph of pH versus the amount of titrant added to a solution?
The point where the magnitude of the slope of the curve is greatest
1 M of a weak acid HZ with a Ka = 10e-8 equilibrates in water according to the equation NZ + H2O –> H3O+ + Z-.
What is the pH of the solution at equilibrium?
Titration curves exhibit an asymptote at very large volumes of added titrant. Which of the following experimental parameters determines the location of this asymptote?
= the pH of hte titrant
Because this question asks about the limiting case of an excess of titrant, you can pick any titrant volume that is in excess of that needed to obtain the equivalence point. Not knowing anything about the species being titrated, a safe bet is to use the case where an infinite amount of titrant has been added to the solution.
In the limit of an infinitely amount of titrant has been added to the solution, the pH should become just the pH of the titrant—the original solution gets diluted away.
We thus expect the asymptotic pH of a titration curve to approach that of the titrant.
Which of the following molecules or ions is an amphoteric species?
A.
H2SO4
B.
HSO4-
C.
SO42-
D.
ClO4-
ALL DIPROTIC INTERMEDIATES can act as acid or base so HSo4-, or HCo3- etc. they can accept a proton or give one away! the starting materials or ending materials can only be an acid or base becasue they can not do both!!!
A.
H2SO4
Amphoteric species can act as either acids or bases. H2SO4 can only act as an acid.
B.
HSO4-
Amphoteric species can act as either acids or bases. HSO4- has the ability to donate its proton (acting as an acid) or to accept a second proton (acting as a base).
C.
SO42-
Amphoteric species can act as either acids or bases. SO42- can only act as a base.
D.
ClO4-
Amphoteric species can act as either acids or bases. ClO4- can only act as a base.
What will be the effect on the pH of 1 L of pure water at 298 K if heat is added to the system?
A.
The pH will decrease.
B.
The pH will increase.
C.
The pH will remain the same.
D.
The pH will increase, then decrease.
A.
The pH will decrease.
The dissociation of water into H+ and OH- is an endothermic process, so increasing the temperature will yield more H+ and OH- ions(products). While the water will still be neutral, both pH and pOH will be lowered. (Note that many students have the misconception that pH + pOH is always equal to 14! This is not true; instead, pH + pOH = pKw, which is equal to 14 only under standard conditions. Raising the temperature increases Kw, which decreases pKw. Since pKw is now lowered, both pH and pOH can decrease, and both do so to the same extent.)
B, C and D are wrong because
The dissociation of water into H+ and OH- is an endothermic process, so increasing the temperature will yield more H+ and OH- ions(products).
Q. 8 Qbank BP 11-25
A student needs to make a buffer solution at a pOH of 4.80. Of the following acids, the student should use:
A.
hydrocyanic acid (Ka = 6.2 × 10-10).
B.
hypochlorous acid (Ka = 3.5 × 10-8)
C.
acetic acid (Ka = 1.8 × 10-5).
D.
hydrofluoric acid (Ka = 7.2 × 10-4).
A. hydrocyanic acid (Ka = 6.2 × 10^-10). CORRECT
To manufacture a buffer, an acid should be selected with a pKa as close to the desired pH as possible. However, watch out – this question asks about a pOH of 4.80, which is a pH of 9.20. Also, the given values are Kas, not pKas. To convert, remember that pKa = -log(Ka). We can estimate negative logs by finding the values that each of these Kas falls between. For hydrocyanic acid, 6.2 × 10^-10 is between 1 × 10^-10 and 1 × 10^-9, making its pKa between 9 and 10.
B.
hypochlorous acid (Ka = 3.5 × 10-8)
The pKa for this acid lies between 7 and 8, which is best suited for a buffer solution with a pOH between 6 and 7.
C.
acetic acid (Ka = 1.8 × 10^-5).
wrong b/c -this would be best acid to make a buffer at pH 4.80, but the question asks about pOH.
D.
hydrofluoric acid (Ka = 7.2 × 10^-4).
wrong b/c- The pKa for this acid lies between 3 and 4, which is best suited for a buffer solution with a pOH between 10 and 11.
