*High Yield GChem: Acids and Bases

Question 1

What type of acid/base makes coordinate covalent bonds?

What are ligands? Chelates?

What is Fe2+?

What are atoms without lone pairs usually classified as? Without H?

Answer 1

Lewis acid/bases makes coordinate covalent bonds bc lewis acids are e- pair acceptors (electrophile) and lewis bases are e- pair donors (nucleophiles)
Lewis bases are also ligands and chelates
Fe2+ has a pos charge and is a lewis acid (wants e-), B would also be an example bc wants more e- so that is can fill octet

Atoms w/o LP are usually not basic
Atoms w/o H can be acids if e- deficient or with large pos charges

Question 2

Name strong acids (if they are not on this list on MCAT they are weak acids)
Name THE weak acid

Answer 2

HBr, HI, HCl, HClO4, H2SO4, HNO3

The weak acid = HF

Question 3

Acid-ionization/acid-dissociation constant? So related to this what would we see for a strong acid?

Base ionization/base-dissociation constant?

Answer 3

Ka
If Ka > 1 then products are favored and strong acid
The larger the Ka value the stronger the acid; the smaller the Ka value, the weaker the acid

Kb, the larger the Kb, the stronger the base; the smaller the Kb value, the weaker the base

Question 4

What is the Ka equation for HA (aq) + H20 (l) -> H3O+ (aq) + A- (aq)
Kb? B(aq) + H2O(l) -> HB+ (aq) + OH-(aq)

Answer 4

Ka = [H30+][A-]/[HA]

Kb = [HB+][OH-]/[B]

Question 5

**Strong bases

weak bases?

Answer 5

Group I hydroxides
Some group II hydroxides (Sr(OH)2, Ba(OH)2, Ca(OH)2)
Group I oxides
Metal amides (ex. NaNH2)

• OH-
• OR-
• NH2-
• NR2-
• H-
• R-
• O2- = dibasic
• can be dibasic if paired with a 2+ metal cation (ZnEt2)

weak bases include ammonia and amines and conjugate bases of weak acids

Question 6

The conjugate base of a strong acid is a___(strong, weak non existant) base
The conjugate base of a weak acid is a____ (strong, weak non existant) base

Answer 6

The conjugate base of a strong acid has no basic properties
The conjugate base of a weak acid is weak base, the weaker the acid, the stronger the conjugate base

The same idea applied to bases

1) The conjugate acids of a strong base has no acidic properties in water
2) The conjugate acid of a weak base is weak acid and the weaker the base the stronger the conjugate acid
ex. conjugate acid of NH3 is NH4+ which is a weak acid

Question 7

Polyprotic

Amphoteric
is water amphoteric?

what is the conjugate base of a weak polyprotic acid (characteristic)

What happens every time a polyprotic acid donates a proton

Answer 7

Polyprotic -> More than one proton to donate
Amphoteric -> When a substance can act as an acid or base (amino acids, H2O) , the conjugate base of a weal polyprotic acid is always amphoteric, bc it can either donate or accept another proton
Water is amphoteric (can lose or gain H)
every time a polyprotic acid donates a proton, the resulting species will be a weaker acid then predecessor

Question 8

Water is _____

What is autionization (self-ionization of water)? What is the ion-product constant of water?

Answer 8

Water is amphoteric (can lose or gain H)

H20 (l) + H2O (l) -> H3O+ (aq) + OH- (aq)

Question 9

*Kw is know as the ion product of water, write equation and what this value is at 25ºC
What is the concentration of H3O+ and OH- ions in pure water?

If we add 0.002 HCl to water to create a 1 liter solution, how many moles of H3O+ will it make? How many moles of OH-?

Answer 9

Kw = [H3O+][OH-]
Kw = 1 * 10^-14

Kw = x^2
1 * 10^-14 = x^2
x = 1 * 10^-7

0.002 moles of H3O+ bc it is strong acid so will completely dissociate 
Kw = [H3O+][OH-]
[OH-] = 1 * 10^-14 / [H3O+]
[OH-] = 1 * 10^-14 / [0.002]
[OH-] = 5 * 10^-12

Question 10

pH equation? and rewritten? pOH equation?
Other pH and pOH combined equation?
*What is the pH of 1L of
2.9 * 10^-4 M Ca(OH)2

Answer 10

pH = -log[H+]
[H+] = 10^-pH
pOH = -log[OH-]
pH + pOH = 14

pOH = -log(2 * 2.9 * 10^-4) 
pOH = -(-4) 
3 < pOH < 4
14-3 > pH > 14-4
11 > pH > 10

Question 11

What is the pH if [H+] = 10^-7

Answer 11

pH = 7

Question 12

If [H+] = 6.2 * 10^-5 what is the pH?

Answer 12

In general, if y is a number b/w 1 and 10, and you’re told that [H+] = y * 10 ^-n, where n is a whole number then pH will be b/w n-1 and n
so pH is b/w 4 and 5 (but Im pretty sure if just peep exponent and guess 5, you would be correct

Question 13

*The _____ the pKa value, the stronger the acid

The _____ the pKb: the lower the pKb value, the stronger the base

Answer 13

lower

lower

Question 14

What equation is true for any acid-base pair?

Answer 14

KaKb = Kw = 1 *10^-14

Question 15

Equation relating pKa and pKb?

Answer 15

pKa + pKb = 14

Question 16

What is Ka equation for strong acids when have to do problems with it

Answer 16

Ka = [H30+][A-]/[HA] BUT don’t need to plug into denominator bc acid so strong that it all dissociates into product, basically no reactants but this is not the case for weak acids

Question 17

**Weak acid calculations

Let’s say 0.2 mol of HCN added to H2O to create a 1 liter solution, what is the pH? Ka = 4.9 * 10 ^-10

Answer 17

Can make an ICE table see pg.241 for this example
answer is that can do

Ka = [H+][CN-]/[HCN] = x^2/(0.2 - x)
x represents the mole dissociated which is very small (can eliminate x from denominator when K < 10^-4 which is mostly the case for MCAT) 
so real equation you use is 
Ka = x^2/ 0.2 = 4.9 * 10^-10
and get x = 1 * 10^-5 so pH is about 5

Question 18

How would you solve square root of 4.6 * 10^-5?

Answer 18

move decimal over to get
46 * 10^-6
then square root of 46 times square root of 10^-6 and get
about 7 * 10^-3 (remember you add exponents when multiply in scientific notation

Question 19

Neutralization rxn

exo or endo

Answer 19

No matter how weak an acid or base is, when mixed with an equimolar amount of strong base or acid, we can expect complete neutralization
All neutralization rxns have the same value for exothermic “heat of neutralization”

Question 20

A salt is a _____ compound
I SKIPPED PAGES 243, 244, 245
An acidic salt contains an ion that is a _____
A basic salt contains an ion that is a ______
Knowing this, what do we have to check to know if salt acidic or basic?

MgBr2 is a basic or acidic or neutral salt?
NH4Cl?
FeCl3?
Li2CO3?

Answer 20

ionic
An acidic salt contains an ion that is a weak acid
A basic salt contains an ion that is a weak base

For acids -> check the cations (makes sense bc want to give up H)
- Group I and II cations essentially not acidic
-Stronger acid than water for example: NH4+, Be2+, Cu2+, Zn2+, Al3+, Cr3+, Fe3+
So when see super pos thing thats a hint it could be acid

For bases -> check the anions

• Cl-, Br-, I- are essentially not basic SO**conjugate base of a strong acid)
• Stronger base than water for example conjugate base of a weak acid

MgBr2 -> neutral
NH4Cl -> acidic
FeCl3 -> acidic
Li2CO3 -> basic

Question 21

What is a buffer? What equation is linked to this?

Answer 21

Buffers are mixtures of conjugate acid/base pairs, minimize pH changes
Resists changing pH when a small amount of acid or base is added. The buffering capacity comes from the presence of a weak acid and its conjugate base (or a weak acid and its conjugate base) in roughly equal concentrations

Henderson-Hasselbach
pH = pKa + log([WB]/[WA])
If more WA, then pH lower than pKa

** want weak acid and weak base together so if have weak acid, then use CONJUGATE BASE, if have weak base, then have CONJUGATE ACID**

pg. 246 for an example

Question 22

What is the Henderson-Hasselbalch equation for acid? for base? What does it tell you?

What does the ideal buffer look like?

Peep the equation and you can see that when pH < pKa, then the acid is protonated (WA dominates in the equation)
When pH > pKa, then acid is deprotonated (WB dominates in the equation)

Which of the following compounds could be added to a solution base of HCN to create a buffer?
a) HNO3
b) CaCl2
c) NaCN
d) KOH
As hydrogen ions are added to an acidic buffer solution, what happens to the concentrations of undissociated acid and conjugate base?

Answer 22

pH = pKa + log ([conjugate base]/[weak acid])

pOH = pKb + log ([conjugate acid]/[weak acid])

This equation tells you the buffer pH

[Weak acid] = [conjugate base], so pH = pKa

Answer is C bc looking for conjugate base, CN-

As hydrogen ions are added the conjugate base A- reacts with added H+ to form HA, so the conjugate base decreases and the undissociated acid increases

Question 23

What is an indicator?

Answer 23

A weak acid that undergoes a color change when it’s converted to its conjugate base

Question 24

Indicators
What pKa is good range for indicators?
What equation can you use for indicators?

Answer 24

pKa of +/- 1 range

Ex. Ka = [H3O+][A-]/[HA]
then rearrange
[H3O+]/Ka = [HA]/[A-] so if see a lot of [HA] may see one color and if a lot of [A-] a lot of other color

Question 25

What is acid-base titration?

What would one not use for titration?

Answer 25

Experimental technique used to determine the identity of an unknown weak acid (or weak base) by determining its pKa (or pKb). Also used to determine concentration of any acid or base solution

Add titrant (strong acid or base) or known identity and concentration to solution of unknown base or acid (one never titrates a acid with another acid/ base with base)

titrant is slowly added

Question 26

**NaOH + HF -> Na+ + F- + H2O
When the amount of titrant is 0, the pH is just the pH of the original, pure solution (OG acid or base, NOT the one being added on X-axis
In diagram HF is titrated with NaOH (added to solution) so as HF added, the pH will _____
Buffer region?
Equivalence point? Is this example basic or acidic here? How can you tell that about this ex and others (for OG weak acid titrated with strong base, weak base titrated with strong acid, for strong acid titrated with strong base or strong base titrated with strong acid)? So what can the pH at the equivalence point tell us?
Where is half equivalence point? What does it tell us?

Answer 26

increase
Buffer region is plato, pH changes gradually bc HF was getting neutralized

Equivalence point = drastic pH increase at which moles OH- added = moles H+ initially present, when equal concentrations of titrant and sample -> [H+] = [OH-]
in this ex. just enough NaOH was added to completely neutralize all HF (all HF reacted with NaOH, so we just have F- and Na+)
This solution is basic bc Marky said that at this point since HF is weak acid, all of it has reacted and it’s conjugate base is left over so basic solution

For weak acid titrated with strong base -> the equivalence point will occur pH > 7
For weak base titrated with strong acid -> the equivalence point will occur pH < 7
For strong acid titrated with strong base or strong base titrated with strong acid pH = 7
The pH at the equivalence point can tell us weather the acid or base we are titrating is weak or strong

Half-equivalence point is middle of flat buffer region, it is half as much as titrant
It represents when in this ex [HF] = [F-] and henderson-hasselbach tells us the
pH = pKa of HF

Question 27

The number of equivalence points is equal to:

What does the weak base titration curve look like

Answer 27

the number of ionizable hydrogens the acid can donate

Weak base curve follows same general pattern expect its flipped bc at volume of 0 it starts at a higher pH

Question 28

What formula can one use with titration and to tell how much base to add to acidic solution or how much acid to add to basic solution in order to cause complete neutralization?

Answer 28

a * [A] * Va = b * [B] * Vb
a -> number of acidic hydrogens per formula unit
b -> how many H+ ions the base can accept

The “MaV formula”

Question 29

Acetylsalicylic acid, also known as aspirin, possesses a pKa of 3.5. What is the pH of a 1 M solution of acetylsalicylic acid?

A. 1.8
B. 3.5
C. 5.3
D. 7

Answer 29

5.3
Find Ka
Make ICE table 
solve for x
then plug into pH equation and rearrange

Question 30

What is the isoelectric point?

Answer 30

average of 2 half equivalence points (which are each at buffer plato part of graph bc they rep pKa of OG substance there, thing being added to)
It is the zwitter ion, when aa is neutral, pH = pI

Question 31

What is an indicator?

What is the end point of the titration curve?

Answer 31

An indicator is a weak acid that is one color in its protonated form and other in its deprotonated form, since it is a weak acid it as a pKa associated with it and when the pKa is equal to the pH of the solution it is in, it will start to deprotonate, so you select an indicator with a pKa that’s close to your target pH and see when you have added enough of your titrant when you observe a color change, so using indicators allows us to monitor titrations w/o machines and see exact moment when color change occurs, some indicators are better than others, ex the brom. blue has pKln = 7
equivalence point is moment when color change occurs

*Within 1 unit of the pKa of the indicator, the end point represents the point of color change of the indicator
end point
The point in the titration process which is indicated by color change of the indicator is called endpoint.

Question 32

What is the pOH of a mixture made by adding 150 mL of 0.20 M sodium hydroxide to 50 mL of 0.20 M hydrochloric acid?

A. 13.0
B. 12.5
C. 1.5
D. 1.0

Answer 32

Can see ipad goodnotes for work or ask Marky

Question 33

What happens to buffer pH if water added?

Answer 33

Nothing!

Question 34

A 2 M solution of the potent organic pollutant sodium methylthiolate (NaSCH3) is titrated to its endpoint with 2 M HCl. Which of the following best approximates the pH of the resultant solution? (pKb of SCH3– = 3.6)

A. –log (1 × 10–3.6)
B. –log (1 × 10–5.2)
C. –log (2 × 10–5.2)
D. –log (1 × 10–1.8)

Answer 34

B.
A 2 M solution of the potent organic pollutant sodium methylthiolate (NaSCH3) is titrated to its endpoint with 2 M HCl. –log (1 × 10–5.2) best approximates the pH of the resultant solution. (pKb of SCH3– = 3.6)

The result of the titration will be a 1 M solution of methylthiol, HSCH3 since the initial 2 M solution has been doubled in volume during the titration. Since we know the pKb of its conjugate base SCH3– is 3.6, then we know the pKa of HSCH3 is 10.4. Therefore, at equilibrium:

10–10.4 = (x)(x)/(1–x)

Where x = [H+] = [SCH3–], at equilibrium. As the resultant value of x is negligibly small compared to 1, we ignore x in the denominator and say that x = [H+] = [SCH3–] = (10–10.4)(0.5) giving 1 × 10–5.2. The answer –log (2 × 10–5.2) would be correct if the resultant solution was 2 M in concentration. The answer –log (1 × 10–1.8) is the result of using the pKb of the methylthiolate anion in the calculation rather than the pKa of HSCH3, and –log (1 × 10–3.6) would be the result of a similar a calculation using this pKb value, while not taking the square root.

Question 35

A 2 M solution of the potent organic pollutant sodium methylthiolate (NaSCH3) is titrated to its endpoint with 2 M HCl. Which of the following best approximates the pH of the resultant solution? (pKb of SCH3– = 3.6)

A. –log (1 × 10–3.6)
B. –log (1 × 10–5.2)
C. –log (2 × 10–5.2)
D. –log (1 × 10–1.8)

Answer 35

B.
A 2 M solution of the potent organic pollutant sodium methylthiolate (NaSCH3) is titrated to its endpoint with 2 M HCl. –log (1 × 10–5.2) best approximates the pH of the resultant solution. (pKb of SCH3– = 3.6)

The result of the titration will be a 1 M solution of methylthiol, HSCH3 since the initial 2 M solution has been doubled in volume during the titration. Since we know the pKb of its conjugate base SCH3– is 3.6, then we know the pKa of HSCH3 is 10.4. Therefore, at equilibrium:

10^–10.4 = (x)(x)/(1–x)

Where x = [H+] = [SCH3–], at equilibrium. As the resultant value of x is negligibly small compared to 1, we ignore x in the denominator and say that x = [H+] = [SCH3–] = (10^–10.4)(0.5) giving 1 × 10–5.2. The answer –log (2 × 10^–5.2) would be correct if the resultant solution was 2 M in concentration. The answer –log (1 × 10^–1.8) is the result of using the pKb of the methylthiolate anion in the calculation rather than the pKa of HSCH3, and –log (1 × 10^–3.6) would be the result of a similar a calculation using this pKb value, while not taking the square root.

Question 36

Addition of sodium acetate to a solution of acetic acid will cause the pH to:

A. increase due to the common ion effect.
B. remain constant because sodium acetate is a buffer.
C. remain constant because sodium acetate is neither acidic nor basic.
D. decrease due to the common ion effect.

Answer 36

A.
Addition of sodium acetate to a solution of acetic acid will cause the pH to increase due to the common ion effect.

Sodium acetate is a basic compound, because acetate is the conjugate base of acetic acid, a weak acid (“the conjugate base of a weak acid acts as a base in water”). The addition of a base to any solution, whether it is buffered or not, will increase the pH. The answer is “increase due to the common ion effect.”

Question 37

The resulting solution made from the combination of 50 mL of 1.0 M LiOH with 50 mL of 1.0 M HBr will be identical in all respects to 100 mL of:

A. 0.5 M LiBr.
B. 1 M LiBr.
C. a saturated solution of LiBr(s).
D. 2 M LiBr.

Answer 37

A.
The resulting solution made from the combination of 50 mL of 1.0 M LiOH with 50 mL of 1.0 M HBr will be identical in all respects to 100 mL of 0.5 M LiBr.

The best way to approach this question is to ignore OH– and H+, because they will neutralize each other when mixed. Thus, in this question we are mixing 50 mL of 1.0 M Li+ and 50 mL of 1.0 M Br–. Since the solution volume is doubling, the concentration of each ion is expected to decrease by a factor of 2. Therefore, the correct choice is “0.5 M LiBr”.

Question 38

A titration of which of the following aqueous hydrogen halides with NaOH will show an equivalence point above pH = 7?

A. HBr (aq)
B. HCl (aq)
C. HI (aq)
D. HF (aq)

Answer 38

D.
A titration of HF with NaOH will show an equivalence point above pH = 7.

HF is the only weak acid of the hydrogen halide series, thus, when titrated with a strong base, it has a basic equivalence point. All the other choices have equivalence points at or very near pH = 7.

Question 39

Enough HF (Ka = 7.4 × 10–4) is added to water to create a pH = 2.1 solution. The addition of which of the following would have the least impact on the pH of this solution?
  	A.  NH4F 
  	B.  PbF2 
  	C.  Na2CO3
  	D.  NaF

Answer 39

B.
Enough HF (Ka = 7.4 × 10–4) is added to water to create a pH = 2.1 solution. The addition of PbF2 would have the least impact on the pH of this solution.

Both NaF and NH4F are soluble salts, and will decrease the solubility (acidity) of HF due to the common ion effect, so both of these answer choices can be eliminated. Na2CO3 is a soluble salt and CO32– is the conjugate base of a weak acid, so it will act to increase the pH. PbF2 is essentially insoluble in water, and therefore will have the least effect on the pH, making it the best answer.

Question 40

Which one of the following statements concerning 50 mL of basic solution composed of 0.2 M NaH2PO4 and 0.2 M Na2HPO4 is correct?

A. The concentration of Na+ is 0.4 M.
B. The pH of the solution will not change with the addition of 20 mL of water.
C. The addition of 1 mL of 0.1 M HCl to the reaction will result in a pH of 2.
D. With the addition of ammonia, a precipitate should be formed.

Answer 40

B.
The correct statement concerning 50 mL of basic solution composed of 0.2 M NaH2PO4 and 0.2 M Na2HPO4 is the pH of the solution will not change with the addition of 20 mL of water. This solution is a buffer. The pH of a buffer does not change with the addition or removal of water.
see pics 6/10/21 for what I think the dissociation of each is

Question 41

A 2 M solution of the potent organic pollutant sodium methylthiolate (NaSCH3) is titrated to its endpoint with 2 M HCl. Which of the following best approximates the pH of the resultant solution? (pKb of SCH3– = 3.6)

Answer 41

For the first question, when the base is titrated to its endpoint, it is all converted to the conjugate acid, HSCH3. For a pair of conjugates, pKa + pKb = 14. Therefore, the pKa of HSCH3 must be equal to 10.4. Since pKa = -log(Ka), you can put a negative sign in front of the pKa and turn it into an exponent to get Ka = 10^-10.4. Once you have the Ka of the weak acid, there are two equivalent ways to get the pH of a weak acid solution - use whichever is easier for you. One is to set up an ICE table, keeping in mind that Ka is an equilibrium constant for acid dissociation and is equal to [H+][A-]/[HA]. The other is to use the equation pH = -1/2log(Ka[WA]). For this question, be careful with the concentration - we doubled the volume when we added the HCl solution, so [HSCH3] = 1M. Then, pH = -1/2log(10^-10.4*(1)), or (-1/2)(-10.4). The pH of the solution should be 5.2, which makes sense since it is a solution of a weak acid.

Question 42

A mixture of acidic wastes is found to be 2 M H2SO4, 0.5 M HI, and 3 M HNO3. What volume of 5 M NaOH solution will be required, per liter of acidic waste, to completely neutralize the acid?

A. 3.0 L
B. 1.5 mL
C. 1.1 L
D. 1.5 L

Answer 42

For the second question, keep track of how much H+ there is to neutralize. The 2M H2SO4 makes 4M H+ since it is diprotic, the 0.5M HI makes 0.5M H+, and the 3M HNO3 makes 3M H+, so there is 7.5M H+ total. Since there is 1 liter of acidic solution, that means there is 7.5 mol of H+ to neutralize, and we need 7.5 mol OH- ions. NaOH dissociates completely, so we can treat it as an equivalent of OH-. 7.5 mol OH- / (5 mol/L OH-) = 1.5 L, so we need 1.5 L of the 5M NaOH solution.