Chemistry 1 Flashcards
State four uses of acids and four uses of base
What is mercerized cotton in uses of bases
What is saponification reaction in uses of bases
Acids: Sulfuric acid (H2SO4) is used for paints and pigments,detergents,fertilizers such as ammonium phosphate Acetyl salicylic acid Tannic acidic Citric acid Ascorbic acid HCL in detergents Malic acid
Many alkaloids (morphine, cocaine, caffeine among others) are organic bases (bitter taste, many are toxic). - Basesareused • inmedicine(antacids,alkaloids), • foodpreparation(bakingsoda), • cleaningagents(ammonia), • textiles(mercerization), • fertilizers(ammonia), many basic oxides (MgO) are heterogeneous catalysts.
Mercerized Cotton: after Treating normal cotton that has α-cellulose with alkali (strong base such as NaOH)
gives normal cotton fiber strength and lustrous appearance (pearl cotton-a mercerized cotton yarn for needlework (as embroidery) ) and making it β-cellulose.
Saponification Reaction
- Bases feel soapy on fingers (react with skin lipids to form soap).
Triglyceride fat or oil + sodium hydroxide ( which is a base ) gives glycerol plus three soap molecules
Saponification is a process that involves the conversion of fat, oil, or lipid, into soap and alcohol by the action of aqueous alkali. Soaps are salts of fatty acids, which in turn are carboxylic acids with long carbon chains.
Saponification can be defined as a “hydration reaction where free hydroxide breaks the ester bonds between the fatty acids and glycerol of a triglyceride, resulting in free fatty acids and glycerol,” which are each soluble in aqueous solutions.
During saponification, ester reacts with an inorganic base to produce alcohol and soap. The reaction of an ester in the presence of a base to give sodium salt of carboxylic acid and alcohol is known as saponification and it is used in the preparation of soap.
Saponification is an exothermic chemical reaction—which means that it gives off heat—that occurs when fats or oils (fatty acids) come into contact with lye, a base. In this reaction, the triglyceride units of fats react with sodium hydroxide or potassium hydroxide and are converted to soap and glycerol.12 Jan 2020
State the definitions of an acid and base based on arrhenius theory,Lewis theory,bronsted Lowry
What is an important feature of bronsted lowry acid base reactions?
Why is arrhenius theory limited
According to Arrhenius, the acids are the hydrogen-containing compounds which give H+ ions or protons on dissociation in water and bases are the hydroxide compounds which give OH− ions on dissociation in water.
Arrhenius Acids and Bases:(oldest & simplest definition) Arrhenius acid = a substance that directly yields H+
•
In aqueous solution, protons (H+) bond through covalent and hydrogen bonds to water molecules (forming H3O+).
ions when dissolved in water.
Arrhenius base = a substance that directly yields OH- ions when dissolved in water.
In the Lewis theory of acid-base reactions, bases donate pairs of electrons and acids accept pairs of electrons. A Lewis acid is therefore any substance, such as the H+ ion, that can accept a pair of nonbonding electrons. In other words, a Lewis acid is an electron-pair acceptor.
According to Bronsted-Lowry theory, acid is a substance which donates an H+ ion or a proton and forms its conjugate base and the base is a substance which accepts an H+ ion or a proton and forms its conjugate acid.
Bronsted Lowry
An acid is a proton donor, any species which donates a H+.
A base is a proton acceptor, any species which accepts a H+.
An acid-base reaction can now be viewed from the standpoint of the reactants AND the products.
An acid reactant will produce a base product and the two will constitute an acid - base conjugate pair.
Proton Transfer is an essential feature of Brønsted acid-base reactions
Example, NH3 (base, H+ acceptor) + H2O (acid, H+ donor) gives NH4 (acid, H+ donor) + OH (base, H+ acceptor)
The NH3 accepted a proton from H2O to become NH4
This reaction is reversible so if NH4 is going back to become NH3,it donates an H+ to OH which accepts it
That’s how you’ll know which is the acceptor and which is the donor
The Arrhenius definition is the narrowest view of acids and bases and Lewis theory is the broadest definition
We know that aqueous NH3 tests basic, but it does not satisfy the narrow definition given of an Arrhenius base..
Arrhenius theory is known to be the most limited of the three theories since it requires the solutions to be aqueous. It only applies to substances that produce Hydrogen ions (H+) or hydroxide ions (OH−). 1- An acid is expected to be an acid in any solvent.
According to the theory of Arrhenius, the reaction between HCl and NH3 is not considered as acid-base reaction because none of these species gives H+ and OH− ions in water.
No… according to Arrhenius a base is one which dissolves in water to give OH- ions.
But instead NH3 is a Bronsted and Lewis base.
It is a Bronsted base because it is a proton acceptor, and a Lewis base because it is an electron pair donor
We know that aqueous NH3 tests basic, but it does not satisfy the narrow definition given of an Arrhenius base.
• Weneedabroaderdefinition! • What’smissing?
Arrhenius Brønsted Lewis
One obvious answer is the role of water:
NH3 (aq) + H2O (l) ⇆ NH4+ (aq) + OH- (aq)
Better, but even in the absence of water…….
NH3 (aq) + HCl (aq) ⇆ NH4+ (aq) + Cl- (aq)
What do these reactions have in common?
What is a conjugate acid base pair
How will you identify it
In this equation what’s the conjugate acid base pair
HNO2(Aq) + H2O(l) ➡️⬅️ NO2(aq) + H3O
1) What is the conjugate acid of each of the following: (a) HS- (b) H2NNH2(hydrazine or N2H4) (c) SO42-
2) What is the conjugate base of each of the following: (a) HS- (b) H2NNH2 (c) H2SO4
Conjugate acid-base pair = two species whose + formulas differ by a single H
Examples: HCN/CN- H2O/H3O+ H2O/OH-
Conjugate acid = a species with one more H+ than its conjugate base
Conjugate base = a species with one fewer H+ than its conjugate acid
a conjugate acid–base pair consists of two substances that differ only by the presence of a proton (H⁺). A conjugate acid is formed when a proton is added to a base, and a conjugate base is formed when a proton is removed from an acid.
The formula of the conjugate acid is the formula of the base plus one hydrogen ion. … Acid1 is HCl its conjugate base is base1 hydroxide ion is base 2 and its conjugate acid (water) is acid2
The conjugate acid has one H+ more than it’s counterpart substance HNO2-acid NO2-conjugate base H2O-base H3O-conjugate acid (H+ was added to it)
- a.H2S
b. Hydrazinium or N2H5+
c. HSO4
2.a. The conjugate base of hydrogen sulfide anion, HS− , is sulfide ion, S2− .
b.
c. HSO4
What are strong acids and weak acids
Look at a picture in the chem slides on examples of acids and their conjugates
Strong acids are strong electrolytes:
100% + −
HNO (aq) + H O(l) “””“→ H O (aq) + NO (aq) 3233
Weak acids are weak electrolytes:
~1% + -
HF(aq) + H O(l) !!!→ H O (aq) + F (aq) 2 ←!!!3
Some substances have negligible acidity in water: CH3OH(aq) + H2O(l) !!→ ~noreaction
• The stronger an acid, the weaker its conjugate base.
• The stronger a base, the weaker its conjugate acid.
Strong acids are completely dissociated in water. Their conjugate bases are quite weak. • Weak acids only dissociate partially in water. Their conjugate bases are weak bases. • Strong bases are completely dissociated in water. Their conjugate acids have negligible acidity.
Acid-Base Reactions (strong acid)
• In any acid-base reaction, the equilibrium will favor the reaction that moves the proton to the stronger base (weak acid is formed).
• Consider HCl (strong acid) in water:
HCl(aq) (acid (1) ) + H2O(l) base (2) (⎯⎯→ H3O+(aq) (acid 2) + Cl-(aq)(base 1)
H2O is a much stronger base than Cl-, so the equilibrium lies so far to the right that K»_space;>1.
Acid-Base Reactions (weak acid)
• In any acid-base reaction, the equilibrium will favor the reaction that moves the proton to the stronger base. – This still holds true.
base (2)
CH3CO2H (aq) + H2O (l) acid (1)
acid (2)
H3O+ (aq) + CH3CO2- (aq) base (1)
Acetate is a stronger base than H2O, so the equilibrium favors the left side (K < 1).
What is Autoionization of Water
Can water be acid?
How is water autonized using logs
Autoionization of Water
• Aswehaveseen,waterisamphoteric.
• Inpurewater,afewmoleculesactasbasesandafewactas acids…..This is referred to as autoionization.
H2O + H2O ⬅️➡️ OH + H3O
• The equilibrium constant for this reaction is:
Kw = [H+][OH-] = [H3O+][OH-] = 1.0x10-14 at 25 °C
Can Water Be an Acid?
Consider the consequences:
H2O(l) H+(aq) + OH-(aq) The equilibrium constant for this reaction is:
Kw = [H+][OH-] Remember to
= 1.0 x 10-14 at 250C
leave out pure liquids and solids
A change in [H+] causes an inverse change in
[OH-].
In an acidic solution, [H+] > [OH-] In a basic solution, [H+] < [OH-] In a neutral solution, [H+] = [OH-]
Autoionization of water: using Logs
H2O(l) H+(aq) + OH-(aq) The equilibrium constant for this reaction is:
Kw = [H+][OH-] = 1.0 x 10-14 at 25 °C
-log(Kw) = -log([H+][OH-]) = -log( 1.0 x 10-14)
pCl used in environmental science for chlorine concentration.
pKw = -log([H+]) + -log([OH-]) = 14 pKw =pH+pOH=14
pOH = -log10[OH-]
What is the pH scale
Remember: Quick Math Review (from appendix A)
• If 10n=A, the logarithm base 10 of A is n, or log10 A = n.
• Becauseoftherelationshipsbetweenlogsand exponentials.
– log1=0
– log(AxB) = log(A) + log(B) & log(A/B) = log(A) - log(B) – Analogous to 10A+B = 10A x 10B & 10A-B = 10A / 10B
– LogAy =ylogA
– log10 A = ln(A) / ln(10) or ln(A) = log(A) / log(e)
Significant Figures: 3.4 x 10-4 can be written as 100.54 x 10-4 Thus, log(100.54 x 10-4 ) = 0.54 + (-4) = -3.46.
Notice only digits after the decimal are significant!
The reverse process gives 10-3.46 = 3.4 x 10-4
(3.46 in this context only has 2-significant figures!)
What is the pH of a solution with [H+]= 4.5 × 10-12 M ??
Is this solution acidic or basic?
What is the pH of a 0.048 M CsOH aqueous solution?
16.4 The pH Scale
To conveniently handle the fact that [H+] spans
many orders of magnitude a logarithmic scale is often used:
pH = -log10[H+]
What is the pH of a neutral solution? [H+] = 1.0x10-7 M
pH = -log10[1.0x10-7] = 7.00
Note: In a logarithm, only the numbers to the right of the decimal place are significant figures. So since the concentration has 2 significant figures, the pH has two decimal places.
pH = -log (4.5 × 10-12 M)
= 11.35
⇒ 2 significant decimal places
Is this solution acidic or basic?
The pH of an acidic solution is less than 7.
The pH of a neutral solution is 7.
The pH of a basic solution is greater than 7.
pH = 11.35 ; the solution is basic
What is the pH of a 0.048 M CsOH aqueous solution?
pKw =pH+pOH=14 pH = pKw – pOH =
= 14 - pOH
How is pH experimentally measured
What colour does Red Cabbage Anthocyanin turn jnto in an acidic solution and in a basic solution
Which indicator is best suited to differentiate a solution that is slightly basic from one that is slightly acidic?
when the reactant base is stronger than the product base than the reaction lies on the right
when the product is the stronger base than the reactants than the equilibrium lies on the left.
True or false
Less accurate measurement of pH observe color
– Litmus paper (disc. in 14th century) • “Red” paper turns blue above
~pH = 8
• “Blue” paper turns red below ~pH = 5
– or an indicator.
pH meter used for accurate measurement of pH
measure voltage
Phenolphthalein is often used as an indicator in acid–base titrations. For this application, it turns colorless in acidic solutions and pink in basic solutions.
Red cabbage contains a water-soluble pigment called anthocyanin that changes color when it is mixed with an acid or a base. The pigment turns red in acidic environments with a pH less than 7 and the pigment turns bluish-green in alkaline (basic), environments with a pH greater than 7.
Indicator: bromthymol blue and phenolphthalein but more of bromothymol blue
What are strong acids and strong bases
You should memorize strong acids and bases. Memorize the strength of common acids For a strong acid, 100% ionized in H2O For a weak acid, Partially ionized in H2O For negligible substances ,Not ionized in H2O
Strong Acids
• You will recall that seven common strong acids are HCl, HBr, HI, HNO3, H2SO4, HClO3, and HClO4.
• These are, by definition, strong electrolytes and exist totally as ions, H+ ions and anions, in aqueous solution.
• For the monoprotic strong acids, [H3O+] = [acid]
Strong Bases
• Strong bases are the soluble hydroxides, which are the alkali metal (group 1A: Na, K) and heavier alkaline earth metal hydroxides (group 2A: Ca2+, Sr2+, and Ba2+).
• Again, these substances dissociate completely in aqueous solution.
Strength vs. Quantity vs. Concentration strong much concentrated
weak. little dilute
For pH,
Strength ⇒ chemical property
Quantity ⇒ mass or volume
• Concentration ⇒ quantity per unit volume
Note: strong coffee ⇒ concentrated coffee
strong acid does not mean concentrated acid
(non-scientific vs. scientific language)
What are weak acids
What does a smaller Ka mean?
What does a stronger acid mean for a reaction?
Explain quantitative analysis of acid and base equilibria
Forageneralizedaciddissociation, HA (aq) + H2O (l) A- (aq) + H3O+ (aq) the equilibrium expression would be Kc = [H3O+] [A-] divided by [HA] • Thisequilibriumconstantiscalledtheacid dissociation constant, Ka. • Wewilluse‘p’again: pKa =−log(Ka)
Quantitative Analysis of Acid/Base Equilibria
To be quantitative, remember that acid dissociation reactions are just
equilibrium reactions:
Strong acids dissociate completely into ions in water: HA(g or l) ➡️ H +(aq) + A-(aq)
Ka»_space; 1
Weak acids dissociate very slightly into ions in water:
HA(aq) ⬅️➡️H+(aq) +A-(aq)
Ka «_space;1
The Acid-Dissociation Constant (Ka):
[H+][A-] divided by [HA]
stronger acid means products favored means larger Ka
smaller Ka means reactants favored means weaker acid
Solve this
Calculate the concentrations of HNO2, H+ and NO2- in 0.20 M HNO2.(so you’re finding the equilibrium concentrations of each substance so depending on the substance concentration you’re to find,you find it)
How many ways can this be solved
How do you know how to test the validity of approximation method?
If this method fails what method do you use?
For better understanding watch the ICE tables made easy by Justin Mueller and
https://chemistrytalk.org/ice-table-chemistry/ and Equilibrium calculations by Chads prep
Solution: HNO (aq) !!→ H+(aq) + NO- (aq) I 0.20.. ~0. 0 C. -x. +x. +x E. 0.20-x. +x. +x
The +x is always on the side with 0 and the side with a known value for the I or intial concentration always has the negative sign attached to it.
So in this example HNO has a value so the negative sign was put there but for the H and NO ,there were no values or even if only the H had no value,the side with no value always had the positive even if it’s only one substance on that side with no value
I means initial concentration
C means Chnage in concentration
E means Equilibrium concentration
For C,look at the coefficient of each substance and write it down with an c
Let x = [H+]
Ka =[H+][NO-] divided by [HNO ] =x squared divided by 0.20-x= 4.5 × 10 to the power -4
How do we solve this algebraic equation?
Option #1: Graphing Calculator
Ka = (x)multiplied by(x) divided by (0.20 - x) = 4.5 × 10to the power -4
expand: x2 = (9.0 × 10-5) - (4.5 × 10-4)x
rearrange: x2 + (4.5 × 10-4) x - (9.0 × 10-5) = 0
solve: Enter into calculator
Option #2: The quadratic formula
rearrange:
substitute:
expand: x2 = (9.0 × 10-5) - (4.5 × 10-4) x
Rearrange: x2 (ax squared)+ (4.5 × 10-4) x(bx) - (9.0 × 10-5) (c)= 0
Substitute: x= -b± square root of (b2-4ac ) divided by 2a
Solve : answer x = 9.26 × 10-3 and - 9.71 × 10-3
- 9.71 × 10-3 is Mathematically correct, but not chemically possible.
So putting the positive answer into the equation,
[H+]=[NO2-] = x = 9.26×10-3 or 9.3×10-3M
Therefore 0.2-x = [HNO2] = 0.1903 or 0.19 M
(also note –log[H+] = -log (9.3 x 10-3) = 2.01 = pH)
Option #3: Approximation Method
Ka = (x)(x) = 4.5 × 10-4 (0.20 - x)
If: “x” is small compared to 0.20.
(as expected based on chemical principles)
Then: (0.20 - x) ≈ 0.20
Or (x)(x) divided by 0.20 is ≈ “approximately equal to” 4.5 × 10-4
x2 ≈ 0.20×4.5×10-4 = 9.0×10-5
x ≈ ±9.487×10-3 or 9.5×10-3
Note: 9.5 × 10-3 and 9.3 × 10-3 differ slightly.
Approximation Method: 5% Test
How do we test an approximation’s validity?
Less that 5% of the concentration that you are comparing to.
Example:
We assumed: (0.20 - x) ≈ 0.20
Is this true? Is “x” negligible compared to 0.20?
Check:
How large is “x” compared to 0.20?
(x divided by 0.20)multiplied by 100
= (‘9.5×10-3 divided by 0.20)multiplied 100
= 4.75%
4.75% ≤ 5% ⇒ PASSES TEST
If the test fails? Use the quadratic equation
Solve this
A 0.050 M solution of a weak acid has a pH of 3.56. (a) Calculate Ka for this acid.
(b) What is the percent ionized or dissociated
To understand better,watch pH calculations for weak acids and bases by chads prep
HA ➡️⬅️ H + A
I 0.050. 0. 0
C -10 to the power -3.56 +10 to the power-3.56. +10 to the power -3.56
E. 0.0497 0.000275 0.000275
Ka= [H+][A-] divided by [HA]
Ka= [0.000275]squared divided by [0.0497]
= 1.5 x 10 to the power -6
b. percent ionization
=( “[H+]equilibrium divided by [HA] initial )× 100%
=”0.000275 divided by 0.05 × 100%
= 0.55%
What are polyprotic acids
Scientists use the acid dissociation constant (Ka) to determine the strength of acids.
Why is the successive Ka decreased in polyprotic acids reactions
Phosphoric acid is considered a polyprotic acid.
A polyprotic acid is an acid that has more than one ionizable hydrogen (H) atom.
In other words, they have more than one acidic proton, so they can undergo more than one dissociation reaction Examples of polyprotic acids include H2SO4 and H3PO4. Let’s look at the dissociation of H3PO4(aq). Phosphoric acid has three acidic protons, and therefore, its ionization happens in three steps.
Notice in this case, that these three dissociation steps are part of a single reaction.
Ionization occurs stepwise, and each step has its own Ka. Note: (i) Ka1 > Ka2 > Ka3 Always true for polyprotic acids, i.e., each ionization step is more difficult because it is more difficult to remove H+ from a molecule as its negative charge increases.
Some Polyprotic Acids H2S,Hydrogen sulfide H2SO4,Sulfuric acid H2SO3,Sulfurous acid H3PO4,Phosphoric acid H2C2O4,Oxalic acid H2CO3,Carbonic acid H2C3H2O4,Malonic acid
Calculate the overall equilibrium constant for oxalic acid.
H2C2O4⇌2H++C2O2−4
K1=5.6E-2
K2=5.4E-5
Solution
The calculation is straightforward:
Koverall=K1K2
=3.0E-6
Example: Oxalic Acid
H2 C2 O 4( a q ) !! → H + ( a q ) + H C2 O4 - ( a q )
HC 2O4 -(aq) !!→ H+(aq) + C 2O4-(aq)
Ka1 = [H+][HC2O4] divided by [H2C2O4]
= 6.5×10 to the power -2
Ka 2= [H+][C2O4] divided by [HC2O4]
=6.1×10 to the power -5
Calculate the concentrations of all (be careful with this) species in 0.20 M H2SO3(remember this is a polyprotic acid so it will dissociate twice) . (Ka1 = 1.7 × 10-2 and Ka2 = 6.4 × 10-8)
Solution: Strategy: work stepwise
1st Step: H SO (aq) !!→ H+(aq) + HSO -(aq
I 0.20 0.0 0.0
C-x +x+x
E 0.20-x +x +x
Let x=[H+]
Ka = [H+][HSO3- ] divided by [H2SO3]
= (x)(x) divided by 0.20-x
= 1.7 × 10-2
After expand and solve using quadratic equation,
x = +0.0504
Putting x back into the equation
[H+] = [HSO3- ] = x = 0.0504 M (round later)
[H2SO3] = 0.20-x = 0.20-0.0504 = 0.15M
2nd Step or second dissociation:
HSO- (aq) !!→ H+(aq) + SO2-(aq)
Let z =[SO32-]
I 0.0504 0.0504 0.0
C -z. +z. +z
E 0.0504-z. 0.0504+z. +z
Ka2
= [H+][SO3 to the power 2-] divided by[ HSO3 to the power -1]
= (0.0504 + z)(z) divided by (0.0504 - z) = 6.4 × 10-8
z = 6.4×10-8
If you were using approximation
approximation: 0.0504 + z ≈ 0.0504 - z ≈ 0.050
z = 6.4×10-8
Putting it back into the equation
[H2SO3] = 0.20 – 0.0504 = 0.15 M
[H+] = 0.0504 + z ≈ 0.0504 or 0.050 M
[HSO3- ] = 0.0504 - z ≈ 0.0504 or 0.050 M [SO32-] = z = 6.4x10-8(means raised to the power -8)M
Note: Stepwise method suitable:
If: Ka1»_space; Ka2
Otherwise: a simultaneous solution required
For H2SO3
Ka1=1.7×10 to the power -2»_space; Ka2=6.4×10 to the power -8
often true
What are strong bases What is base ionization constant,Kb The smaller the Kb? The larger the Kb? What are the two major categories of weak bases
Strong bases are strong electrolytes: ! ! !
NaOH(s) → Na+(aq) + OH-(aq)
Weak bases are weak electrolytes:
NH3 (aq) + H 2O(l) !!→ NH4+(aq) + OH-(aq)
Base Ionization Constant,
Kb=[NH4+ ][OH- ] divided by [NH3 ]
Kb = 1.8×10 to the power -5
Note: Water (solvent) omitted (pure liquid).
Most bases are weak bases.
• Be familiar with the common strong bases.
e.g., NaOH, KOH, and Ba(OH)2
• Kb is a measure of base strength:
smaller Kb ⇒ weaker base
larger Kb ⇒ stronger base
for all strong bases, Kb»_space;> 1
• Two major categories of weak bases:
1) Neutral molecules with lone pairs of electrons,
e.g., NH3, CH3NH2 (often with a trivalent, – N – )
2) Conjugate bases of weak acids, e.g., HCO3-, F-
Calculate the pH of 0.10 M CH3NH2.
(Kb = 4.4×10-4)
Take note of how the equation is written
CH3NH2 (aq)+H2O(l)!!→CH3NH3+(aq)+OH-(aq)
0.10. 0.0 0.0(forOH)
Kb= CH3NH3+][OH-] divided by CH3NH2
=(x)(x) divided by 0.10-x
Kb = 4.4×10 to the power -4
Use quadratic equation
x = 6.417×10-3M
Putting it back into the equation
[OH-] = x = 6.42×10-3M pOH = -log(6.42 x 10-3) = 2.19 pH + pOH = 14.00 pH = 14.00- pOH=14.00 -2.19 pH = 11.81
What is Relationship Between Ka and Kb
The Ka of hypochlorous acid, HClO, is 3.0 x 10-8. Write the Kb equilibrium expression for ClO-, and compute Kb.
Watch a video to understand
What is pKa?
Example 8: The pKa of HF is 3.17. What is the value of Kb for the F- ion?
[H+][OH-] = Kw
Kw = Ka for the weak acid × Kb for the conjugate base
H+ (aq) + ClO- (aq) ←”” HClO(aq) “ for
1/2Ka
H2O ( l ) ←” “ H + ( a q ) + O H - ( a q ) for Kw
ClO- (aq) +H2O(l) ←!! HClO(aq) + OH (aq)
Reference books often list pKa and pKb values for weak acids and bases rather than Ka and Kb values pKa = -log(Ka) pKb = -log(Kb) It follows from, Kw = Ka × Kb That, pKw = pKa + pKb = 14.00
14 – 3.17 = pKb
= 10.83
Kb = 10-10.83= 1.5 × 10 to the power -11
Learn a bit of logarithms to help you understand
Salts may be acidic, basic or neutral.
Give an example each
How can we predict whether a salt is acidic or basic?
What will happen to the conjugate base of a strong acid ?
Conjugate base of a weak acid?
Cation of a strong base?
Conjugate acid of a weak base ?
Acid Base Properties of Salt Solutions • Salts may be acidic, basic or neutral. Examples: • acidic salts: NH4NO3, NH4Cl, CrCl3 • basic salts: NaCN, KF • neutral salts: NaBr, KCl
Q: How can we predict whether a salt is acidic or basic?
Ans: Decide if and how its ions react with water.
Hydrolysis of ions = the reaction of ions with water to produce H+(aq) or OH-(aq) ions.
Hydrolysis of Ions
• The conjugate base of a strong acid will not hydrolyze. Examples: Cl-, NO3-, ClO4-
ClO4-(aq) + H2O(l)➡️no reaction
• The conjugate base of a weak acid will hydrolyze.
Examples: F-, CN-, HCO3-, CO32-
F-(aq) + H2O(l) ⬅️➡️HF(aq) + OH-(aq)
• The cation of a strong base will not hydrolyze. Ex. Na+, K+, Ba2+
K+(aq) + H2O(l)➡️no reaction
• The conjugate acid of a weak base will hydrolyze.
Ex. NH4+, CH3NH3+, NH3OH+
CH3NH3+(aq) + H2O(l) ⬅️➡️CH3NH2(aq) + H3O+(aq)
How do you Predict acidity or basicity of salts in aqueous solution
Memorize the common weak acids,based,strong acids and strong bases in the slides
What’s a quicker way of finding out if a salt is acidic base or neutral
(1) Identify the ions formed in an aqueous solution.
H2O + - NaCN(s) !!!→ Na (aq) + CN (aq)
(2) Assume cation and anion react with water (H+,OH-).
Na+(aq) + H2O(l) !!→ NaOH(aq) (strong base) + H+(aq)
CN-(aq)(weak base) +H2O(l) !!→ HCN(aq) (weak acid) + OH-(aq) 2 ←!!
(3) Determine which of the reactions is realistic in solution (or both are).
1) Identify the ions formed in an aqueous solution. 4 !!! 4
NH4Cl(s) → NH4 +(aq) + Cl-(aq)
(2) Assume cation and anion react with water (H+–OH-).
NH4+(aq)(Weak acid ) + H2O(l) !!→ H3O+(aq) + NH3(aq)(Weak base)
Cl−(aq) +H2O(l) ““→ HCl(aq)(Strong acid) + OH−(aq)
(3) Determine which of the reactions is realistic in solution (or both are).
A quicker way of deciding whether a salt is acidic, basic or neutral …
Binary salts have two “parents” – an acid and a base, i.e., acid + base → salt + water
(1) Work backwards and identify a salt’s “parents.”
Example-NaF is made up of NaOH (strong) and HF(weak)
So it’s a Basic salt
NH4Br is made up of NH3(weak) and HBr (strong) so it’s an acidic salt
KNO3 is made up of KOH(strong) and HNO3(strong)
So it’s a neutral salt
(2) Label each acid or base strong or weak.
(3) Only those cations or anions whose conjugate is weak will hydrolyze.
Ex. X- ⇒ OH- Basic
M+ ⇒ H+
Acidic
What if both ‘parents’ of a salt are weak?
NH4CN
= weak acid
⇒ NH4+ is a weak acid and CN- is a weak base.
weak base = NH3
Weak acid=HCN
So for NH4CN no matter how you put it both ways are both weak so you use this
Data: Ka (NH4+) = 5.6 x 10-10
Kb (CN-) = 2.0 x 10-5
Ans: If Kb(CN-) > Ka(NH4+) ⇒ basic solution
If Kb(CN-) < Ka(NH4+) ⇒ acidic solution If Kb(CN-) ≈ Ka(NH4+) ⇒ ~neutral
Conclusion: Kb > Ka ⇒ basic solution
For an ampiphrotic ion,NaHS
strong base =NaOH H2S= weak acid
⇒ HS- is a weak base but also is a weak acid!
H2S(aq) + OH-(aq)➡️⬅️HS-(aq) + H2O(l) ➡️⬅️S2- (aq)+ H3O+(aq) Data: Ka (HS-) ~ 10-19 Ka (H2S) ~ 10-7 Kb (HS-) = 10-14/10-7= 10-7 Conclusion: Kb > Ka ⇒ basic solution
What is hydrolysis of metals
What does it depend on
What factors determine acid strength
This explains the acidity of aqueous metal ion solutions. Not significant for alkali metal or alkaline earth metal ions
Dependence on Cation Charge and Size
Ø The pull of electron density toward the cation weakens the polar O—H bond of the water molecule and allows the transfer of an H+ ion to a nearby water molecule.
• As a result, hydrated cations tend to be acidic, with their acidities increasing with increasing charge and decreasing size
Consider the binary acid HX, HX(aq) !!→ H+(aq) + X-(aq) ←!! Three major factors determine the strength of this acid: (1) H–X bond polarity: δ+ • amorepositiveH ⇒ amoreacidicH (2) H–X bond strength: • a weaker H–X bond ⇒ a more acidic H (3) Conjugate base stability: • amorestableX- ⇒ amoreacidicH Other factors: solvent, polarizability,.
For binary acids of the same group,
(weakest) HF «< HCl < HBr < HI (strongest)
Here the bond strength is more important than the bond polarity or electro negativity difference
For binary acids of the same period ,
(weakest) CH4 < NH3 «_space;H2O < HF (strongest)
Note: Herebondpolarityisclearlymoreimportant than bond strength.
The more polar the H-X bond and/or the weaker the H-X bond, the more acidic the compound.
• So acidity increases from left to right across a row and from top to bottom down a group in the periodic table
What is an oxyacid and what determines its strength
What are Lewis acids and bases
Identify the Lewis acids and Lewis bases in the following reactions:
a) H+ + OH-➡️⬅️H2O
(b) Cl- + BCl3⬅️➡️BCl4
(c) K+ + 6H2O➡️⬅️K(H2O)6
(Hint: Look for electron pair acceptors (acids) and donors (bases).)
In oxyacids
(an -OH is bonded to another atom, Y)
the more electronegative Y is,
the more acidic the acid.
Example inHCLO, the OH is bonded to CL
For a series of oxyacids, acidity increases with the number of oxygens.
Resonance in the conjugate bases of carboxylic acids stabilizes the base and makes the conjugate acid more acidic.
Lewis Acids and Bases
Acid = an electron pair acceptor Ex: H+, BF3 & CO2
Base = an electron pair donor Ex: OH-, NH3 & NF3
Note: ALewisaciddoesnothavetocontainH+ions. A Lewis base does not have to contain OH- ions.
Lewis acids are defined as electron-pair acceptors.
• Atoms with an empty valence orbital can be Lewis acids.
Lewis bases are defined as electron-pair donors.
• Lewis bases can interact with things other than protons, H+, however (Brønsted base).
SOLUTION:
(a) H+ + OH-
(b) Cl- + BCl3
H2O BCl4-
Solve:
If the pH = 2.00 for an HNO3 solution, what is the concentration of HNO3? A. 0.10 M B. 0.20 M C. 0.010 M D. 0.020 M E. 0.0010 M
Practice Problem 2 If the pH = 10 for a Ca(OH)2 solution, what is the concentration of Ca(OH)2? A. 1.0 x 10–10 M B. 5.0 x 10–11 M C. 1.0 x 10–2 M D. 5.0 x 10–5 M E. 2.0 x 10–2 M
Practice Problem 3 A 1.0 M solution of a weak acid with pKa = 8.0 will have a pH of about \_\_\_. A. pH=2 B. pH=4 C. pH=6 D. pH=8 E. pH=10
What are the rules for assigning oxidation state
The oxidation state of chromium in the dichromate ion, Cr2O72-, is _______.
a. –1 b. +3 c. +6 d. +7
Rules for assigning oxidation state (number):
1. An element in the pure state is zero.
Al(metal), C(coal), Cl2 (gas) all equal 0.
2. A monoatomic ion equals its charge.
Na+ +1 S2- -2 Fe3+ +3
3. Sum of ox states for all atoms in a neutral formula is zero.
AlCl3: Al3+ =+3 Cl─ =-1
(+3) + 3(-1) = 0
4. Sum of ox states of all atoms in polyatomic ion = ion charge.
ClO3─ : O = -2 Cl = +5
(+5) + 3(-2) = (+5) + (-6) = -1
charge that would be present on an atom if the electrons were assigned to the more electronegative atom with convention: Group 1A Li, Na, K +1 (NaCl) Group 2A Mg, Ca +2 (CaCO3) Hydrogen H. +1 (HNO3) Fluorine F -1 (CF4) Oxygen O -2 (H2O) Group 7A (usually Cl, Br, I) -1 (AgBr) Group 6A -2 (CdS) Group 5A -3 NH3
Priority tracks with electronegativity.
O is the most electronegative the oxidation number of O is -2 7 O: -2x7 = -14
overall charge -2 =>
Cr2 must have +12 => single Cr +6
What is electrochemistry,an electrochemical cell
How does this cell work?
What is a charge,current,potential,potential difference,resistance,current ,capacitance and state their units
Electrochemistry - relationship between chemical change and electrical work
• Electrochemical cell - systems that incorporate a redox reaction to produce or utilize electrical energy.
Such system …
– does work by releasing energy from a spontaneous
reaction to produce electricity – voltaic cell
– does work by absorbing energy from a source of electricity to drive a nonspontaneous reaction – electrolytic cell
Charge (Q) – fundamental property – properties of elementary particles (protons and electrons) the measure of charge is coulomb (C) : 1.60219 x 10-19 C for a single electron Faraday’s constant – (Avagadro’s number) x (electron charge) 6.0228 x 1023 (e-/mol e-) * 1.60219 x 10-19 (C/e-) = 96485 (C/mol e-) Current (I) – rate of charge flow along conductor path (either e- in metal or ions in solution) measured in Amperes (A) : 1 A = 1 C/s Potential – the potential energy of a unit charge at a particular point in a circuit defined as the potential energy involved in moving a charge of 1 C to that point from infinity (difficult to measure). Potential difference (Voltage (V)) – difference in potential between two points 1 V defined as 1 J of work required to move 1 C Resistance (R) – obstruction of current flow. Ohm’s law (V = I×R) R has units of ohms () Capacitance (C) – tendency of a device with two conductors separated by an insulator to absorb charge when the voltage between the conductors is changed. units are Farads (F) : Q = C ΔV: also Q = I Δt
What is oxidation
What is reduction
What is an oxidizing agent
Reducing agent
For redox reactions The total number of electrons gained by atoms/ions of oxidizing agent always equals the total number of electrons lost by the atoms/ions of the reducing agent. True or false
Zn + Cu2+ ➡️Zn2+ + Cu The reducing agent in the reaction above is \_\_\_\_\_\_\_. And \_\_\_\_\_\_\_\_\_ is reduced. a. Zn a. Zn b. Cu2+ c. Zn2+ d. Cu b. Cu2+ c. Zn2+ d. Cu
How do you balance redox reactions. State the two methods ?
of Definitions (Review of Chapter 4)
Oxidation - loss of electrons (OIL) Reduction - gain of electrons (RIG)
Oxidizing agent - species that does the oxidizing - taking electrons from substance being oxidized.
Oxidizing agent is reduced in the process.
Reducing agent - species that does the reducing - giving electrons to the substance being reduced.
Reducing agent is oxidized.
The total number of electrons gained by atoms/ions of oxidizing agent always equals the total number of electrons lost by the atoms/ions of the reducing agent.
Answers: reducing agent is Zn and Cu2+ is reduced
A reducing agent reduces other substances and loses electrons; therefore, its oxidation state increases. An oxidizing agent oxidizes other substances and gains electrons; therefore, its oxidation state decreases
Two Methods of Balancing Redox Equations:
•Oxidation number method – Ch. 4 (section 4.4)
•Half-reaction method – Ch. 20
•Divides the overall redox reaction into oxidation and reduction half-reactions
Example: It is known that acidic Cr2O72− (dichromate ) solutions oxidize Zn metal to Zn2+ forming Cr3+. Where do H2O and H+ appear in the balanced equation?
(Use the half-reaction method to solve.)
Using the half reaction method Complete and balance the equation: Cr2O72−(aq) + Zn(s) Cr3+(aq) + Zn2+(aq) (acidic)
Exercise: Complete and balance the equation: Cr2O72−(aq) + Zn(s) Cr3+(aq) + Zn2+(aq) (acidic)
Step 1. Separate into two half-reactions. Cr2O72− ➡️Cr3+
Zn(s) ➡️Zn2+
Step 2a. Balance all elements except O and H.
Cr2O72− 2Cr3+
Zn(s) Zn2+
Step 2b. Balance O using H2O.
Cr2O72− 2 Cr3+ + 7 H2O
Zn(s) Zn2+
Step 2c. Balance H using H+.
14H+ + Cr2O72− 2Cr3+ + 7H2O
Zn(s) Zn2+
Step 2c. Balance charge using e−.
6e− + 14H+ + Cr2O72− 2Cr3+ + 7H2O
(Left: +14 −2 = +12 Right: +6 +0 = +6 add 6 e− to the left)
Zn(s) Zn2+ + 2e−
(Left: = 0 Right: = +2 add 2e− to the right)
Step 3. Multiply by factors to equate e− loss and gain.
(6e− + 14H+ + Cr2O72− 2Cr3+ + 7H2O )multiplied by 1
(Zn(s) Zn2+ + 2e− )multiplied by 3
3Zn(s) 3Zn2+ + 6e−
Step 4. Add balanced half-reactions.
6e− + 14H+ + Cr2O72− 2Cr3+ + 7H2O
3Zn(s) 3Zn2+ + 6e−
14H+ + Cr2O72− + 3Zn 3Zn2+ + 2Cr3+ + 7H2O
Step 5. Double check mass and charge balance. left-14H = 14H-right 2Cr = 2 Cr 7O=7O 3Zn = 3Zn \+14 −2 +0 = +6 +6 +0 The equation is balanced!
Complete and balance the equation:
MnO4−(aq) + Br−(aq) MnO2(s) + BrO3−(aq) (basic)
Using half reaction method
Do free electrons appear anywhere
in the balanced equation for a redox reaction? If yes, where?
a) Yes, on the left-hand side of the balanced equation for a redox reaction.
b) Yes, on both sides of the balanced equation for a redox reaction.
c) No, the electrons do not appear in half-reactions and they do not appear in a balanced equation for a redox reaction.
d) No, the electrons cancel in adding half-reactions to form a balanced equation for a redox reaction.
Steps 1 - 4. Balance as if it were acidic.
2H+ + 2MnO4− + Br− 2MnO2 + BrO3− + H2O
Step 4b. Neutralize the H+ by adding OH– to both sides.
Add 2 OH– to both sides.
2H2O + 2MnO4− + Br− 2MnO2 + BrO3− + H2O + 2OH−
(2H+ +adding the 2OH have the 2H2O)
Step 4c. Cancel H2O if necessary.
H2O + 2MnO4− + Br− 2MnO2 + BrO3− + 2OH−
Answer is d
Balance the following skeleton reaction in acidic solution.
Identify the oxidizing and reducing agents.
ClO3-(aq) + I-(aq)I2(s) + Cl-(aq)
Oxidizing agent (taking electrons) is I- Reducing agent (giving electrons) is ClO3-
