Class 7: Weak Acids & Bases – Combining Quantitative & Qualitative

Question 1

Write a Ka expression for a chemical equation, its equilibrium ratio, and draw the curved arrows to show the proton transfer.

Answer 1

• Chemical equation:
  CH3COOH(aq) + H2O(l) ⇌ CH3COO-(aq) + H3O+(aq)
• Ka expression:
  Ka = [H3O+][CH3COO-] / [CH3COOH]
• Equilibrium ratio:
  Q = [H3O+][CH3COO-] / [CH3COOH]
  If Q < Ka, reaction favors products
  If Q > Ka, reaction favors reactants
• Curved arrows for proton transfer:

     H
     |   O=C-O-H   +   H-O-H
     |              |
     CH3            H
         ⇌
     H
     |   O=C-O-   +   H-O-H
     |         |
     CH3       H
      
  CH3COOH      H2O
        ➝       ➝    
  CH3COO-    H3O+

So in summary:
- Write Ka with equilibrium concentrations
- Compare Q to Ka to predict direction
- Use curved arrows to show proton transfer

Question 2

Write a Kb expression for a chemical equation, its equilibrium ratio, and draw the curved arrows to show the proton transfer.

Answer 2

Here are concise bullet points on the key concepts and process for writing Ka/Kb expressions and using curved arrows:

• Concepts:
  
  • Ka measures acid strength (proton donor ability)
  • Kb measures base strength (proton acceptor ability)
  • Ka and Kb are related by Ka x Kb = Kw
• Writing Ka expression:
  
  • Identify acid (HA) and conjugate base (A-)
  • Ka = [H3O+][A-] / [HA]
• Writing Kb expression:
  
  • Identify base (B) and conjugate acid (BH+)
  • Kb = [BH+][OH-] / [B]
• Equilibrium ratio (Q):
  
  • Calculate Q with equilibrium concentrations
  • Compare Q to K to predict direction
• Curved arrows:
  
  • Track movement of electron pairs
  • From bonds breaking (reactants)
  • Into new bonds forming (products)
  • Reveals proton transfer mechanism

So in essence:
- Identify acid/base and conjugates
- Write Ka/Kb expression correctly
- Use Q vs K to predict direction
- Curved arrows show electron flow/proton transfer

Question 3

Calculate the pH of a solution of a weak acid or a weak base.

Answer 3

• For a weak acid (HA):
  
  • pH = -log[H3O+]
  • [H3O+] = √(Ka*Ca) (Ca = initial acid concentration)
  • Assumes negligible contribution from water ionization
• For a weak base (B):
  
  • pOH = -log[OH-]
  • [OH-] = √(Kb*Cb) (Cb = initial base concentration)
  • pH = 14 - pOH
• General steps:
  1) Write out dissociation reaction
  2) Identify Ka or Kb value
  3) Use initial concentration (Ca or Cb)
  4) Calculate [H3O+] or [OH-]
  5) Take -log to get pH or pOH
  6) Convert pOH to pH if needed
• Approximations:
  
  • Ignore water ionization if acid/base is not too weak
  • Assume х ≈ √(KaCa) or х ≈ √(KbCb)

So in essence:
- Use Ka/Kb and initial concentration
- Calculate [H3O+] or [OH-]
- Convert to pH or pOH
- Make appropriate approximations

Question 4

Calculate equilibrium concentrations of all species for a weak acid or weak base hydrolysis in water.

Answer 4

• For a weak acid (HA) hydrolysis:
  
  • HA + H2O ⇌ A- + H3O+
  • Ka = [H3O+][A-]/[HA]
  • [H3O+] = [A-] = x
  • [HA] = Ca - x
  • Substitute and solve for x using Ka and initial Ca
• For a weak base (B) hydrolysis:
  
  • B + H2O ⇌ BH+ + OH-
  • Kb = [BH+][OH-]/[B]
  • [OH-] = [BH+] = x
  • [B] = Cb - x
  • Substitute and solve for x using Kb and initial Cb
• General steps:
  1) Write out hydrolysis reaction
  2) Define equilibrium concentrations in terms of x
  3) Substitute into Ka or Kb expression
  4) Solve quadratic for x
  5) Calculate other concentrations from x
• Assumptions:
  
  • Neglect auto-ionization of water
  • Use approximation if x &laquo_space;Ca or Cb

So in summary:
- Set up equilibrium concentrations with x
- Substitute into Ka/Kb expression
- Solve quadratic for x
- Calculate remaining concentrations from x

Question 5

Interconvert the Ka value of a weak acid and the Kb value of its weak conjugate base. Understand the inverse relationship between these two values.

Answer 5

• Ka and Kb are related by the ion product of water (Kw):
  
  • Ka x Kb = Kw = 1.0 x 10^-14 at 25°C
• Converting Ka to Kb:
  
  • For HA/A- pair: Ka = [H3O+][A-]/[HA]
  • Kb for A-/HA pair = Kw/Ka
• Converting Kb to Ka:
  
  • For B/BH+ pair: Kb = [BH+][OH-]/[B]
  • Ka for BH+/B pair = Kw/Kb
• Taking logarithms:
  
  • pKa + pKb = pKw = 14.00 at 25°C
• Inverse relationship:
  
  • As Ka increases (stronger acid), Kb decreases (weaker conjugate base)
  • As Kb increases (stronger base), Ka decreases (weaker conjugate acid)

So in essence:
- Use Kw to interconvert between Ka and Kb
- Take log to relate pKa and pKb
- Stronger acid pairs with weaker conjugate base
- Stronger base pairs with weaker conjugate acid

Understanding and applying this quantitative inverse relationship is key.

Question 6

Apply the concept of conjugate base stability and how it relates to acid strength.

Answer 6

• Conjugate base stability principle:
  
  • The more stable the conjugate base, the stronger the parent acid
• Factors affecting conjugate base stability:
  
  • Charge delocalization (resonance)
  • Electronegativity
  • Inductive effects
• Charge delocalization:
  
  • Delocalized conjugate bases are more stable
  • e.g. Carboxylates (R-COO-) are stabilized by resonance
  • Stronger acids have stabilized conjugate bases
• Electronegativity:
  
  • More electronegative atoms stabilize negative charge better
  • e.g. F- is more stable than Cl-, Br-, I-
  • HF is strongest, HI is weakest binary acid
• Inductive effects:
  
  • Electron-withdrawing groups stabilize conjugate base
  • e.g. Cl-CH2-COO- more stable than CH3-COO-
  • Stronger inductive effect, stronger acid

So in summary:
- Stable conjugate base = stronger parent acid
- Delocalization, electronegativity, inductive effects stabilize bases
- Quantify and apply these factors to predict relative acid strengths

Question 7

Define and identify a strong electrolyte (no equilibrium here).

Answer 7

• Definition:
  
  • A strong electrolyte is a compound that dissociates completely into ions when dissolved in water
• Properties:
  
  • 100% ionization in aqueous solution
  • Does not involve an equilibrium process
  • Conducts electricity very well
• Examples of strong electrolytes:
  
  • Soluble ionic compounds (NaCl, KNO3, etc.)
  • Strong acids (HCl, H2SO4, HNO3, etc.)
  • Strong bases (NaOH, KOH, etc.)
• Identification:
  
  • Ionic compounds with high lattice energies
  • Acids from the set of six strong acids
  • Bases containing OH- or other very weak conjugate acids
• Key point:
  
  • No equilibrium expression needed
  • Electrolyte is 100% dissociated in solution

So in essence, strong electrolytes fully ionize and dissociate into their component ions in aqueous solution, with no equilibrium involved. This allows excellent electrical conductivity.

Question 8

Write a Ka expression for a chemical equation, its equilibrium ratio, and draw the curved arrows to show the proton transfer.

Answer 8

Ka=[Products]/[Reactants] — pH=pKa+log[base/acid]
pH=pKa+log([prob deprot]/[prob prot])

Question 9

Write a Kb expression for a chemical equation, its equilibrium ratio, and draw the curved arrows to show the proton transfer.

Answer 9

Ka=[Products]/[Reactants]

Question 10

Acids with a higher Ka have conjugate bases with a lower Kb

Answer 10

STRONG Arrhenius acid has a Ka greater than 1 and its conjugate base has a Kb of less than 10^-14 (basticity ignored)

Weak Arrhenius acid is between 10^-14 and 1, same with the conjugate base (BOTH WEAK)

Super weak Bronsted-Lowry acid is less that 10^-14. So weak that its acidity can be ignored, but its base is a STRONG base
Ex: HCl is such a strong acid that Cl- is not considered a base, but rather a spectator ion
STRONG ACIDS: HCl HBr, HNO3, H2SO4, HClO4

Question 11

Water concentration constant (55.5M).
This means it can be removed
from the equilibrium expression and incorporated into
the equilibrium constant.

Question 12

The Ka of acetic acid is 1.8x10-5
. What is the pH of a 0.50 M solution of this acid?

Answer 12

Make ice table
Find x
plug into pH=-log[H+]

Question 13

b. The pH of 0.50 M chloroacetic acid is 1.58. What is the Ka of this acid?

Answer 13

Make ice table
find x by plugging in 1.58 into pH=-log[H+] and find H+

x^2/(0.5-x)=Ka

Question 14

Ka for chloroacetic acid is
much greater than Ka for
acetic acid. Therefore,
given equimolar
solutions, the [H+] for
chloroacetic acid is much greater than that of acetic acid.
The conjugate base of chloroacetic acid is more stable than that of acetic acid. They both have
the same resonance stabilization, but the electronegative chlorine inductively removes electron
density from the negatively charged oxygens.

Question 15

a. A stronger weak acid has a HIGHER / LOWER Ka than a weaker weak acid.
b. A stronger weak acid has a HIGHER / LOWER pKa than a weaker weak acid.
c. A stronger weak acid has a MORE / LESS stable conjugate base than a weaker weak acid.

Answer 15

Higher
Lower
More

Question 16

The stronger the weak acid the WEAKER / STRONGER its weak conjugate base.

Answer 16

Weaker

Question 17

Consider the acid-base reaction below. Which are favored (i.e., present in the higher concentration)
at equilibrium—the products or the reactants? Explain your answer.
CH3COOH (aq) + ClCH2COO- (aq) ⇄ CH3COO- (aq) + ClCH2COOH (aq)

Answer 17

The reactants are favored because the conjugate base of chloroacetic acid is more stable than
the conjugate base of acetic acid, so ∆H > 0. Assuming that ∆S° for the reaction is small (equal
numbers and types of reactants and products makes this a reasonable assumption) this means
that DG° for the reaction is positive and the reactants are favored at equilibrium.

Question 18

HBr is a strong acid, and therefore dissociates completely. This means the initial concentration, 0.45
M, will also be the final concentration of H+ (interchangeable with H3O+
); this is the value we need
to solve for pH:
pH = - log[H3O+
] à pH = - log(0.45) à pH = 0.35
HCN is a weak acid, meaning it only dissociates partially. Therefore, we need to make an ICE table to
calculate how much H+ dissolves:
HCN H2O CN- H3O+
Initial 0.45 M 0 M 0 M
Change - x + x + x
Equilibrium 0.45-x x x
We now use our equilibrium expression with the Ka value for HCN (provided on the first page of this
worksheet):
Ka = 4.9 x 10-9 = x2
/(0.45-x) à x = 4.7 x 10-5 = [H3O+
]
pH = - log[H3O+
] à pH = - log(4.7 x 10-5
) à pH = 4.3

Answer 18

pH ultimately depends on hydronium concentration. Although we have the same initial
concentration of each acid, the degree to which they dissociate is the cause of the pH
discrepancy. Strong acids dissociate completely, while weaker acids only dissociate partially.
Therefore, at equal starting concentrations, a strong acid will yield a lower pH (more
hydronium ions) than a weak acid.

Question 19

Qualitatively explain under what set of conditions a weak acid could have a lower pH than a strong acid.

Answer 19

pH is dependent upon concentration of H+ ions. A weak acid solution could have a
lower pH than a strong acid if the weak acid’s concentration is much higher than
that of the strong acid. We know that weak acids dissociate partially, while strong
acids dissociate completely. At a very high concentration of weak acid, it is possible
that the small amount of H+ that dissociates could exceed the amount of H+
released by a complete dissociation of a smaller amount of strong acid.

Question 20

Calculate the pH of a 0.0015 M aqueous solution of HCl and the pH of a 0.80 M aqueous
solution of acetic acid. For each, calculate the percent dissociation. Explain how these
calculations either support or refute your explanation in part a.

Answer 20

HCl strong acid so we can just put that in the ph equations

pH=-log[H3O+]
pH=-log[0.0015]. pH=2.82

Acetic acid is a weak acid so it only dissociates partially. Therefore, we need to make an ICE
table to calculate how much H+ dissolves:

CH3COOH H2O CH3COO- H3O+
Initial 0.80 M 0 M 0 M
Change - x + x + x
Equilibrium 0.80-x x x

(there’s nothing in H2O column)

We now use our equilibrium expression with the Ka value for CH3COOH (provided on the first page
of this worksheet):
Ka = 1.8 x 10-5 = x2
/(0.80-x) à x = 3.8 x 10-3 = [H3O+]
pH = - log[H3O+]
pH = - log(3.8 x 10-3) -> pH = 2.42
Percent dissociation = [H3O+
]/[CH3COOHinitial] x100 = (3.8 x 10-3
)/(0.80) x100 = 0.48%

Question 21

intensive
or extensive property.

Answer 21

(i) Ka = intensive, because it is a constant value for the dissociation
equilibrium expression of a weak acid. We use the same Ka value
regardless of the given amount of acid.
(ii) pH = extensive, because pH is dependent upon the concentration of H+
formed by the reaction.
(iii) percent dissociation = intensive for strong acids and bases (always around
100%), but extensive for weak species, because this value depends on the
initial amount of weak acid or base present. It may seem like the percent
dissociation would be constant for any initial concentration, but
mathematically this is not the case: try plugging in a different initial
concentration for acetic acid in #2 part b and see what happens to the
percent dissociation…

Question 22

What are the [H3O+], [OH-] and pOH of a solution with a pH of 3.47?

Answer 22

Here are the equations we need:
pH = - log[H3O+]
pH + pOH = 14
pOH = - log[OH-]

Now, let’s solve for the desired quantities by using these equations:
3.47 = -log[H3O+] -> 10-^3.47 = H3O+
-> [ H3O+] = 3.39 x 10-4

3.47 + pOH = 14 -> 14 – 3.47 = pOH -> pOH = 10.53

10.53 = - log[OH-] -> 10^-10.53 = [OH-] -> [OH] = 2.95 x 10-11

Question 23

Sodium hypochlorite solution is sold as “chlorine bleach.” It is
potentially dangerous because of the basicity of the ClO- ion.
What is the concentration of ClO- in bleach if the pH of the
solution is 10.75?

Answer 23

ClO- (aq) + H2O(l) <-> HClO (aq) + OH- (aq)

Give pH
Since looking at basic reaction, calculate pOH

pH + pOH = 14 -> 10.75 + pOH = 14 -> pOH = 3.25
pOH = - log[OH-] -> 3.25 = -log[OH-] -> [OH-] = 5.623 x 10-4 M

Ice table
ClO- H2O HClO OH
Initial
[ClO-] 0 0
Change
- x + 5.623 x 10-4 + 5.623 x 10-4
Equilibrium [ClO-]
- 5.623 x 10^-4 5.623 x 10^-4 5.623 x 10^-4

(there’s nothing in H2O column)

Kw = Ka x Kb = 1 x 10^-14
Ka of HClO is 3.5x10-8
, which we then plug into the equation:
(3.5x10^-8) x Kb = 1 x 10^-14 -> Kb = 2.86 x 10^-7

Now, let’s combine this newly found Kb value with the information from our ICE table:
Kb = [HB+][OH-]/[B] -> 2.86 x 10^-7 = (5.623 x 10^-4)^2 /([ClO-] - 5.623 x 10^-4) -> [ClO-] = 1.11 M

Question 24

Codeine (C18H21NO3) is a derivative of morphine that is used as an
analgesic and narcotic. It is a base. It was once used in cough syrups,
but is now available only by prescription because of its addictive
properties. If the pH of a 1.7x10-3 M solution of codeine is 9.59,
calculate its Kb.

Answer 24

C18H21NO3 (aq) + H2O (l) <-> C18H22NO3 + (aq) + OH- (aq)

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

Since we have a weak base, we can use an ICE table to calculate Kb. To find the equilibrium
concentration of OH-
, we can calculate the pOH and then use the -log equation to solve for
[OH-]:

pH + pOH = 14 -> 9.59 + pOH = 14 -> pOH = 4.41
pOH = - log[OH-] -> 4.41 = - log[OH-] -> 10^-4.41 = [OH-] -> [OH-] = 3.89 x 10^-5

Ice table (nothing in H20 column)
C18H21NO3 H2O C18H22NO3
+ OH
Initial 1.7x10-3 0 M 0 M
Change - x + x + x
Equilibrium 1.7x10^-3 - x 3.89 x 10^-5 3.89 x 10^-5

Now we have all the values we need to plug into the Kb expression to solve for Kb:
Kb = [BH+][OH-]/[B] -> Kb = [3.89 x10-5]^2/[1.7x10-3 – 3.89 x 10-5] -> Kb = 9.1 x 10^-7

Question 25

The autoionization of water, as represented by the below equation, is known to be
endothermic. Which of the following correctly states what occurs as the temperature of water
is raised?

H2O(l) + H2O(l) ⇆ H3O+
(aq) + OH–(aq)

Answer 25

c) The pH of the water decreases, and the water remains neutral.

The reaction above is endothermic, meaning we can consider heat as a reactant. Increasing the amount of heat, a reactant, will shift the reaction towards the products. This means we will have more H3O+ (aq), which will cause the pH to decrease (because pH is equal to the -log of [H3O+ ]). This being said, the water remains neutral because for each additional H3O+ formed, an OH- is formed as well.

Question 26

the weak acid solution with the largest Ka will have the lowest pH

Question 27

Consider a 0.35 M solution of each acid. Provide a qualitative or structural explanation
for which solution would have the lowest pH. Your argument should focus on the
structure of the conjugate bases and how those structures affect the position of the
equilibrium.

Answer 27

We know that stronger acids have lower-energy conjugate bases. The energy-lowering
characteristic that applies most strongly to the conjugate bases in this question is Induction. Size
and electronegativity do not contribute, because the hydrogen is being taken away from oxygen
in all four of the weak acids. Resonance is present along the carboxylic acid/carboxylate anion,
however it is not a differentiating contributor because it is the same for all four conjugate bases.
Induction – which is stabilization due to the withdrawal of negative charge along sigma bonds by
electronegative atoms - is what makes the difference here. Let’s look at each conjugate base
(from left to right) in the chart on the next page: