29 Weak Acid and Weak Base Calculations (M15Q5)

Example 2

Determination of K_a from Equilibrium Concentrations
Acetic acid is the principal ingredient in vinegar (Figure 1); that’s why it tastes sour. At equilibrium, a solution contains [CH₃CO₂H] = 0.0787 M and [H₃O⁺] = [CH₃CO₂^–] = 0.00118 M. What is the value of K_a for acetic acid?

Solution
We are asked to calculate an equilibrium constant from equilibrium concentrations. At equilibrium, the value of the equilibrium constant is equal to the reaction quotient for the reaction:

Check Your Learning
What is the equilibrium constant for the ionization of the HSO₄^– ion, the weak acid used in some household cleansers:

In one mixture of NaHSO₄ and Na₂SO₄ at equilibrium, [H₃O⁺] = 0.027 M; [HSO₄^–] = 0.29 M; and [SO₄^2-] = 0.13 M.

Example 3

Determination of K_b from Equilibrium Concentrations
Caffeine, C₈H₁₀N₄O₂ is a weak base. What is the value of K_b for caffeine if a solution at equilibrium has [C₈H₁₀N₄O₂] = 0.050 M, [C₈H₁₀N₄O₂H⁺] = 5.0 × 10⁻³ M, and [OH⁻] = 2.5 × 10⁻³ M?

Solution
At equilibrium, the value of the equilibrium constant is equal to the reaction quotient for the reaction:

Check Your Learning
What is the equilibrium constant for the ionization of the HPO₄^2- ion, a weak base:

In a solution containing a mixture of NaH₂PO₄ and Na₂HPO₄ at equilibrium, [OH⁻] = 1.3 × 10⁻⁶ M; [H₂PO₄^–] = 0.042 M; and [HPO₄^2-] = 0.341 M.

Example 4

Determination of K_a or K_b from pH
The pH of a 0.0516 M solution of nitrous acid, HNO₂, is 2.34. What is its K_a?

Solution
We determine an equilibrium constant starting with the initial concentrations of HNO₂, H₃O⁺, and NO₂^– as well as one of the final concentrations, the concentration of hydronium ion at equilibrium. (Remember that pH is simply another way to express the concentration of hydronium ion.)

We can solve this problem with the following steps in which x is a change in concentration of a species in the reaction:

pH → [H₃O⁺]_eq → x (in ICE table) → Calculate [HNO₂]_eq and [NO₂^–]_eq → K_a

We can first create an ICE table for this reaction:

Next, we can use the pH to find [H₃O⁺]_eq, which is “x” in our ICE table:

[H₃O⁺]_eq = 10^-pH = 10^-2.34 = 0.0046 M = x

Finally, we calculate the value of the equilibrium constant using the data in the table:

K_a = [latex]\frac{[H_{3}O^{+}][NO_{2}^{-}]}{[HNO_{2}]}[/latex] = [latex]\frac{(0.0046)(0.0046)}{(0.0470)}[/latex] = 4.5 × 10^-4

Check Your Learning.
The pH of a solution of household ammonia, a 0.950 M solution of NH_3, is 11.612. What is K_b for NH₃.

Example 5

Equilibrium Concentrations in a Solution of a Weak Acid
Formic acid, HCO₂H, is the irritant that causes the body’s reaction to ant stings (Figure 2).

What is the concentration of hydronium ion and the pH in a 0.534-M solution of formic acid?

Solution

1. Determine x and equilibrium concentrations. The equilibrium expression is:
   HCO₂H(aq)  +  H₂O(l)   ⇌   H₃O⁺(aq)  +  HCO₂^–(aq)

   The concentration of water does not appear in the expression for the equilibrium constant, so we do not need to consider its change in concentration when setting up the ICE table.

   The table shows initial concentrations (concentrations before the acid ionizes), changes in concentration, and equilibrium concentrations follows (the data given in the problem appear in color):

   	HCO2H(aq) + H2O(l)  ⇌  H3O+(aq) + HCO2–(aq)
   I	0.534	–	0	0
   C	– x	–	+ x	+ x
   E	0.534 – x	–	x	x

2. Solve for x and the equilibrium concentrations. At equilibrium:
   K_a = 3.0 × 10^-4 = [latex]\frac{[H_{3}O^{+}][HCO_{2}^{-}]}{[HCO_{2}H]}[/latex] = [latex]\frac{(x)(x)}{0.534 - x}[/latex]

Now solve for x. Because the initial concentration of acid is reasonably large and K_a is very small, we assume that x << 0.534, which permits us to simplify the denominator term as (0.534 − x) = 0.534. This gives:

Solve for x as follows:

To check the assumption that x is small compared to 0.534, we calculate:

x is less than 5% of the initial concentration; the assumption is valid.

We find the equilibrium concentration of hydronium ion in this formic acid solution from its initial concentration and the change in that concentration as indicated in the last line of the table:

The pH of the solution can be found by taking the negative log of the [H₃O⁺], so:

Check Your Learning
Only a small fraction of a weak acid ionizes in aqueous solution. What is the percent ionization of acetic acid in a 0.100-M solution of acetic acid, CH₃CO₂H?

(Hint: Determine [CH₃CO₂^–] at equilibrium.) Recall that the percent ionization is the fraction of acetic acid that is ionized × 100, or [latex]\frac{[CH_{3}CO_{2}^{-}]}{[CH_{3}CO_{2}H]_{initial}}[/latex] × 100.

Example 6

Equilibrium Concentrations in a Solution of a Weak Base
Find the concentration of hydroxide ion in a 0.25 M solution of trimethylamine, a weak base:

Solution
This problem requires that we calculate an equilibrium concentration by determining concentration changes as the ionization of a base goes to equilibrium. The solution is approached in the same way as that for the ionization of formic acid in Example 5. The reactants and products will be different and the numbers will be different, but the logic will be the same:

1. Determine x and equilibrium concentrations. The table shows the changes and concentrations
   

   	(CH3)3N(aq) + H2O(l)  ⇌  (CH3)3NH+(aq) + OH–(aq)
   I	0.25	–	0	0
   C	– x	–	+ x	+ x
   E	0.25 – x	–	x	x

    

2. Solve for x and the equilibrium concentrations. At equilibrium:
   K_b = [latex]\frac{[(CH_{3})_{3}NH^{+}][OH^{-}]}{[(CH_{3})_{3}N]}[/latex] = [latex]\frac{(x)(x)}{0.25 - x}[/latex] = 6.2 × 10^-5

   If we assume that x is small relative to 0.25, then we can replace (0.25 − x) in the preceding equation with 0.25. Solving the simplified equation gives:

   x = 3.9 × 10^-3

   This change is less than 5% of the initial concentration (0.25), so the assumption is justified.

   Recall that, for this computation, x is equal to the equilibrium concentration of hydroxide ion in the solution (see earlier tabulation):

   [OH^–] = ~0 + x = x = 3.9 × 10^-3 M

   Then calculate pOH as follows:

   pOH = -log[OH^–] = -log(3.9 × 10^-3) = 2.40

   Using the relation introduced in the previous section of this chapter:

   pH + pOH = pK_w = 14.00

   permits the computation of pH:

   pH = 14.00 – pOH = 14.00 – 2.40 = 11.60
3. Check the work. A check of our arithmetic shows that K_b = 6.2 × 10⁻⁵.

Check Your Learning
(a) Show that the calculation in Step 2 of this example gives an x of 3.9 × 10⁻³ and the calculation in Step 3 shows K_b = 6.2 × 10⁻⁵.

(b) Find the concentration of hydroxide ion in a 0.0325 M solution of ammonia, a weak base with a K_b of 1.77 × 10⁻⁵. Calculate the percent ionization of ammonia, the fraction ionized × 100, or [latex]\frac{[NH_{4}^{+}]}{[NH_{3}]}[/latex] × 100

Example 7

Equilibrium Concentrations in a Solution of a Weak Acid
Sodium bisulfate, NaHSO₄, is used in some household cleansers because it contains the HSO₄^– ion, a weak acid. What is the pH of a 0.50 M solution of HSO₄⁻?

Solution
We need to determine the equilibrium concentration of the hydronium ion that results from the ionization of HSO₄⁻ so that we can use [H₃O⁺] to determine the pH. As in the previous examples, we can approach the solution by the following steps:

1. Determine x and equilibrium concentrations. This table shows the changes and concentrations:
   

   	HSO4–(aq)  +  H2O(l)   ⇌   H3O+(aq)  +  SO42-(aq)
   I	0.50	–	0	0
   C	– x	–	+ x	+ x
   E	0.50 – x	–	x	x

2. Solve for x and the concentrations. As we begin solving for x, we will find this is more complicated than in previous examples. As we discuss these complications we should not lose track of the fact that it is still the purpose of this step to determine the value of x.

   At equilibrium:

   K_a = 1.1 × 10^-2 = [latex]\frac{[H_{3}O^{+}][SO_{4}^{2-}]}{[HSO_{4}^{-}]}[/latex] = [latex]\frac{(x)(x)}{0.50 - x}[/latex]

   If we assume that x is small and approximate (0.50 − x) ≈ 0.50, we find:

   1.1 × 10^-2 = [latex]\frac{(x)(x)}{0.50}[/latex]

   x = 7.4 × 10^-2 M

   When we check the assumption, we calculate:

   [latex]\frac{x}{0.50}[/latex] = [latex]\frac{7.4 \times 10^{-2}}{0.50}[/latex] = 0.15 = 15%

   The value of x is not less than 5% of 0.50, so the assumption is not valid. We need the quadratic formula to find x.

   The equation:

   K_a = 1.1 × 10^-2 = [latex]\frac{(x)(x)}{0.50 - x}[/latex]

   gives

   (1.1 × 10^-2)(0.50 – x) = x²

   x² + 1.1 × 10^-2x – 5.5 × 10^-3 = 0

   This equation can be solved using the quadratic formula. Solving for x gives a negative root (which cannot be correct since concentration cannot be negative) and a positive root:

   x = 6.9 × 10^-2

   Now determine the hydronium ion concentration and the pH:

   [H₃O⁺] = 0 + x = 0 + 6.9 × 10^-2 M = 6.9 × 10^-2 M

   The pH of this solution is:

   pH = -log[H₃O⁺] = -log(6.9 × 10^-2) = 1.16

Check Your Learning
(a) Show that the quadratic formula gives x = 6.9 × 10⁻².

(b) Calculate the pH in a 0.010-M solution of caffeine, a weak base:

(Hint: It will be necessary to convert [OH⁻] to [H₃O⁺] or pOH to pH toward the end of the calculation.)