Question

You are given an Erlenmeyer flask containing 100.00 mL of 0.185 M B- and a buret containing 0.141 M HClO4. B- is a weak base with Kb of 5.5 x 10-6. Calculate the pH of the solution (to 2 decimal places) after titrating the following volumes of HClO4:

What is the pH after 0.0 mL HClO₄ is added?

What volume of HClO₄ is added at the equivalence point (to 0.1 mL)?

What is the pH at the equivalence point?

What is the pH after 97.0 mL HClO₄ is added?

What is the pH after 196.0 mL HClO₄ is added?

Answer 1

1) When 0.0 mL HClO₄ is added, we have only B^-. Consider the ionization of the base B^- as below.

B^- (aq) + H₂O (l) <=====> BH (aq) + OH^- (aq)

Since OH^- is formed, we work with K_b.

K_b = [BH][OH^-]/[B^-]

=====> 5.5*10^-6 = (x)(x)/(0.185 – x)

where x = equilibrium concentration of BH = equilibrium concentration of OH^- due to 1:1 nature of the ionization.

Since K_b is small, we can assume x << 0.185 M and write

5.5*10^-6 = x²/(0.15)

=====> x² = 1.0175*10^-6

=====> x = 1.0087*10^-3

Therefore, [OH^-] = 1.0087*10^-3 M and pOH = -log [OH^-] = -log (1.0087*10^-3 M) = 2.996.

We know that

pH + pOH = 14; therefore,

pH = 14 – pOH

= 14 – 2.996 = 11.004 ≈ 11.00 (ans).

2) Millimoles of B^- = (100.00 mL)*(0.185 M) = 18.5 mmole.

The neutralization reaction is given as

B^- (aq) + HClO₄ (aq) ---------> BH (aq) + ClO₄^- (aq)

As per the stoichiometric equation,

1 mole B^- = 1 mole HClO₄.

Therefore, millimoles of HClO₄ required to reach the equivalence point = 18.5 mmole.

Volume of HClO₄ required to reach the equivalence point = (18.5 mmole)/(0.141 M) =131.2057 mL ≈ 131.2 mL (ans).

3) At the equivalence point, we have BH, which is the conjugate acid of the weak base B^-. BH ionizes as below.

BH (aq) + H₂O (l) <======> B^- (aq) + H₃O⁺ (aq)

Since H₃O⁺ is formed, we work with K_a, the acid ionization constant of BH.

Given K_b = 5.5*10^-6, we know that

K_a*K_b = K_w

=====> K_a = K_w/K_a

=====> K_a = (1.0*10^-14)/(5.5*10^-6) [K_w = 1.0*10^-14]

=====> K_a = 1.818*10^-9

Millimoles of BH formed = millimoles of B^- neutralized = 18.5 mmole.

Total volume of the solution = (100.00 + 131.2) mL = 231.2 mL.

[BH] at equilibrium = (millimoles of B^-)/(volume of solution)

= (18.5 mmole)/(231.2 mL)

= 0.0800 M.

Write down the expression for the acid dissociation constant as

K_a = [B^-][H₃O⁺]/[BH]

=====> 1.818*10^-9 = (x)(x)/(0.0800 – x)

Since K_a is small, we assume x << 0.0800 M and thus,

1.818*10^-9 = x²/(0.0800)

=====> x² = 1.4544*10^-10

=====> x = 1.20598*10^-5

Therefore, [H₃O⁺] = 1.20598*10^-5 M and pH = -log [H₃O⁺] = -log (1.20598*10^-5 M)

= 4.918 ≈ 4.92 (ans).

4) Given K_b = 5.5*10^-6,we have

pK_b = -log (K_b)

= -log (5.5*10^-6)

= 5.259

Millimoles of HClO₄ added = millimoles of B^- neutralized = millimoles of BH formed = (97.0 mL)*(0.141 M) = 13.677 mmole.

Millimoles of B^- neutralized = (18.5 – 13.677) mmole = 4.823 mmole.

Since the volume of the solution is constant for both BH and B^-, we have,

[BH]/[B^-] = (millimoles of BH)/(millimoles of B^-)

= (13.677 mmole)/(4.823 mmole)

= 2.836.

Use the Henderson-Hasslebach equation for base as below.

pOH = pK_b + log [BH]/[B^-]

= 5.259 + log (2.836)

= 5.259 + 0.453

= 5.712

Therefore, pH = 14 – pOH

= 14 – 5.712

= 8.288 ≈ 8.29 (ans).

5) Millimoles of HClO₄ added = (196.0 mL)*(0.141 M) = 27.636 mmole.

Millimoles of HClO₄ required for neutralization = 18.5 mmole.

Millimoles of HClO₄ retained at equilibrium = (27.636 – 18.5) mmole = 9.136 mmole.

Since HClO₄ is a strong acid, HClO₄ will decide the pH of the solution.

Total volume of the solution = (100.00 + 196.0) mL = 296.00 mL.

[HClO₄] at equilibrium = (millimoles of HClO₄)/(volume of solution in mL)

= (9.136 mmole)/(296.00 mL)

= 0.03086 M.

pH = -log [HClO₄]

= -log (0.03086 M)

= 1.5106 ≈ 1.51 (ans).