Question

2. The equilibrium constant of a Weak Acid.

A: pH of 25mL 0.10 M unknown acid = 2.04

a) [H]?

b) [A]?

c) [HA(aq)]?

d) Keq?

B: pH of 25mL 0.20 M unknown acid + 5mL of 0.1M NaOH = 2.79

a) [H]?

b) [A]?

c) [HA(aq)]?

d) Keq?

C: pH of 15mL of 0.1M NaOH plus 30mL of Unknown Acid = 3.65

a) [H]?

b) [A]?

c) [HA(aq)]?

d) Keq?

I know the answers, but want to know how to do it. Please show me step-by-step how you get each answer.

Answer 1

A) The pH of the acid is given as pH = 2.04.

(a) We define pH as pH = -log [H⁺]; therefore, [H⁺] = antilog (-pH) = antilog (-2.04) = 9.12*10^-3 M (ans).

(b) Assume a monoprotic acid HA which dissociates as

HA (aq) <======> H⁺ (aq) + A^- (aq)

Note the 1:1 nature of dissociation; therefore, [H⁺] = [A^-] = 9.12*10^-3 M (ans).

(c) Assume the acid is weak; therefore, we can assume the extent of ionization to be very low. Consequently, [HA] = 0.10 M (ans). This becomes true also because [H⁺] << [HA].

(d) K_eq = [H⁺][A^-]/[HA] = (9.12*10^-3)*(9.12*10^-3)/(0.1) = 8.31744*10^-4 ≈ 8.32*10^-4 (ans).

B) We have pH = 2.79.

(a) [H⁺] = antilog (-pH) = antilog (-2.79) = 1.62*10^-3 M (ans).

(b) Here [A^-] is not the same as [H⁺] since there is no 1:1 dissociation. Rather, HA reacts with NaOH as below.

HA (aq) + NaOH (aq) ------> NaA (aq) + H₂O (l)

As per the stoichiometric equation,

1 mole HA = 1 mole NaOH = 1 mole NaA.

Moles NaA (A^-) formed = moles NaOH added = (5 mL)*(0.1 M) = 0.5 mmole.

Volume of the solution = (25.0 + 5.0) mL = 30.0 mL; therefore, [A^-] = (0.5 mmole)/(30.0 mL) = 0.017 M (ans).

c) Millimoles of HA added initially = (25.0 mL)*(0.20 M) = 5.0 mmole.

Millimoles HA neutralized = millimoles NaOH added = 0.5 mmole; therefore, millimoles HA retained at equilibrium = (5.0 – 0.5) mmole = 4.5 mmole.

[HA] = (4.5 mmole)/(30.0 mL) = 0.15 M (ans).

d) K_eq = [H⁺][A^-]/[HA] = (1.62*10^-3)*(0.017)/(0.15) = 1.836*10^-4 (ans).

C) We need to know the initial concentration of HA, which you have missed out.