Question

1)The Ka of a monoprotic weak acid is 7.59 × 10-3. What is the percent ionization of a 0.106 M solution of this acid? 2)The Ka value for acetic acid, CH3COOH(aq), is 1.8× 10–5. Calculate the pH of a 1.80 M acetic acid solution.Calculate the pH of the resulting solution when 2.00 mL of the 1.80 M acetic acid is diluted to make a 250.0 mL solution.

Answer 1

1)The Ka of a monoprotic weak acid is 7.59 × 10-3. What is the percent ionization of a 0.106 M solution of this acid?

Let the acid dissociates as follows:

HA + H₂O = H₃O⁺ + A^-

Then you do an ICE table H₂O is left out because it is a pure liquid

                              HA        +       H₂O          = H₃O⁺            +           A^-

Initial: .                 0.106                                    0                               0

change:                  -x                                      +x                             +x

equilibrium: .        0.106-x +x                              +x

K_a= [H₃O⁺][A^-] / [HA]

7.59 × 10^-3 = [x]²/ [0.106 - x]

x² + (7.59 × 10^-3)x – 8.05 × 10^-4 = 0

Solving the quadratic equation, we get

x = 0.025

Now, to find the percent ionization of the weak acid:

([H⁺] @ equilibrium / [HA] initial ) x 100 = (0.025 /0.106) x 100 = 23.58 %.

2.) Part -I

CH₃COOH(aq)    CH3COO^-(aq) + H⁺(aq)

Ka = [H⁺] [CH₃COO^-] / ([CH₃COOH]-x)

Since, [H⁺] = [CH₃COO^-] = x

(-x) is excluded as Ka & conc. differs >factor 100

Ka = x²/ [HA]

1.8 x 10^-5 = x² / 1.8

x² = 3.2 x 10^-5

x = 5.66 x 10^-3

[H+] = x = 5.66 x 10^-3

pH = -log[H+] = - log (5.66 x 10^-3) = 2.25

Part -II

Now, M1V1 = M2V2

(1.80 M)(0.002 L)=(M2) (0.250L)

M2 = 0.0144

Now,

1.8 x 10^-5 = x² / 0.0144

x² = 2.6 x 10^-7

x = 5.1 x 10^-4

So, [H⁺] = x = 5.1 x 10^-4

Hence pH = - log[H⁺] = - log (5.1 x 10^-4) = 3.29