Question

1. Determine the molarity of an acetic acid solution, [HC2H3O2], with a pH of 2.87 (Ka for HC2H3O2 = 1.7 x 10^-5).

2. What is the initial concentration of an NH3 solution, [NH3], with a pH of 11.57 and a percent ionization value of 5%?

Answer 1

1) Let molarity of CH3COOH is M.

pH = 2.87 thus [H+] = 10^(-pH)

[H+] = 10^(-2.87) = 1.35*10^-3 M

Hence, x = [H+] = 1.35*10^-3 M

CH3COOH ------> CH3COO- + H+ , Ka = 1.7*10^-5

M 0 0 (at t=0)

M - x x x (at equilibrium)

Ka = x * x/(M-x)

Ka = x^2 / (M - x)

Since, CH3COOH is weak acid. It will dissociate very less. Thus, (M - x) = M (approx)

Ka = x^2 / M

M = x^2 / Ka = (1.35*10^-3)^2 / 1.7*10^-5

M = 0.108 M .....Answer

2)

pH = 11.57

pOH = 14 - pH = 14 - 11.57 = 2.43

[OH-] = 10^(-pOH) = 10^(-2.43) = 3.715*10^-3 M

Let initial concentration of NH3 is Y.

NH3 + H2O -----> NH4 + + OH-

[OH-] = Y * 5/100

3.715*10^-3 * 100 / 5 = Y

Y = 0.0743 M ....Answer

Let me know if any doubts.