Question

Determine the Concentration of H+
, OH-
, pH, and pOH of the following solutions:

1. 0.0100 M
2. 0.0200 M HOCl [ HOCl(aq) + H2O(l) ⇄ H3O+(aq) + OCl-(aq); Ka = 3.0 x 10 -8]

Answer 1

HOCl is a weak acid that dissociates according to the following reaction:

HOCl(aq) + H₂O(l) ⇄ H₃O⁺(aq) + OCl^-(aq).

If the equilibrium concentration of H₃O⁺ is x, then the equilibrium constant K_a can be expressed in function of x:

$K_a = \frac{[H_3O^+][OCl^-]}{[HOCl]} = \frac{x^2}{c(HOCl)-x}$ .

As K_a is very small, the can suggest that x<<c(HOCl). Then, the equilibrium concentration of [H₃O⁺] (you can also use the [H⁺] notation) is $\sqrt{K_ac(HOCl)}$ :

1. c(HOCl)=0.0100 M : 1.73·10^-5 M
2. c(HOCl)=0.0200 M : 2.45·10^-5 M.

Therefore, the pH of these solutions is:

1. c(HOCl)=0.0100 M : pH = -log([H₃O⁺]) = 4.76
2. c(HOCl)=0.0200 M : pH = 4.61.

The pOH of water solutions can be found from the ionic product K_W:

$$pH + pOH = 14$$

1. c(HOCl)=0.0100 M : pOH = 14 - 4.76 = 9.24
2. c(HOCl)=0.0200 M : pOH = 14 - 4.61 = 9.39.

Finally, the equilibrium concentration of OH^- in these solutions:

1. c(HOCl)=0.0100 M : [OH^-] = 10^-9.24 = 5.77·10^-10 M
2. c(HOCl)=0.0200 M : [OH^-] = 10^-9.39 = 4.08·10^-10 M.

Answer:

1. c(HOCl)=0.0100 M : [H⁺]=1.73·10^-5 M, pH = 4.76, pOH = 9.24, [OH^-] = 5.77·10^-10 M
2. c(HOCl)=0.0200 M : [H⁺]=2.45·10^-5 M, pH = 4.61, pOH = 9.39, [OH^-] = 4.08·10^-10 M