Answer to Question #190351 in Chemistry for Kenan

Question

Calculate the concentration of H, OH and pH and pOH of an aqueous solution of calcium hydroxide with a concentration of 0.005 mol / dm3 assuming complete dissociation

Answer 1

Solution:

calcium hydroxid - Ca(OH)₂

Dissociation equation for Ca(OH)₂:

Ca(OH)₂ → Ca²⁺ + 2OH^-

According to the equation above: C_M[Ca(OH)₂] = [OH^-]/2

[OH^-] = 2 × C_M[Ca(OH)₂] = 2 × 0.005 M = 0.01 M

[OH^-] = 0.01 M

We can convert between [OH^-] and pOH using the following equation:

pOH = - log[OH^-]

pOH = - log(0.01) = 2

pOH = 2

For any aqueous solution at 25^∘C:

pH + pOH = 14

pH = 14 - pOH = 14 - 2 = 12

pH = 12

We can convert between pH and [H⁺] using the following equation:

pH = - log[H⁺]

[H⁺] = 10^-pH = 1×10^-12 M

[H⁺] = 1×10^-12 M

Answer:

[H⁺] = 1×10^-12 M;

[OH^-] = 0.01 M;

pH = 12;

pOH = 2.