Answer to Question #99054 in Chemistry for Mia G

Question

Calculate the [H3O+], [OH], pH, and pOH for a solution prepared by dissolving 3.1g KOH(s) (FW=56g/mol) unto enough water to give a volume of 0.500L.

Answer 1

Answer on Question #99054 – Chemistry – Other

Task:

Calculate the [H₃O⁺], [OH^-], pH, and pOH for a solution prepared by dissolving 3.1g KOH(s) (FW=56 g/mol) unto enough water to give a volume of 0.500L.

Solution:

1) Find the molar concentration of the KOH:

C_m = n/V = m / (M*V).

Then,

C(KOH) = m(KOH) / (M(KOH)*V) = 3.1 / (56*0.5) = 3.1 / 28 = 0.1107 (mol/L).

2) dissociation reaction of KOH:

KOH = K⁺ + OH^-;

KOH is strong electrolyte. Then,

C(KOH) = [OH^-] = 0.1107 mol/L;

3)pOH = -log[OH^-] = 0.96;

pOH = 0.96.

4) pOH + pH = 14;

pH = 14- pOH = 14 - 0.96 = 13.04;

pH = 13.04.

5) pH = -log[H₃O⁺];

[H₃O⁺] = 10^-pH = 9.12*10^-14 mol/L;

[H₃O⁺] = [H⁺] = 9.12*10^-14 mol/L.

Answer: [H₃O⁺] = 9.12*10^-14 mol/L ; [OH^-] = 0.1107 mol/L; pH = 13.04; pOH = 0.96.