Solutions - Chemistry

Question

A solution has 42.7 grams of KOH dissolved into 692.8 mL. How much water in mL must be added to reduce the concentration of the solution to 0.073 M?

Answer 1

First calculate the Molarity of the KOH solution when 42.7 g KOH is dissolved in 692.8 mL of water.

mol KOH = 42.7 g KOH x (1 mol KOH / 56.11 g KOH) = 0.761 mol KOH

L of solution = 692.8 mL x (1 L / 1000 mL) = 0.6928 L

Molarity = mol / L = 0.761 mol / 0.6928 L = 1.10 M KOH

Next use the dilution equation to solve for the final volume since the concentration is decreasing from 1.10 M to 0.073 M. We will also assume that we are using the entire volume of the solution, 0.6928 L, to decrease the concentration.

M_iV_i = M_fV_f

V_f = M_iV_i / M_f

V_f = (1.10 M x 0.6928 L) / 0.073 M = 10.4 L x (1000 mL / 1 L) = 10,400 mL

Amount of water needed to decrease the concentration to 0.073 M = 10,400 - 692.8 = 9,707 mL