What mass of KOH would need to be dissolved in 500.0 mL of water to produce a solution with a pH of 12.40?

Answer 1

A pH of 12.4 indicates a pOH of 14-12.4 = 1.6. As pOH = -log[OH^-], the [OH^-] is the antilog of

- 1.6 = 0.0251 M/L.

The number of grams of OH^- required would be

0.0251 moles / L x 17.0 g/mole x 1L/1000 mL x 500 mL = 0.213 g OH^-. The mass of KOH needed then is molar mass KOH / molar mass OH^- x g OH^-.

56.1 g / 17.0 g x 0.213 g OH^- = 0.703 g KOH.

I only used 3 sig figs for the calculations. Always check math.