Assuming complete dissociation, what is the pH of a 4.52 mg/L Ba(OH)2 solution? pH=

Question

Assuming complete dissociation, what is the pH of a 4.52 mg/L Ba(OH)2 solution? pH=

Answer 1

Ba(OH)2 dissociates into Ba²⁺ + 2OH^1-.

For every mole of Ba(OH)2 you get 2 moles of OH^1-.

We need to calculate the molar concentration of Ba(OH)2.

4.52 mg is equal to 0.00452 g/L, so the # of moles is 0.00452g / 171.3 g/mole = 2.64 x 10^-5 M.

The OH^1- concentration is 2 x 2.64 x 10^-5 M = 5.28 x 10^-5 M.

the [H¹⁺] concentration is 10^-14/5.28 x 10^-5 = 1.89 x 10^-10 M.

pH = - log 1.89 x 10^-10 = 9.72