Question

What are the concentrations of OH– and H in a 0.00086 M solution of Ba(OH)2 at 25 °C? Assume complete dissociation.
[OH^-]=

[H⁺]=

Answer 1

Ba(OH)2 -----> Ba+2 + 2OH-

now on complete dissociation...

since one mole Ba(OH)2 gives 2 mole OH-

hence [OH-] = 2 * [Ba(OH)2] = 2 * 0.00086 = 1.72 * 10^-3

since [OH-] * [H+] = 10^-14

[H+] = 10^-14/[OH-] = 5.81 * 10^-12