Answer to Question #202797 in Chemistry for Shaza Shihadeh

Question

The solubility product (Ksp) is 0.005 for the reaction below,

Ba (OH)2 (s) ⇌ Ba2+ (aq) + 2OH- (aq)

If the initial concentration of Ba (OH)2 is [Ba(OH)2] = 2M, what is the concentration of

[H3O+] at equilibrium?

Answer 1

According to the equation:

K_sp = [Ba²⁺]_eq[OH^-]²_eq

Lets assume that [Ba²⁺] = x. Then, [OH^-] = 2x. Finally:

K_sp = x(2x)²

K_sp = 4x³

x³ = K_sp / 4

From here:

x = (K_sp / 4)^1/3 = (0.005 / 4)^1/3 = 0.107

As a result, [OH^-] = 2 × 0.107 = 0.214

Finally:

[H₃O⁺] = 1 × 10^-14 / [OH^-] = 1 × 10^-14 / 0.214 = 4.7 × 10^-14 M

Answer: 4.7 × 10^-14 M