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?
According to the equlibrium equation:
Ksp = [Ba2+]eq[OH-]2eq
Let assume that [Ba2+] = x. Then, [OH-] = 2x. Finally:
Ksp = x(2x)2
Ksp = 4x3
x3 = Ksp / 4
From here:
x = (Ksp / 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
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