Answer to Question #169630 in Chemistry for Jane Virgin

Question

What is the maximum possible concentration of Ba²⁺ ions in a NaOH solution if the concentration of OH^– is maintained at 0.100 M? The solubility product constant of Ba(OH)₂ is K_sp = 5.0 × 10^–3.

•  A. 0.32 M
•  B. 0.020 M
•  C. 0.50 M
•  D. 0.080 M
•  E. 0.125 M

Answer 1

Solution:

NaOH → Na⁺ + OH^–

According to the equation above: C(NaOH) = [Na⁺] = [OH^–]

Hence, [OH^–] = C(NaOH) = 0.100 M

Ba(OH)₂ → Ba²⁺ + 2OH^–

Ba(OH)₂ contains 2 hydroxide ions (OH^–), therefore: [OH^–] = 2 × 0.100 M = 0.200 M

K_sp = [Ba²⁺] × [OH^–]²

5.0×10^–3 = [Ba²⁺] × (0.200)²

[Ba²⁺]_max = (5.0×10^–3) / (0.04) = 0.125 M

[Ba²⁺]_max = 0.125 M

Answer: E. 0.125 M