Answer to Question #169631 in Chemistry for Johj Juah

Question

An ionic compound A₂B dissociates in water according to the equation

A₂B(s) ⇌ 2A⁺(aq) + B^2–(aq)

If the concentration of B^2– ions in a saturated solution of A₂B is 3.2 × 10^–3 M, then what is the solubility product constant (K_sp) of this compound?

•  A. 8.2 × 10^–7
•  B. 3.3 × 10^–8
•  C. 1.3 × 10^–7
•  D. 5.6 × 10^–7
•  E. 2.2 × 10^–7

Answer 1

Solution:

An ionic compound A₂B dissociates in water according to the equation:

A₂B(s) ⇌ 2A⁺(aq) + B^2–(aq)

[B^2–] = 3.2×10^–3 M

[A⁺] = 2 × [B^2–] = 2 × 3.2×10^–3 M = 6.4×10^–3 M (according to the equation above)

K_sp = [A⁺]² × [B^2–]

Hence,

K_sp = (6.4×10^–3)² × (3.2×10^–3) = 1.31×10^–7 = 1.3×10^–7

K_sp = 1.3×10^–7

Answer: C. 1.3×10^–7