Question

In-room temperature, the solubility of AgBr is aqua is 7.1X10^-7M and that of Baf2 is 6.3X10^-3M, ask the solubility product constants of these two salts.

Answer 1

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① Ag Br⁻(s) ⇌ Ag⁺(ag) + Br⁻(ag)

The Ksp expression is:

$$K_{SP} = [Ag^+] [Br^-] = 7.1 \times 10^{-7} \times 7.1 \times 10^{-7}$$ $$K_{SP} = \boxed{50.41 \times 10^{-14}} \quad \text{☑}$$

② BaF₂(s) ⇌ Ba²⁺(ag) + βF⁻(ag)

The Ksp expression is:

$$K_{SP} = [Ba^{2+}] [F^-]^2 = 1.5 \times 10^{-6} \times 4 \times (1.5 \times 10^{-6})^2 = 13.5 \times 10^{-18}$$ $$K_{SP} = \boxed{135 \times 10^{-19}} \quad \text{☑}$$