Q.18 The solubility product of AgBr is 5.2×10–13. Calculate its solubility in mol dm–3 and g dm–3. (Molar mass of AgBr = 187.8 g mol–1)
Best answer
Answer: S = 7.21 × 10–7 mol dm–3 = 1.354 × 10–4 g dm–3
AgBr(s) ⇌ Ag+(aq) + Br–(aq); if solubility is S mol dm–3, [Ag+] = [Br–] = S.
\(K_{sp} = S^2 \Rightarrow S = \sqrt{5.2\times10^{-13}} = \sqrt{52\times10^{-14}} = 7.211\times10^{-7}\ \text{mol dm}^{-3}\)
Solubility in g dm–3 = S × molar mass \(= 7.211\times10^{-7}\times187.8 = 1.354\times10^{-4}\ \text{g dm}^{-3}\)
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