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Exercises · 6.58

Q.The solubility of Sr(OH)2 at 298 K is 19.23 g/L of solution. Calculate the concentrations of strontium and hydroxyl ions and the pH of the solution.

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Convert solubility to molar solubility ss, then [Sr2+]=s=0.158 M[\text{Sr}^{2+}] = s = 0.158\ \text{M}, [OH−]=2s=0.316 M[\text{OH}^-] = 2s = 0.316\ \text{M}, and pH=13.5\text{pH} = 13.5.

1. Molar solubility. M(Sr(OH)2)=87.6+2(16.0+1.0)=121.6 g mol−1M(\text{Sr(OH)}_2) = 87.6 + 2(16.0 + 1.0) = 121.6\ \text{g mol}^{-1}

s=19.23121.6=0.158 mol L−1s = \frac{19.23}{121.6} = 0.158\ \text{mol L}^{-1}

2. Dissociation.

Sr(OH)2(s)⇌Sr2+(aq)+2 OH−(aq)\text{Sr(OH)}_2(s) \rightleftharpoons \text{Sr}^{2+}(aq) + 2\,\text{OH}^-(aq) …

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