The for is — Ionic Equilibrium Chemistry Question
Question
The $K_{sp}$ for $\text{AgBrO}_3$ is
Answer: D
💡 Solution & Explanation
The dissolution of the sparingly soluble salt is $\text{AgBrO}_3\text{(s)} \rightleftharpoons \text{Ag}^+\text{(aq)} + \text{BrO}_3^-\text{(aq)}$. Because it is a 1:1 salt, the concentration of $\text{BrO}_3^-$ equals the original concentration of $\text{Ag}^+$, which is $2.0 \times 10^{-4}\text{ M}$. The solubility product is $K_{sp} = [\text{Ag}^+][\text{BrO}_3^-] = (2.0 \times 10^{-4})(2.0 \times 10^{-4}) = 4.0 \times 10^{-8}$. Therefore, correct answer is D.
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