Equilibrium in Solution

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Physical Chemistry › Equilibrium in Solution

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$Ca(OH)_{2(s)}\rightleftharpoons Ca_{(aq)}^{2+}+2OH_{(aq)}^{-}$ $K_{sp}=5.5\times10^{-6}$

A saturated aqueous solution of calcium hydroxide is prepared. Given the information above, how will adding of $1M\ NaOH$ affect the solubility and the $K_{sp}$ of the solution?

The solubility will decrease; the $K_{sp}$ will remain unchanged.

The solubility will increase; the $K_{sp}$ will remain unchanged

The solubility will remain unchanged; the $K_{sp}$ will remain unchanged

The solubility will decrease; the $K_{sp}$ will decrease

The solubility will increase; the $K_{sp}$ will increase

Explanation

Adding contributes ions into the solution, driving the reaction to the left, and decreasing the solubility of the calcium hydroxide. This is known as the common-ion effect. The $K_{sp}$ remains unchanged. This is due to the fact that only a change in temperature can bring about a change in an equilibrium constant.

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