Question

what is the solubility of ca(oh)₂ in 0.480 m ba(oh)₂? k_sp for ca(oh)₂ is 6.50 × 10⁻⁶. assume ba(oh)₂ dissolves completely.

Explanation:

Step1: Write the dissolution equation and \(K_{sp}\) expression

The dissolution of \(Ca(OH)_2\) is \(Ca(OH)_2(s)
ightleftharpoons Ca^{2 +}(aq)+2OH^{-}(aq)\). The \(K_{sp}\) expression is \(K_{sp}=[Ca^{2 +}][OH^{-}]^{2}\).

Step2: Determine the concentration of \(OH^{-}\) from \(Ba(OH)_2\)

Since \(Ba(OH)_2\) dissolves completely (\(Ba(OH)_2 = Ba^{2+}+2OH^{-}\)), and \([Ba(OH)_2]=0.480M\), then \([OH^{-}]=2\times0.480M = 0.960M\) (from \(Ba(OH)_2\)). Let the solubility of \(Ca(OH)_2\) be \(s\), so \([Ca^{2 +}]=s\) and the total \([OH^{-}]=0.960 + 2s\). But since \(K_{sp}\) is small, \(2s\ll0.960\), so \([OH^{-}]\approx0.960M\).

Step3: Calculate the solubility \(s\)

Substitute into \(K_{sp}\) expression: \(6.50\times 10^{-6}=s\times(0.960)^{2}\). Then \(s=\frac{6.50\times 10^{-6}}{(0.960)^{2}}\).

$$s=\frac{6.50\times 10^{-6}}{0.9216}\approx7.05\times 10^{-6}M$$

Answer:

The solubility of \(Ca(OH)_2\) is approximately \(7.05\times 10^{-6}M\)