Chemistry: The Molecular Nature of Matter and Change 7th Edition

$(a) Solubility = 1.4 \times 10^{-5}M$ $(b) Solubility = 1.4 \times 10^{-7}M$
(a) 1. Write the $K_{sp}$ expression: $BaCrO_4(s) \lt -- \gt 1Ba^{2+}(aq) + 1CrO{_4}^{2-}(aq)$ $2.1 \times 10^{-10} = [Ba^{2+}]^ 1[CrO{_4}^{2-}]^ 1$ 2. Considering a pure solution: $[Ba^{2+}] = 1S$ and $[CrO{_4}^{2-}] = 1S$ $2.1 \times 10^{-10}= ( 1S)^ 1 \times ( 1S)^ 1$ $2.1 \times 10^{-10} = 1S^ 2$ $\sqrt [ 2] {2.1 \times 10^{-10}} = S$ $1.4 \times 10^{-5} = S$ - This is the molar solubility value for this salt. ------- (b) 3. Write the $K_{sp}$ expression: $BaCrO_4(s) \lt -- \gt 1Cr{O_4}^{2-}(aq) + 1Ba^{2+}(aq)$ $2.1 \times 10^{-10} = [Cr{O_4}^{2-}]^ 1[Ba^{2+}]^ 1$ $2.1 \times 10^{-10} = (1.5 \times 10^{-3} + S)^ 1( 1S)^ 1$ 4. Find the molar solubility. Since 'S' has a very small value, we can approximate: $[Cr{O_4}^{2-}] = 1.5 \times 10^{-3}M$ $2.1 \times 10^{-10}= (1.5 \times 10^{-3})^ 1 \times ( 1S)^ 1$ $2.1 \times 10^{-10}= (1.5 \times 10^{-3})^ 1 \times ( 1S)^ 1$ $2.1 \times 10^{-10}= 1.5 \times 10^{-3} \times ( 1S)^ 1$ $\frac{2.1 \times 10^{-10}}{1.5 \times 10^{-3}} = ( 1S)^ 1$ $1.4 \times 10^{-7} = ( 1S)^ 1$ $1.4 \times 10^{-7} = S$