Question
The solubility product of barium chromate is $2.4 \times$ $10^{-10}$, the maximum concentration of barium nitrate possible without precipitation in a solution of $6 \times$ $10^{-4} \mathrm{M} \mathrm{K}_{2} \mathrm{CrO}_{4}$ is(1) $4 \times 10^{\top} \mathrm{M}$(2) $1.2 \times 10^{10} \mathrm{M}$(3) $6 \times 10^{4} \mathrm{M}$(4) $3 \times 10^{4} \mathrm{M}$
Step 1
Step 1: The solubility product of barium chromate is given by the equation: \[BaCrO_4 \rightarrow Ba^{2+} + CrO_4^{2-}\] The solubility product constant, $K_{sp}$, is given by: \[K_{sp} = [Ba^{2+}][CrO_4^{2-}]\] Show more…
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