Answer:
The solubility of silver chromate in 1 M potassium chromate solution is \(5.29\times 10^{-7} mol/L\)
.
Explanation:
The dissociation of silver chromate is written as:
\(Ag_2CrO_4\rightleftharpoons 2Ag^{+}+CrO_4^{-}\)
The expression for solubility constant for this reaction will be,
\(K_{sp}=[Ag^{+}]^2[CrO_4^{-}]\)
\(K_{sp}=(2S)^2\times (S)\)
\(K_{sp}=1.12\times 10^{-12}\)
The dissociation of \(KCrO_4\) is written as:
\(K_2CrO_4\rightleftharpoons 2K^++CrO_4^{-}\)
1.0M 2 × 1.0M 1.0M
The new expression for solubility constant will be silver chromate will be:,
\(1.12\times 10^{-12}=(2s)^2\times (s+1.0)\)
\(1.12\times 10^{-12}=(4S^2)\times (s+1.0)\)
\(1.12\times 10^{-12}=4S^3+4.0S^2\)
By solving the term, we get:
\(s=5.29\times 10^{-7}M\)
The solubility of silver chromate in 1 M potassium chromate solution is \(5.29\times 10^{-7} mol/L\).