Question 17.SE.11: Solid silver chromate is added to pure water at 25 ºC, and s...
Solid silver chromate is added to pure water at 25 ºC, and some of the solid remains undissolved. The mixture is stirred for several days to ensure that equilibrium is achieved between the undissolved Ag_2CrO_4(s) and the solution. Analysis of the equilibrated solution shows that its silver ion concentration is 1.3 × 10^{-4} M. Assuming that the Ag_2CrO_4 solution is saturated and that there are no other important equilibria involving Ag^+\, or \,CrO_4²^ – ions in the solution, calculate K_{sp} for this compound.
Analyze We are given the equilibrium concentration of Ag^+ in a saturated solution of Ag_2CrO_4 and asked to determine the value of K_{sp} for Ag_2CrO_4.
Plan The equilibrium equation and the expression for K_{sp} are
\begin{aligned}& Ag _2 CrO _4(s) \rightleftharpoons 2 Ag ^{+}(a q)+ CrO _4^{2-}(a q) \\& K_{s p}=\left[ Ag ^{+}\right]^2\left[ CrO _4^{2-}\right]\end{aligned}
To calculate K_{sp}, we need the equilibrium concentrations of Ag^+ and CrO_4^{2-}. We know that at equilibrium \left[ Ag ^{+}\right]=1.3 \times 10^{-4}\,M. All the Ag^+ and CrO_4^{2 -} ions in the solution come from the Ag_2CrO_4 that dissolves. Thus, we can use \left[ Ag ^{+}\right] to calculate \left[ CrO _4^{2-}\right].
Solve From the chemical formula of silver chromate, we know that there must be two Ag^+ ions in solution for each CrO_4^{2 -} ion in solution. Consequently, the concentration of CrO_4^{2 -} is half the concentration of Ag^+ :
\left[ CrO _4^{2-}\right]=\left(\frac{1.3 \times 10^{-4}\,\cancel{mol\,Ag ^{+}}}{L}\right)\left(\frac{1 \,mol\, CrO _4^{2-}}{2\, mol\,Ag ^{+}}\right)=6.5 \times 10^{-5}\,M
and K_{sp} is
K_{s p}=\left[ Ag ^{+}\right]^2\left[ CrO _4^{2-}\right]=\left(1.3 \times 10^{-4}\right)^2\left(6.5 \times 10^{-5}\right)=1.1 \times 10^{-12}
Check We obtain a small value, as expected for a slightly soluble salt. Furthermore, the calculated value agrees well with the one given in Appendix D, 1.2 × 10^{-12}.