At \(25^{\circ} \mathrm{C}\), the reaction $$\mathrm{CaCrO}_{4}(s) \rightleftharpoons \mathrm{Ca}^{2+}(a q)+\mathrm{CrO}_{4}^{2-}(a q)$$ has an equilibrium constant \(K_{c}=7.1 \times 10^{-4}\). What are the equilibrium concentrations of \(\mathrm{Ca}^{2+}\) and $\mathrm{CrO}_{4}{ }^{2-}\( in a saturated solution of \)\mathrm{CaCrO}_{4} ?$

Set up an ICE table to represent the initial concentrations of the reactants and products, the changes in their concentrations as the reaction proceeds, and their concentrations at equilibrium. In this case, since we are dealing with a saturated solution of \(\mathrm{CaCrO}_{4}\), the initial concentration of \(\mathrm{CaCrO}_{4}\) can be represented as "s", and the concentrations of \(\mathrm{Ca}^{2+}\) and \(\mathrm{CrO}_{4}^{2-}\) as 0. ICE Table: | | Initial | Change | Equilibrium | |---------------|------------|--------------|-------------------| | \(\mathrm{CaCrO}_{4}\) | s | -x | s - x | | \(\mathrm{Ca}^{2+}\) | 0 | +x | x | | \(\mathrm{CrO}_{4}^{2-}\) | 0 | +x | x |

Write the equilibrium constant expression, \(K_c\), for the reaction in terms of the concentrations of the reactants and products at equilibrium. \[K_c = \frac{[\mathrm{Ca}^{2+}][\mathrm{CrO}_{4}^{2-}]}{[\mathrm{CaCrO}_{4}]}\]

Substitute the equilibrium concentrations of the reactants and products from the ICE table into the equilibrium constant expression. \[K_c = \frac{(x)(x)}{(s - x)}\] Since the dissolution of a solid is negligible compared to the concentration of the aqueous ions, we can assume that the change in concentration of \(\mathrm{CaCrO}_4\) is negligible. Thus, \((s - x) \approx s\).

Given that the equilibrium constant \(K_c = 7.1 \times 10^{-4}\), substitute this value into the equilibrium constant expression and solve for the unknown x value. \[7.1 \times 10^{-4} = \frac{x^2}{s}\] Now we need to express 's' in terms of 'x'. Since the solution is saturated, the concentration of \(\mathrm{Ca}^{2+}\) and \(\mathrm{CrO}_{4}^{2-}\) are stoichiometrically equal, hence we can write: \[s = x\] Substituting s as x in the equilibrium expression, we get: \[7.1 \times 10^{-4} = \frac{x^2}{x}\]

Solve for x to find the equilibrium concentrations of \(\mathrm{Ca}^{2+}\) and \(\mathrm{CrO}_{4}^{2-}\). \[x = 7.1 \times 10^{-4} \, \text{mol/L}\] Therefore, the equilibrium concentrations of \(\mathrm{Ca}^{2+}\) and \(\mathrm{CrO}_{4}^{2-}\) in the saturated solution are both \(7.1 \times 10^{-4} \, \text{mol/L}\).