Given a solution containing Ca^{2+} at 2.00 imes 10^{-2} ext{ M} and CrO_4^{2-} at 3.00 imes 10^{-2} ext{ M}, what is the value of the reaction quotient Q for the precipitation of CaCrO_4?

• A. Q = (2.00 imes 10^{-2})^2 imes (3.00 imes 10^{-2})^2 = 3.60 imes 10^{-6}
• B. Q = (2.00 imes 10^{-2}) / (3.00 imes 10^{-2}) = 0.667
• C. Q = (2.00 imes 10^{-2}) + (3.00 imes 10^{-2}) = 5.00 imes 10^{-2}
• D. Q = (2.00 imes 10^{-2}) imes (3.00 imes 10^{-2}) = 6.00 imes 10^{-4}

Answer: (D)

Verified step by step guidance

1. Identify the ions involved in the precipitation reaction: calcium ion (Ca^{2+}) and chromate ion (CrO_4^{2-}). The solid precipitate formed is calcium chromate (CaCrO_4).
2. Write the expression for the reaction quotient Q based on the dissolution equilibrium of CaCrO_4. Since CaCrO_4 dissociates into Ca^{2+} and CrO_4^{2-}, the reaction quotient is given by the product of their molar concentrations: \(Q = [\text{Ca}^{2+}] \times [\text{CrO}_4^{2-}]\).
3. Substitute the given concentrations into the expression: \([\text{Ca}^{2+}] = 2.00 \times 10^{-2} \text{ M}\) and \([\text{CrO}_4^{2-}] = 3.00 \times 10^{-2} \text{ M}\).
4. Calculate the product of the concentrations without performing the final multiplication: \(Q = (2.00 \times 10^{-2}) \times (3.00 \times 10^{-2})\).
5. Interpret the value of Q in the context of precipitation by comparing it to the solubility product constant (K_{sp}) of CaCrO_4 to determine if precipitation will occur.