In a solution containing 2.00 × 10^{-2} M Sr^{2+} and 1.50 × 10^{-3} M CrO_4^{2-}, what is the value of the reaction quotient Q for the precipitation of SrCrO_4? (Q = [Sr^{2+}][CrO_4^{2-}])

• A. 1.33 × 10^{-1}
• B. 3.00 × 10^{-5}
• C. 2.00 × 10^{-2}
• D. 1.50 × 10^{-3}

Answer: (B)

Verified step by step guidance

1. Identify the expression for the reaction quotient Q for the precipitation reaction of SrCrO_4. Since SrCrO_4 dissociates into Sr^{2+} and CrO_4^{2-}, the reaction quotient is given by the product of their ion concentrations: \(Q = [\text{Sr}^{2+}][\text{CrO}_4^{2-}]\).
2. Write down the given concentrations from the problem: \([\text{Sr}^{2+}] = 2.00 \times 10^{-2} \; M\) and \([\text{CrO}_4^{2-}] = 1.50 \times 10^{-3} \; M\).
3. Substitute the given concentrations into the expression for Q: \(Q = (2.00 \times 10^{-2})(1.50 \times 10^{-3})\).
4. Multiply the numerical values and apply the rules of exponents for powers of ten: multiply the coefficients and add the exponents of 10.
5. Interpret the value of Q in the context of precipitation: compare Q to the solubility product constant (K_{sp}) of SrCrO_4 to determine if precipitation will occur (though this step is beyond just calculating Q).