A solution contains 0.10 mol/L BaCl2. What is the minimum volume (in L) of 0.20 mol/L Na2SO4 solution that must be added to 100.0 mL of the BaCl2 solution to initiate the precipitation of BaSO4? Assume complete mixing and that BaSO4 begins to precipitate when the product of [Ba^{2+}] and [SO_4^{2-}] exceeds the K_{sp} of BaSO4 (K_{sp} = 1.1 imes 10^{-10}).

• A. 2.2 imes 10^{-6} L
• B. 1.1 imes 10^{-5} L
• C. 0.10 L
• D. 5.5 imes 10^{-7} L

Answer: (D)

Verified step by step guidance

1. Identify the ions involved in the precipitation reaction: BaCl\_2 dissociates to give Ba^{2+} ions, and Na\_2SO\_4 dissociates to give SO\_4^{2-} ions. The precipitation occurs when BaSO\_4 forms, which depends on the ion product of [Ba^{2+}] and [SO_4^{2-}].
2. Write the expression for the solubility product constant (K_{sp}) of BaSO\_4: \(K_{sp} = [Ba^{2+}][SO_4^{2-}] = 1.1 \times 10^{-10}\)
3. Calculate the initial moles of Ba^{2+} in the 100.0 mL (0.100 L) BaCl\_2 solution using its concentration: \(moles\ of\ Ba^{2+} = 0.10\ mol/L \times 0.100\ L\)
4. Let the volume of Na\_2SO\_4 solution added be \(V\) liters. The moles of SO_4^{2-} added are \$0.20\ mol/L \times V\(. After mixing, the total volume is \)(0.100 + V)\( liters. Calculate the concentrations after mixing: \)[Ba^{2+}] = \frac{moles\ of\ Ba^{2+}}{0.100 + V}\( \)[SO_4^{2-}] = \frac{0.20 \times V}{0.100 + V}$
5. Set up the equation for the ion product at the point of precipitation: \([Ba^{2+}][SO_4^{2-}] = K_{sp}\) Substitute the expressions for concentrations and solve for \(V\): \(\left(\frac{0.10 \times 0.100}{0.100 + V}\right) \times \left(\frac{0.20 \times V}{0.100 + V}\right) = 1.1 \times 10^{-10}\) Simplify and solve this equation to find the minimum volume \(V\) of Na\_2SO\_4 solution required to start precipitation.