The solubility product of \( \text{BaSO}_4 \) at 298 K is \( 1.10 \times 10^{-10} \). Its molar solubility in a \( 0.01\text{ M }\text{BaCl}_2 \) solution compared to that in pure water is reduced by a factor of approximately:

In pure water, \( S_0 = \sqrt{K_{sp}} = \sqrt{1.10 \times 10^{-10}} \approx 1.049 \times 10^{-5}\text{ M} \). In \( 0.01\text{ M }\text{BaCl}_2 \), \( [\text{Ba}^{2+}] \approx 0.01\text{ M} \), so \( S' = \frac{1.10 \times 10^{-10}}{0.01} = 1.10 \times 10^{-8}\text{ M} \). Factor \( = \frac{S_0}{S'} = \frac{1.049 \times 10^{-5}}{1.10 \times 10^{-8}} \approx 953 \).