$K_{s p}$ of $PbCl X_{2}$ is $1.6 \times 1 0^{- 5}$ . By what factor is its solubility in pure water greater than its solubility in $0.10 M$ $NaCl$ (neglect $s$ in comparison with $0.10$ )? Give the ratio $s_{w a t er} / s_{N a Cl}$ to two decimals.

Numerical answer type, tolerance 0.03.

Answer: 9.92

In water

s = (K_{s p} /4)^{1/3} = 1.6 \times 1 0^{- 2} M

. In

0.10 M

NaCl

s = K_{s p} / (0.10)^{2} = 1.6 \times 1 0^{- 3} M

. Ratio

= 1.6 \times 1 0^{- 2} /1.6 \times 1 0^{- 3} = 9.92