$50 mL$ of $0.01 M$ $AgNO X_{3}$ is mixed with $50 mL$ of $0.02 M$ $NaCl$ . Taking $K_{s p} (AgCl) = 1.7 \times 1 0^{- 10}$ and assuming precipitation is essentially complete, find the equilibrium $[Ag X^{+}]$ in units of $1 0^{- 8} mol L^{- 1}$ (two decimals).

Numerical answer type, answer in x 10^-8 mol/L, tolerance 0.02.

Answer: 3.4 x 10^-8 mol/L

^{+}

= 0.50

mmol and Cl

^{-}

= 1.00

mmol, so

0.50

mmol

Cl X^{-}

is left in

100 mL

[Cl X^{-}] = 0.005 M

. Then

[Ag X^{+}] = K_{s p} / [Cl X^{-}] = 3.4 \times 1 0^{- 8} M