What minimum total concentration of ammonia (mol per litre of solution) is needed to just dissolve $0.010 mol$ of AgCl in $1.0 L$ (no volume change)? $K_{s p} (AgCl) = 1.8 \times 1 0^{- 10}$ , $K_{f}$ of $[Ag (NH X_{3}) X_{2}] X^{+} = 1.6 \times 1 0^{7}$ . Give the answer to two decimals.

Numerical answer type, answer in mol/L, tolerance 0.01.

Answer: 0.21 mol/L

On complete dissolution

[[Ag (NH X_{3}) X_{2}] X^{+}] = [Cl X^{-}] = 0.010 M

, so

[Ag X^{+}] = K_{s p} / [Cl X^{-}] = 1.8 \times 1 0^{- 8} M

. From

K_{f}

[NH X_{3}]^{2} = 0.010/ (K_{f} [Ag X^{+}]) = 0.0347

[NH X_{3}] = 0.1863 M

. Total

= 0.1863 + 2 (0.010) = 0.2063 \approx 0.21 M