A solution is $0.010 M$ in $Mg X^{2 +}$ and $0.10 M$ in $NH X_{3}$ ( $K_{b} = 1.8 \times 1 0^{- 5}$ ). What is the minimum concentration of $NH X_{4} Cl$ (in mol/L, three decimals) that must be added to prevent precipitation of $Mg (OH) X_{2}$ ( $K_{s p} = 1.2 \times 1 0^{- 11}$ )? Neglect volume change and any contribution of $NH X_{3}$ ionisation to $[NH X_{4} X^{+}]$ .

Numerical answer type, answer in mol/L, tolerance 0.002.

Answer: 0.051962 mol/L

No precipitate requires

[OH X^{-}]^{2} \leq K_{s p} / [Mg X^{2 +}] = 1.2 \times 1 0^{- 9}

, so

[OH X^{-}]_{ma x} = 3.464 e - 05 M

. For the buffer,

[OH X^{-}] = K_{b} [NH X_{3}] / [NH X_{4} X^{+}]

, hence

[NH X_{4} X^{+}] = 1.8 \times 1 0^{- 5} \times 0.10/3.464 e - 05 = 0.0520 M

(

\approx 0.052