How much \(NH_3\) must be added to \(0.004 M\) \(Ag^+\) solution to prevent the precipitation of \(AgCl\) when \(Cl^-\) reaches \(0.001 M\)?
\(K_{sp}\) for \(AgCl\) is \(1.8 \times 10^{-10}\) and \(K\) for \([Ag(NH_3)_2]^+\) is \(5.9 \times 10^{-8}\)
Answer & explanation
Correct answer: option 3
The correct answer is option 3. \(0.044\, \ mol/L\).
\([Ag^+] = \frac{K_{sp}}{[Cl^-]} = \frac{1.8 \times 10^{-10}}{0.001} = 1.8 \times 10^{-7}M\)
\(NH_3\) is added to keep the conc. pf \(Ag^+\) below \(1.8 \times 10^{-7}\) to prevent precipitation.
\([Ag(NH_3)_2]^+\) at above limiting condition \(= 0.004 - 1.8 \times 10^{-7} \approx 0.004M\)
\([Ag(NH_3)_2]^+ ⇌ Ag^+ + 2NH_3\)
\(K_d = \frac{[Ag^+][NH_3]^2}{[Ag(NH_3)_2]^+}\)
or, \(5.9 \times 10^{-8} = \frac{[1.8 \times 10^{-7}][NH_3]^2}{[0.004]}\)
or, \([NH_3] = 0.036M\)
\([NH_3]_{Total} = [NH_3]_{Free} + [NH_3]_{Complexed}\)
or, \([NH_3]_{Total} = 0.036 + 2 \times 0.004 = 0.044\, \ mol/L\)