Question

Calculate how many grams of silver chloride will dissolve in 1.0 L of a 1.0 M NH₃ solution through formation of the complex ion Ag(NH₃)₂⁺.

Short answer

The mass of AgCl that dissolves is 6.7 g.

Step-by-step solution

The complex ion equilibria

When transition metal ions react with certain anions or neutral molecules, a coordinate covalent bond exists between them. The formation of equilibria also occurs.

The equilibria involved in this reaction

The key equilibria involved in this reaction are shown below:

$AgCI\left(s\right)\square A{g}^{+}\left(aq\right)+C{I}^{-}\left(aq\right)K_{sp}=1.6\times {10}^{-10}$ ---(1)

$A{g}^{+}\left(aq\right)+2N{H}_3\left(aq\right)\square Ag{{\left(N{H}_3\right)}_2}^{+}\left(aq\right)K_f=1.7\times {10}^7$---------(2)

Calculation

Add (1) and (2), gives

$AgCI\left(s\right)+2N{H}_3\left(aq\right)\square Ag{{\left(N{H}_3\right)}_2}^{+}\left(aq\right)+C{I}^{-}\left(aq\right)K=K_{sp}.K_f=2.7\times {10}^{-3}$

Construct ICE table:

role="math" localid="1661861446520" width="553" height="104">AgCIs+2NH3aq□AgNH32+aq+CI-aqK=Ksp.Kf=2.7×10-3I:-1.00M00C:--2x+x+xE:-1.00-2xxx

$\begin{array}{rcl}K& =& \frac{\left[Ag{{\left(N{H}_3\right)}_2}^{+}\right]\left[C{I}^{-}\right]}{{\left[N{H}_3\right]}^2}\\ 2.7\times {10}^{-3}& =& \frac{x^2}{{\left(1.00-2x\right)}^2}\\ 0.0519& =& \frac{x^2}{1.00-2x}\\ 0.0519-0.1038x& =& x\\ x& =& \frac{0.0519}{1.1038}\\ x& =& 0.0470M\end{array}$

The mass of 0.0470 mol (since the volume of the solution is 1.00 L) :

The molar mass of AgCl is 143.3 g/mol.

$\begin{array}{rcl}Mas{s}_{AgCI}& =& 0.0470mol\times 143.3g/mol\\ & =& 6.7g\end{array}$