# Question #28d0a

Apr 27, 2015

You'll dissolve 0.33 moles of silver chloride in that much ammonia.

The idea behind this problem is a little complex, I'll give it that. Basically, you're dealing with two equilibrium reactions that take place at the same time.

The first one is the dissociation of silver chloride in aqueous solution

$\textcolor{b l u e}{\left(1\right)}$: $A g C {l}_{\left(s\right)} r i g h t \le f t h a r p \infty n s A {g}_{\left(a q\right)}^{+} + C {l}_{\left(a q\right)}^{-}$, ${K}_{s p} = 1.8 \cdot {10}^{- 10}$

Because the solubility product constant is so small, the concentrations of both the silver cations and of the chloride anions will be very, very small.

Now you add the ammonia solution and a second equilibrium is estblished (all the species are in aqueous solution)

$\textcolor{b l u e}{\left(2\right)}$: $A {g}^{+} + 2 N {H}_{3} r i g h t \le f t h a r p \infty n s {\left[A g {\left(N {H}_{3}\right)}_{2}\right]}^{+}$, ${K}_{f} = 1.7 \cdot {10}^{7}$

Here's what actually takes place. When the complex ion is formed, the concentration of silver cations present in solution is lowered, since silver cations are being consumed.

This will trigger a shift in equilibrium $\textcolor{b l u e}{\left(1\right)}$ and cause more silver chloride to dissolve.

The net equilibrium constant for this reaction will be a product of the two constants given

${K}_{\text{net}} = {K}_{f} \cdot {K}_{s p}$

${K}_{\text{net}} = 1.7 \cdot {10}^{7} \cdot 1.8 \cdot {10}^{- 10} = 3.06 \cdot {10}^{- 3}$

Use the definitions of the two constants to get

${K}_{s p} = \left[A {g}^{+}\right] \cdot \left[C {l}^{-}\right]$

${K}_{f} = \frac{{\left[A g {\left(N {H}_{3}\right)}_{2}\right]}^{+}}{\left[A {g}^{+}\right] \cdot {\left[N {H}_{3}\right]}^{2}}$

This will be equivalent to

${K}_{\text{net" = K_"f}} \cdot {K}_{s p} = \frac{{\left[A g {\left(N {H}_{3}\right)}_{2}\right]}^{+}}{\cancel{\left[A {g}^{+}\right]} \cdot {\left[N {H}_{3}\right]}^{2}} \cdot \cancel{\left[A {g}^{+}\right]} \cdot \left[C {l}^{-}\right]$

${K}_{\text{net}} = \frac{{\left[A g {\left(N {H}_{3}\right)}_{2}\right]}^{+} \cdot \left[C {l}^{-}\right]}{{\left[N {H}_{3}\right]}^{2}}$

Because the chloride ions are spectators in equilibrium $\textcolor{b l u e}{\left(2\right)}$, their concentration will be equal to that of the silver cations that dissolved from the silver chloride.

And since the silver cations end up in the complex ion, you can assume that

$\left[A g {\left(N {H}_{3}\right)}_{2}^{+}\right] = \left[C {l}^{-}\right] = x$

Since you're dealing with 6.0 moles of ammonia in 1.00 L of solution, $\left[N {H}_{3}\right] = \text{6.0 M}$

This will get you

${K}_{\text{net}} = \frac{x \cdot x}{6} ^ 2 = {x}^{2} / 36 = 3.06 \cdot {10}^{- 3}$

Solving for $x$ will get you

$x = 0.3319 = \left[A g {\left(N {H}_{3}\right)}_{2}\right]$

Since you have a $1 : 1$ mole ratio between the complex ion and the silver cations, i.e. the silver chloride, and 1.00 L of solution, you'll get

${n}_{A g C l} = \textcolor{g r e e n}{\text{0.33 moles}}$ $\to$ rounded to two sig figs.