How do I solve this Ionic Equilibrium problem?
Chemistry
AgNO3 is a strong electrolyte it dissociates completely.
$\ce{AgNO3 -&gt; Ag+ + NO3-}$
If it was just HCN in solution, then most of the $\ce{CN^-}$ species would be as HCN. But since $\ce{AgCN}$ precipitates, more HCN dissociates. So essentially the overall reaction is:
$\ce{HCN + Ag+ -&gt; AgCN}\,\ce{v + H+}$
The $K_{sp}$ of $\ce{AgCN}$ is such that it is reasonable to assume that for all practical purposes that the precipitation is essentially complete.
Assumptions
(1) I assume that the correct answer should have two significant figures which are all that the equilibrium constants have. However I will carry 4 significant figures in the intermediate calculations to try to avoid significant rounding errors.
(2) $\ce{[H^+]} \approx 0.01000$ molar and $\approx 0.01000\ \mathrm{moles}\ \ce{AgCN}\ \text{ppt per liter of the mixture}$ (In other words about 1 ppt error for both)
Mass balance
If we consider 1 liter of the mixture of solution then we also have the following two mass balance equations:

The volume doesn't matter. Thinking of it as 1 liter just makes it simpler to get to the final result.

$0.01000 = \ce{[CN-] + [HCN]} +\ \mathrm{moles\ \ce{AgCN}\ce{v}\tag{1}}$
$0.01000 = \ce{[Ag+] +\ \mathrm{moles}\ \ce{AgCN}\ce{v}\tag{2}}$
$\therefore \ce{[Ag+] = [CN-] + [HCN]}\tag{3} $
Equilibrium equations
$K_{a}(\ce{HCN})=\pu{6.2\times10^{-10}}= \dfrac{[\ce{H+}][\ce{CN-}]}{[\ce{HCN}]}\tag{4}$
$K_{sp}(\ce{AgCN})=\pu{2.2\times10^{-16}} = [\ce{Ag+}][\ce{CN-}]\tag{5}$
dividing (4) by (5) we can get:
$2.819\times10^6 = \dfrac{\ce{[H+]}}{\ce{[Ag+][HCN]}}\tag{6}$
Rearranging (6) and substituting $\ce{[H+] = 0.01000\ \mathrm{molar}}$
$\ce{[HCN] =}\dfrac{3.547\times10^{-9}  }{\ce{[Ag+]}}\tag{7}$
Rearranging (5) we get:
$\ce{[CN-]} = \dfrac{\pu{2.2\times10^{-16}}}{\ce{[Ag+]}}\tag{8}$
Substituting (7) and (8) into (3) we get:
$\ce{[Ag+] =} \dfrac{\pu{2.2\times10^{-16}}}{\ce{[Ag+]}} + \dfrac{3.547\times10^{-9} }{\ce{[Ag+]}}\tag{9} $
Solving:
$\ce{[Ag+]^2} \approx 3.547\times10^{-9}\tag{10} $
$\ce{[Ag+] =} 5.955 \times10^{-5}\tag{11}$
Rounding to two places... $\ce{[Ag+] =} 6.0 \times10^{-5}$
Solving for other species
Substituting for $\ce{[Ag+]}$ in (8) gives
$\ce{[CN-]} = \dfrac{\pu{2.2\times10^{-16}}}{5.955 \times10^{-5}} = 3.694\times 10^{-12}\tag{12}$
Substituting for $\ce{[Ag+]}$ in (7) gives
$\ce{[HCN] =}\dfrac{3.547\times10^{-9}   }{5.955 \times10^{-5}} = 5.956\times10^{-5}\tag{13}$
Argentocyanide Complex
The problem has ignored the argentocyanide complex. Is that reasonable?
$K = 5.6\times10^{18} =\dfrac{\ce{[Ag(CN)2^-]}}{\ce{[Ag+][CN-]^2}}\tag{15}$
rearranging
$\ce{[Ag(CN)2^-]} = \ce{[Ag+][CN-]^2}\times(5.6\times10^{18})=4.794\times10^{-9}\tag{16}$
So it is reasonable to ignore the additional complication.
Checking


You always check your assumptions on this sort of problem. It is easy to go down the wrong rabbit hole.


with $K_\mathrm{a}\ce{(HCN)}$
$\pu{6.2\times10^{-10}} = \dfrac{[\ce{H+}][\ce{CN-}]}{[\ce{HCN}]} = \dfrac{(0.01)(3.694\times 10^{-12})}{5.956\times10^{-5}} = 6.202\times10^{-10}\tag{14}$
mass balance
It makes sense chemically that in the acidic solution $\ce{[HCN]} \gg \ce{[CN-]}$ so that $\ce{[Ag+]} \approx \ce{[HCN]}$. Making that assumption earlier using equation (6) would have reduced the work for this problem.
[H+] = 0.01000 molar and 0.01000 moles AgCN ppt per liter of the mixture
Also assuming that $\ce{[H^+] = 0.01000\ \mathrm{molar}}$ is certainly correct to two significant figures. However I tried to carry 4 significant figures in the intermediate calculations to avoid rounding errors. The assumption isn't that good.
!! PROBLEM !!
The assumptions that: $\quad\quad\ce{[H^+]} \approx 0.01000$ molar  $\quad\quad\approx 0.01000\ \mathrm{moles}\ \ce{AgCN}\ \text{ppt per liter of the mixture}$
are not good to 4 significant figures. It would have been better to use:
$\quad\quad\ce{[Ag+]} \approx \ce{[HCN]}$
$\quad\quad\ce{[H^+] \approx 0.01000 - [HCN] \approx 0.010000 -[Ag+]}$
and get a quadratic equation when simplifying equation (6). Then $\ce{[Ag+]} = 5.938\times10^{-5}$ and $\ce{[H+]} = 9.941\times10^{-3}$.
So the value $\ce{[Ag+]} = 5.956\times10^{-5}$ is off by +0.303%, or 3 ppt, which I think is acceptable, but it is marginal.