Problem :

Equal volumes of $\pu{0.02M}$ $\ce{AgNO3}$ and $\pu{0.02M}$ $\ce{HCN}$ were mixed. Calculate $\ce{[Ag+]}$ at equilibrium assuming no cyano-complex formation. $K_{sp}(\ce{AgCN})= \pu{2.2\times10^{-16}}$, $K_{a}(\ce{HCN})=\pu{6.2\times10^{-10}}$

My solution:
Concentration of $\ce{AgNO3}$ and $\ce{HCN}$ before reacting when mixed will become $\pu{0.01M}$. Since $\ce{AgNO3}$ is a strong electrolyte it dissociates completely. So, initial concentration $[Ag^+]$ is $\pu{0.01M}$. $$\ce{HCN<=>H+ + CN-} $$ After reaction let $[\ce{HCN}]= \pu{0.01-x}$, $[\ce{H+}]=\pu{x}$, $[\ce{CN-}]=\pu{x}$.
Now $\ce{CN-}$ reacts with $\ce{Ag+}$ $$\ce{Ag+ + CN^-<=> AgCN}$$ Initial $[\ce{Ag+}] = \pu{0.01}$, $[CN^-]=\pu{x}$.
After reaction let $[\ce{Ag+}] = \pu{0.01-y}$, $[\ce{CN-}]=\pu{x-y}$, $[\ce{AgCN}] = \pu{y}$.
Now, $K_{a}(\ce{HCN})=\pu{6.2\times10^{-10}}= \frac{[\ce{H+}][\ce{CN-}]}{[\ce{HCN}]}=\frac{x(x-y)}{0.01-x}\approx\frac{x(x-y)}{0.01} $.
And $K_{sp}(\ce{AgCN})=\pu{2.2\times10^{-16}} = [\ce{Ag+}][\ce{CN-}] = (0.01-y)(x-y)$.
Solving these two equations I get $\pu{(0.01-y)= 8.9\times10^{-11}=[\ce{Ag+}]}$.
But the solution in my book says the answer is $\pu{5.9\times10^{-5}}$.
I don't where am I going wrong.

  • $\begingroup$ Yeah, but the question has asked to assume no formation of complex. $\endgroup$
    – Shu-P
    Dec 9, 2017 at 16:51
  • 1
    $\begingroup$ The error is your simplification $0.01-x ≈ 0.01$. $x$ is close to 0.01, you can't ignore it. If you use the non-simplified version you get more or less the correct result. FYI, by writing all the mass balances and solving numerically, you would get: $[ [HCN]=5.939E-5, [CN^-]=3.704E-12, [H^+]=0.009941, [Ag^+]=5.939E-5, [NO_3^-]=0.01]$. $\endgroup$ Dec 9, 2017 at 17:00
  • $\begingroup$ BTW, I was intrigued by Gert's comment, so I solved the system including the $Ag(CN)_2^-$ equilibrium. The result, as far as $[Ag^+]$ is concerned, is the same: $[[HCN]=5.939E-5, [CN^-]=3.704E^-12, [H^+]=0.00994, [Ag^+]=5.939E-5, [NO_3^-]=0.01, [Ag(CN)_2^-]=4.563E-9]$. $\endgroup$ Dec 9, 2017 at 17:10
  • $\begingroup$ @user6376297 thanks for pointing out the mistake $\endgroup$
    – Shu-P
    Dec 9, 2017 at 17:24
  • 1
    $\begingroup$ @MaxW : sorry, this is not correct. The OP knows that $[H^+]$ and $[CN^-]$ are not the same. If you look carefully, you will see that in the final equations from the OP, $[H^+] = x$ and $[CN^-] = x-y$, and the value of $x$ and $y$ is determined by the solution of the system of equations (which is $x=0.009940608815422887, y=0.009940608811718633$ - and you can easily find that $[Ag^+] = 0.01-y = 5.939E-5$). As I pointed out above, the main error was to consider $x$ much smaller than $0.01$, thus wrongly simplifying the first equation. $\endgroup$ Dec 10, 2017 at 8:48

3 Answers 3



AgNO3 is a strong electrolyte it dissociates completely.

$\ce{AgNO3 -> Ag+ + NO3-}$

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:

$\ce{HCN + Ag+ -> AgCN}\,\ce{v + H+}$

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.


(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} $


$\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?

$K = 5.6\times10^{18} =\dfrac{\ce{[Ag(CN)2^-]}}{\ce{[Ag+][CN-]^2}}\tag{15}$


$\ce{[Ag(CN)2^-]} = \ce{[Ag+][CN-]^2}\times(5.6\times10^{18})=4.794\times10^{-9}\tag{16}$

So it is reasonable to ignore the additional complication.


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.


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.

  • 1
    $\begingroup$ Sorry but the solution by the OP, when removing the wrong simplification $0.01 - x ≈ 0.01$, gives more correct figures than the ones in this answer. This answer misleads the OP into thinking that a much more complicated approach is necessary, and actually introduces errors that may result in the answer not being accepted. $\endgroup$ Dec 10, 2017 at 8:57
  • $\begingroup$ @user6376297 - Yes and no. I hate problems like this. It is never clear to me how many significant figures "0.2 M $\ce{AgNO3}$" is supposed to have. I assumed that two significant figures were good enough. In the "Checking" section I made the point that I went down the wrong rabbit hole and a better solution was possible. The point was that I was trying to make was that for this type of problem that checking your assumptions is necessary. I'll edit and make this clearer... $\endgroup$
    – MaxW
    Dec 10, 2017 at 15:34

Complete Analysis

To beat this to death perhaps here is a more complete analysis:


(1) Assume that the correct answer should have two significant figures (which are all that the equilibrium constants have) but carry 4 significant figures in the intermediate calculations to try to avoid significant rounding errors.

(2) AgNO3 is a strong electrolyte it dissociates completely.

$\ce{AgNO3 -> Ag+ + NO3-}$

Thus the $\ce{[NO3^-]}$ can be assumed to be exactly 0.010000 molar.

(3) $\ce{Ag(CN)}$ is very insoluble. So the reaction should produce about 0.01 molar $\ce{H+}$ and about 0.01 moles of $\ce{Ag(CN)}$ per liter of solution.

(4) In a fairly acid solution the remaining cyanide species in solution will be mostly $\ce{HCN}$ with very little $\ce{CN^-}$.


Equilibrium equations

$K_{a}(\ce{HCN})=\pu{6.2\times10^{-10}}= \dfrac{[\ce{H+}][\ce{CN-}]}{[\ce{HCN}]}\tag{E1}$

$K_{sp}(\ce{AgCN})=\pu{2.2\times10^{-16}} = [\ce{Ag+}][\ce{CN-}]\tag{E2}$

dividing (E1) by (E2) gives another equation, but only two of the three equilibrium equations will be independent.

$2.818\times 10^6 = \dfrac{\ce{[H+]}}{\ce{[Ag+][HCN]}}\tag{E3}$

Mass Balance equations

In order to balance charges in the remaining solution:

$$\ce{[Ag^+] + [H^+] = [NO3^-] + [CN^-] }$$

but since $\ce{NO3^-} =0.01000$

$$\ce{[Ag^+] + [H^+] = 0.01000 + [CN^-]\tag{M1}}$$

To mass balance $\ce{CN^-}$ for some volume, V, we have:

$$0.01000V = \ce{[CN-]V + [HCN]V} +\ \mathrm{moles(\ce{AgCN}\ce{v})}$$

To mass balance $\ce{Ag^+}$ for some volume, V, we have:

$$0.01000V = \ce{[Ag+]V} +\ \mathrm{moles(\ce{AgCN}\ce{v})}$$

Subtracting B5 from B4, rearranging, and then dividing by $\ce{V}$ gives:

$$\ce{[Ag+] = [CN-] + [HCN]}\tag{M2}$$

Now adding (M1) and (M2) we get (M3), but again only two of the three equations M1, M2 and M3 are independent.

$$\ce{[H^+] + [HCN] =0.01000}\tag{M3}$$


There are five unknowns ( $\ce{[H+], [Ag+], [HCN], [CN-], AgCN(s)}$ ) in the original system.

Manipulation of the mass balances for $\ce{CN^-}$ and $\ce{Ag^+}$ eliminated the $\ce{AgCN(s)}$ as a variable and left mass balance equation M2.


Now with two equations from E1, E2, and E3 and then two equations from M1, M2 and M3 there are 4 equations and 4 unknowns ( $\ce{[H+], [Ag+], [HCN], [CN-]}$ ) so an exact solution is possible. Rather than to try to tediously solve some messy polynomial equation by hand the idea is to use some notion of chemistry to simplify the equations so that the result can be obtained with at most solving a quadratic equation.

Of course to obtain a numerical solution by computer is possible regardless how messy the equations and the computer doesn't care how tedious the calculation is.


Exact Solution

Starting with the analysis noted for this problem, rearrange equation E1 to:

$K_{a}\ce{[HCN]} = \ce{[H+][CN-]}\tag{E1}$

Now solve for each of the three unknowns in terms of $\ce{[Ag+]}$.

$\ce{[CN-]}$ in terms of $\ce{[Ag+]}$.

Just need to rearrange E2:

$\ce{[CN-]} = \dfrac{K_{sp}}{\ce{[Ag+]}}\tag{E2}$

$\ce{[H+]}$ in terms of $\ce{[Ag+]}$ using M1 and E2.

\begin{align}\\ \ce{[H^+] &= 0.010000 + [CN^-] - [Ag^+]}\tag{M1}\\ \\ \ce{ &= 0.010000 + \dfrac{K_{sp}}{\ce{[Ag+]}} - [Ag^+]}\\ \\ &=\dfrac{K_{sp} + 0.010\ce{[Ag+] - [Ag+]^2}}{\ce{[Ag+]}}\\ \end{align}

$\ce{[HCN]}$ in terms of $\ce{[Ag+]}$ using M2 and E2.

\begin{align}\\ \ce{[HCN] &= [Ag+] - [CN-]}\tag{M2}\\ \\ &= \ce{[Ag+]} - \dfrac{K_{sp}}{\ce{[Ag+]}}\\ \\ &= \dfrac{\ce{[Ag+]^2}-K_{sp}}{\ce{[Ag+]}}\\ \end{align}

Substituting gives $$K_a \cdot\dfrac{\ce{[Ag+]^2}-K_{sp}}{\ce{[Ag+]}} = \dfrac{K_{sp} + 0.010\ce{[Ag+] - [Ag+]^2}}{\ce{[Ag+]}} \cdot \dfrac{K_{sp}}{\ce{[Ag+]}}$$

Final Equation

One form of the final equation is given below. Of course multiplying and dividing by constant would yield other forms.

Using any other combination of equations would yield an equivalent equation.

$$ 0 = \dfrac{K_a}{K_{sp}}\ce{[Ag+]^3} + \ce{[Ag+]^2} - (K_a+0.01)\ce{[Ag+]} - K_{sp} $$

Numerical Solution

Solving with WolframAlpha equation solver using:

$$\text{a==6.2`20e-10, b==2.2`20e-16, N[[(a/b)*x^3+x^2-(a+0.01000)*x-b=0],20]}$$

gives $\ce{[Ag+]} = 5.939\,118\,459\times10^{-5}$ as noted by user user6376297 in a comment under the OP's question.


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