# How to simplify these system of equilibrium equations

What is the pH when $$0.025$$ mol $$\ce{NiCl2}$$ is added to $$\pu{1.00 L}$$ of a $$\pu{0.100 M}$$ solution of $$\ce{HCN}$$, if the formation constant for $$\ce{Ni(CN)4^2-}$$ is $$K_f=\pu{1.0e22}$$, and the $$K_a$$ for $$\ce{HCN}$$ is $$\pu{4.9e-10}$$?

My approach is as follows:

1) $$\ce{\frac{[H+][CN-]}{[HCN]}=4.9*10^{-10}}$$

2) $$\ce{\frac{[Ni(CN)4^{2-}]}{[Ni^2+][CN-]^4}}=1.0*10^{22}$$

3) $$\ce{[H+][OH-]}=10^{-14}$$

4) $$\ce{0.025=[Ni^{2+}] + [Ni(CN)4^{2-}]}$$ ............. (mass balance for $$\ce{Ni^{2+}}$$)

5) $$\ce{[CN-] + 0.05 + [OH-] + 2*[Ni(CN)4^{2-}] = 2*[Ni^{2+}] + [H+]}$$ ........... (charge balance)

6) $$\ce{0.1 = [CN-] + [HCN] + 4*[Ni(CN)4^{2-}]}$$ ............. (mass balance for $$\ce{CN^{-}}$$)

Since the equilibrium constant for the formation of the complex ion is very large, I assume that $$\ce{[Ni(CN)4^{2-}] >> [Ni^{2+}]}$$, so $$\ce{[Ni(CN)4^{2-}] = 0.025 M}$$, from the mass balance for $$\ce{Ni^{2+}}$$.

However, I am not sure what is the next step and what other assumption to make.

• If $\ce{[Ni(CN)4^{2-}] = 0.025 M}$, then what does that imply for mass balance of $\ce{CN-}$ species? – MaxW Dec 24 '19 at 7:04
• One information is missing : What is the volume of the HCN solution ? If the volume is greater that 1 cubic meter, the effect of NiCl2 is negligible. – Maurice Dec 24 '19 at 10:17
• @Maurice the volume of the system is 1.00 L – Cyclopropane Dec 24 '19 at 13:06
• @MaxW it seems to be that [CN-] + [HCN] = 0? – Cyclopropane Dec 24 '19 at 13:07
• "I assume that $\ce{[Ni(CN)4^2−]≫[Ni^2+]}$" $$.$$ That assumption is incorrect. The equilibrium constant seems large, but the exponents are high, so it is misleading. In fact, more of the cyanide is in the form of HCN than in complex with nickel, and there is very little free cyanide. – Karsten Theis Dec 26 '19 at 3:24

Since the equilibrium constant for the formation of the complex ion is very large, I assume that $$\ce{[Ni(CN)4^{2-}] >> [Ni^{2+}]}$$

[comments:] That assumption is incorrect. The equilibrium constant seems large, but the exponents are high, so it is misleading. In fact, more of the cyanide is in the form of HCN than in complex with nickel, and there is very little free cyanide.

OP replied:

How may I make a better assumption to solve this problem?

In the absence of nickel, the pH is around 5 and HCN is the major species, with the concentration of cyanide in the micromolar range (standard weak acid problem). When you add nickel, the concentration of cyanide gets lowered further. Because the free cyanide concentration contributes little to the total, we can combine the two reactions to see whether there is more complex or more HCN and to determine the pH:

$$\ce{4HCN + Ni^2+ <=> Ni(CN)4^2- + 4H+}$$

The equilibrium constant is

$$K_\mathrm{combined} = \pu{5.8e−16}$$

So for this reaction, reactants are favored. HCN and nickel concentrations hardly decrease compared to the initial concentrations. If we call the concentration of the complex $$x$$, we get:

$$K_\mathrm{combined} = \frac{x (4x)^4}{[\ce{HCN}]^4 \cdot [\ce{Ni^2+}]}$$

Solving for $$x$$, we get $$\pu{2.24e−5}$$, confirming that reactants are favored. From this we can calculate the pH as

$$\mathrm{pH} = -\log 4x = 4.05$$

Checking the solution

Upon adding nickel, the pH turned a bit more acidic. That makes sense as a micromolar concentration of nickel complex formed, releasing some more protons from HCN. We can calculate the cyanide concentration from either one of the given equilibrium constants, and verify that it is miniscule. Then, we can verify that all reactions are at equilibrium (at pH 4, hydroxide is a minor species as well, and has little impact on the exact solution).

• Very nice answer! – MaxW Dec 26 '19 at 17:35

My approach is the following. There is exactly enough $$\ce{Ni^{2+}}$$ ion and CN- ions for making up the ion $$\ce{Ni(CN)4^2-}$$. As the stability constant of this complex is huge, and HCN is an extremely weak acid. I would admit that the rare HCN molecules remaining in the solution have no effect on the pH of water, which must be 7.

• If most of the original $\ce{Ni^{2+}}$ is converted to $\ce{Ni(CN)4^2-}$ then the solution must be very acidic. – MaxW Dec 25 '19 at 3:50

I had a different approach to the problem which I have highlighted below

Notes :

1. Initially, I have considered the entire reagents to have reacted and then considered the equilibrium to set in
2. I have considered two equilibria simultaneously, one involving the dissociation of the Nickel complex and also the hydrolysis of the Cyanide ion because it is a strong conjugate base of HCN which is a weak acid

• The reactants are $\ce{NiCl2}$ and $\ce{HCN}$. Assuming complete reaction, the equation would be $$\ce{4HCN + NiCl2 <=> Ni(CN)4^{2-} + 2Cl- + 4H+}$$ – MaxW Dec 26 '19 at 17:08