# Accurate method to calculate the pH of a salt from a weak acid and weak base

Can someone help me with this?

The problem is as follows:

Find the $$\mathrm{pH}$$ of $$\pu{1M}$$ solution of $$\ce{NH4CN}$$. Their respective hydrolysis constants are:

$$K_h = 5.6\times 10^{-10}$$

$$K_h = 2.5 \times 10^{-5}$$

The alternatives given are as follows:

$$\begin{array}{ll} 1.&2.8\\ 2.&3.5\\ 3.&6.4\\ 4.&11.7\\ 5.&12.2\\ \end{array}$$

What I've attempted to do was to establish the following reactions:

$$\begin{array}{lllll} \ce{NH4+ + H2O <=> NH3 + H3O+} &k_{h}=\frac{K_\mathrm{w}}{K_\mathrm{b}}\\ \ce{CN- + H2O <=> HCN + OH-} &k_{h}=\frac{K_\mathrm{w}}{K_\mathrm{a}}\\ \ce{H3O+ + OH- <=> 2H2O}&k_{h}=\frac{1}{K_\mathrm{w}}\\ \end{array}$$

This meant that:

$$K=\frac{K_\mathrm{w}}{K_\mathrm{a}\times K_\mathrm{b}}$$

For the given information it can be calculated:

$$K=\frac{10^{-14}}{\frac{10^{-14}}{5.6\times 10^{-10}}\times\frac{10^{-14}}{2.5 \times 10^{-5}}}=1.4$$

All of this is translated into:

$$\ce{NH4+ + CN- <=> NH3 + HCN}$$

From this equation it can be established the equilibrium equation:

$$K=\frac{\ce{[NH3][HCN]}}{\ce{[NH4+][CN-]}}$$

Returning to the previous equation I'm getting:

$$\begin{array}{cccccc} &\ce{NH4+ + CN- & <=> & NH3 + HCN }\\ i&1&1 & & &\\ c&-x&-x & &+x &+x\\ \hline e&(1-x)&(1-x) & &x &x\\ \end{array}$$

Using this relation with what was established earlier it becomes into:

$$K=\frac{\ce{[NH3][HCN]}}{\ce{[NH4+][CN-]}}$$

$$1.4=\frac{\ce{[NH3][HCN]}}{\ce{[NH4+][CN-]}}$$

$$1.4=\frac{x^2}{(1-x)^{2}}$$

Solving this for $$x$$ results into:

$$x=0.54196$$

and

$$x=6.45804$$

The thing here comes on which to choose for the concentration. I decided that the first seems more logical as it will not produce negative concentration in the denominator.

Therefore the concentration for $$x=0.54196$$

This means:

$$\ce{[NH3]=[HCN]}=0.54196$$

$$\ce{[CN-]=[NH4+]}=1-0.54196= 0.45804$$

Then to get the $$\mathrm{pH}$$ can be established from the hydrolyzation of $$\ce{HCN}$$:

$$\ce{HCN + H2O <=> CN- + H3O+}$$

Then:

$$K=\frac{10^{-14}}{2.5\times 10^{-5}}$$

$$K=\frac{\ce{[CN-][H3O+]}}{\ce{[HCN]}}$$

$$4.0\times 10^{-10}=\frac{\ce{[H3O+]}(0.45804)}{(0.54196)}$$

$$[\ce{H3O+}]=4.73286\times 10^{-10}$$

$$\mathrm{pH}=-\log(4.73286\times 10^{-10})=9.3248$$

Which seems kind of logical that the $$\mathrm{pH}$$ is within the range of an alkaline solution as $$K_\mathrm{b}>K_\mathrm{a}$$.

But since this option does not appear within the alternatives. I'm not convinced whether my solution is right or not.

I've found this equation which relates the $$\mathrm{pH}$$ with the $$\mathrm{p}K_\mathrm{w}$$ and $$\mathrm{p}K_\mathrm{a}$$ and $$\mathrm{p}K_\mathrm{b}$$ and the concentration:

It states:

$$\mathrm{pH}=\frac{1}{2}[\mathrm{p}K_\mathrm{w} - \mathrm{p}K_\mathrm{b} + \mathrm{p}K_\mathrm{a} + \log c]$$

where $$c = \text{concentration}$$ of the salt.

Inserting the given values in the above equation yields:

$$\mathrm{pH}=(0.5)\left[-\log(10^{-14})+\log\left(\frac{10^{-14}}{5.6\times 10^{-10}}\right)-\log\left(\frac{10^{-14}}{2.5\times 10^{-5}}\right)+\log(1)\right]$$

$$\mathrm{pH}=9.3248$$

which also checks with what was obtained previously.

But again none of these seem to check with any of the alternatives given. Could it be that my procedure was wrong?

The thing which it bothers me here is if this could be accepted as a general procedure to calculate the $$\mathrm{pH}$$ of a salt which has a weak acid and weak base ions. Can somebody help me here? And more importantly, how to prove that equation for $$\mathrm{pH}$$? I'm not sure if the hydrolyzation constants given in the problem are right or accurate. Does somebody has access to cross reference and check if those values are correct?

Can somebody help me with that part too?

• You might find some help here chemistry.stackexchange.com/questions/60068/… .and in the link to a related answer in a comment. – porphyrin Jan 25 at 16:44
• Does this answer your question? How to set up equation for buffer reaction? – Sir Arthur7 Jan 25 at 17:10
• @SirArthur7 The existing answer doesn't delve into the specific of hydrolisis of both ion and is more about establishing the equation for a buffer coming from acid and its conjugate salt which is well explained in many textbooks. It doesn't clear out my confusion on whether is it alright to multiply the constants and considering an interaction of acidic and basic ions for make water in the end. – Chris Steinbeck Bell Jan 26 at 5:51
• @porphyrin That's and interesting entry and I've checked into it but it was not exactly what I was looking for. – Chris Steinbeck Bell Jan 26 at 5:52
• @ChrisSteinbeckBell there are currently two votes to close as duplicate, it takes five. If you can edit your question and in the beginning add a link to the ducplicate and explain exactly why your question is not answered there, detailing how an answer to this question needs to differ from what's in those answers, it may help to keep this question open. – uhoh Jan 27 at 0:42  