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:


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:




Solving this for $x$ results into:




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:


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

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

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


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


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

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

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

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:

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


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?

  • $\begingroup$ You might find some help here chemistry.stackexchange.com/questions/60068/… .and in the link to a related answer in a comment. $\endgroup$
    – porphyrin
    Commented Jan 25, 2020 at 16:44
  • 1
    $\begingroup$ Does this answer your question? How to set up equation for buffer reaction? $\endgroup$ Commented Jan 25, 2020 at 17:10
  • 1
    $\begingroup$ @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. $\endgroup$ Commented Jan 26, 2020 at 5:51
  • $\begingroup$ @porphyrin That's and interesting entry and I've checked into it but it was not exactly what I was looking for. $\endgroup$ Commented Jan 26, 2020 at 5:52
  • $\begingroup$ @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. $\endgroup$
    – uhoh
    Commented Jan 27, 2020 at 0:42

1 Answer 1


Ignoring activities, activity coefficients, ionic strength effects, and so on, I get a pH of around 9.3. Given the 1 M concentration, I would not bet a lot of money on this, though. My solution, assuming a more typical solution concentration, is given in the two figures below. Sorry these are figures rather than proper formatting: I have been away for 3 months and am beyond rusty at this point. If the mods wish to delete this answer, they should feel totally free to do so.

Figure 1

Figure 2

There are obviously other ways to do this problem and what I have posted is simply an approximation: any pH problem solution, of this type, that does not explicitly involve the formal concentration of the salt, i.e., "F", always gives me pause. But maybe this helps a bit.

  • $\begingroup$ So in general terms (it is a relief), does this validates the way how I proceeded?. I'm wondering if you take into an account the other aspects (ionic strength, and so on) would it change the value of pH which you have obtained?. $\endgroup$ Commented Jan 26, 2020 at 5:56
  • $\begingroup$ The approximation which you mentioned at the end of your answer, what's its justification?. Can you explain this part please?. Out of curiosity what software or tool did you used to draw your cursive sketch? $\endgroup$ Commented Jan 26, 2020 at 5:57
  • $\begingroup$ In general terms, I think your approach is fine! If all the fancy stuff is included, the pH will probably not be 9.3, but it is a good enough zero order approximation. It really all comes down to setting up all the equilibria expressions, in their full glory, and then solving them. Which is why it is useful to have a good piece of software that can do that number crunching. I do not have such a program, which would be useful, at the very least, to check manually calculated results. The trouble with pH problems is that concentrations need to be in the Goldilocks regime: not too concentrated, $\endgroup$
    – Ed V
    Commented Jan 26, 2020 at 13:57
  • $\begingroup$ not too dilute, but ‘just right’. Otherwise, you have slightly dirty water or need activities, etc. The software was an iPad app called Noteshelf. What I posted was screenshots from that app. They were from the last time (2015) that I taught the second semester general chemistry course. Finally, the simple approximation just recognizes that there can only be one pH in a solution. Nothing deep about it. $\endgroup$
    – Ed V
    Commented Jan 26, 2020 at 14:05

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