I'm trying to calculate the pH of a 1M $\ce{NH_4CH_3COO}$. I know that I'll have these reactions:

$\ce{NH_4CH_3COO <=> NH_4^+ + CH_3COO^-}$
$\ce{NH_4^+ <=> NH_3 + H^+}$
$\ce{CH_3COOH <=> CH_3COO^- + H^+}$

I know the $K_a$s of the last two, so I'm able to calculate the $K$ of the first one (it's $\frac {K_{a1}}{K_{a2}}$), which gives me these equations:

$\frac {K_{a1}}{K_{a2}} = \frac {(x-y)(x-z)} {1 - x}$
${K_{a1}} = \frac {y(y+z)} {x-y}$
${K_{a2}} = \frac {z(y+z)} {x-z}$

But I have only 2 independent equations (the first one is just the ratio of the second and third) and three variables so I'm unable to solve for $[\ce{H^+}]$, which is $y+z$...

What do I do?

  • $\begingroup$ It would be good to know what are your x, y, and z. Also, you don't really have K for the first reaction, nor do you need one. $\endgroup$ May 23, 2017 at 21:13
  • 2
    $\begingroup$ You're missing the conservation of matter constraint. The total amount of ammonium, ammonia, acetate, acetic acid is equal to the amount you started with. $\endgroup$
    – Zhe
    May 23, 2017 at 21:22
  • 1
    $\begingroup$ Are you sure the concentration is 1 M? If so, I think your problem is even more complicated. At this high concentration you should perhaps also consider the activity coefficients of all the protolytes involved in order to make a reasonable calculation. $\endgroup$
    – Bive
    May 23, 2017 at 23:15
  • 1
    $\begingroup$ @Bive I think you would just assume (possibly incorrectly) that concentration effects are not significant. $\endgroup$
    – Zhe
    May 24, 2017 at 0:22

2 Answers 2


Ok, I will follow the assumption proposed by @Zhe above (possible incorrect as he states, but please do not be confused by that).

In order to solve this problem we need two acidity constants:
$\mathrm{p}K_{\mathrm{a}(\ce{NH4+})} = 9.25$ and $\mathrm{p}K_{\mathrm{a}(\ce{CH3COOH})} = 4.76$.

First we state the proton balance (the amount of protons taken up have to be equal to the amount of protons given off in the system): Initially we have $\ce{H2O}$ and $\ce{CH3COONH4}$.

Proton balance: $[\ce{H3O+}] + [\ce{CH3COOH}] = [\ce{OH-}] + [\ce{NH3}]$

At $\mathrm{pH} = 7, [\ce{H3O+}] = [\ce{OH-}] = \pu{10^{-7} M}$:
At this $\mathrm{pH}$ the proton balance can be simplified as $[\ce{CH3COOH}] = [\ce{NH3}]$.

The simplified proton balance will be true only at a $\mathrm{pH}$ that is exactly in the middle of the two $\mathrm{p}K_{\mathrm{a}}$-values. We get $\mathrm{pH} = (4.76 + 9.25)/2 = 7.005 \approx 7$ (only one significant figure is given, since you have stated the concentration as $\pu{1 M}$).


Simple answer

The salt ammonium acetate composed of the anion acetate ion (conjugate base of weak acetic acid)and of the cation ammonium ion(conjugate acid of a weak base ammonia), both cation and anion hydrolyzed in water equally ${(k_a=k_b)}$, so the solution is neutral $${[H3O^+]=[OH^-]=10^{-7}and\ pH=7}$$

I will give a more theoretical answer to this question using equilibrium constant and deriving formula:

Four equilibria are possible in the ammonium acetate solution; the auto ionization of water, the reaction of the cation and anion with water, and their reaction with each other : $$ \begin{array}{ll} \ce{NH_4^+ + H2O <=> H3O+ + NH_3} &\quad\left(K_{a(NH_4^+)} = \frac{K_w}{K_{b(NH_3)}}=\frac{10^{-14}}{10^{-4.74}}=10^{-9.26}\right)\\ \ce{CH_3COO^- + H2O <=> OH^- + CH_3COOH} &\quad\left(K_{b(CH_3COO^-)} =\frac{K_w}{K_{a(CH_3COOH)}}=\frac{10^{-14}}{10^{-4.74}}= 10^{-9.26}\right)\\ \ce{H_3O^+ + OH^- <=> 2H_2O }&\quad\left(\frac{1}{K_w}\right)\\ \ce{NH_4^+ + CH_3COO^- <=> CH_3COOH + NH_3} &\quad\left({K_{eq}}=\right)\\ \end{array} $$

The last equation is the sum of the first three equations, the value of $K_{eq}$ of the last equation is therefore $$K_{eq}=\frac{10^{-9.26}\times10^{-9.26}}{K_w}=3\times10^{-5}$$

Because $K_{eq}$ is several orders of magnitude greater than $ K_{a(NH_4^+)}\ or\ K_{b(CH_3COO^-)} $ , it is valid to neglect the other equilibria and considering only the reaction between the ammonium and acetate ions .Also, the products of this reaction will tend to suppress the extent of the first and second equilibria, reducing their importance even more than the relative values of the equilibrium constants would indicate.

From the stoichiometry of ammonium acetate : $$\ce{[NH_4^+]=[CH_3COO^-]\ and\ [NH_3]=[CH_3COOH]} $$ Then $$K_{eq}=\frac{[CH_3COOH][NH_3]}{[NH_4^+][CH_3COO^-]}=\frac{[CH_3COOH]^2}{[CH_3COO^-]^2} =\frac{K_w}{K_{a(CH_3COOH)}K_{b(NH_3)}}$$

From the acetic acid dissociation equilibrium : $$\frac{[CH_3COOH]}{[CH_3COO^-]}=\frac{[H_3O^+]}{K_{a(CH_3COOH)}}$$ Rewriting the expression for $K_{eq}$ , $$K_{eq}=\frac{[CH_3COOH]^2}{[CH_3COO^-]^2}=\frac{[H_3O^+]^2}{K_{a(CH_3COOH)}^2} =\frac{K_w }{K_{a(CH_3COOH)}K_{b(NH_3)}}$$ Which yields the formula

$${[H_3O^+]}=\sqrt{\frac{K_wK_{a(CH_3COOH)}}{K_{b(NH_3)}}}={\sqrt{\frac{10^{-14}\times10^{-4.74}}{ 10^{-4.74}}} =\ 10^{-7}}$$ $pH=7$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.