I need to calculate the $\mathrm{pH}$ of a salt of weak acid and weak base in an aqueous solution.

Method I used for calculation

For a salt of a weak acid and a weak base (assuming complete dissociation of salt) both the cations and the anions in the salt undergo hydrolysis. When the anions undergo hydrolysis, since they are the conjugate base of the weak acid, they grab $\ce{H+}$ from the water and release $\ce{OH-}$ ions. Similarly, when the cations undergo hydrolysis they release $\ce{H3O+}$ ions.

These $\ce{H+}$ ions and $\ce{OH-}$ ions may react to give water. If these ions have the same concentration, the solution will have $\mathrm{pH}$ equal to $7$. If there is any excess of any of these ions after reaction, then the $\mathrm{pH}$ of the solution may be greater than or less than $7$. But my answer didn't match up with what is given in my text.

Textbook answer

Formula given in my textbook is

$$\mathrm{pH} = \frac{1}{2}\left(\log K_\mathrm{b} - \log K_\mathrm{a} - \log K_\mathrm{w}\right)$$

where $K_\mathrm{b}$ is dissociation constant of base, $K_\mathrm{a}$ is the dissociation constant of acid, and $K_\mathrm{w}$ is the ionic product of water.

When I try to derive this formula I need to make these assumptions:

  1. The degree of hydrolysis of both the anionic and the cationic parts of the salt is equal.
  2. The $\ce{H+}$ in the solution is contributed by the dissociation of weak acid.

But how can this be possible? The dissociation of the weak acid formed from salt hydrolysis has already been suppressed by the available concentration of its conjugate base (anion part of salt).


1 Answer 1


Another approach, I believe, is this:

Definitions \begin{align} \mathrm{pH} &= \frac{1}{2}\left(\log K_\mathrm{b} - \log K_\mathrm{a} - \log K_\mathrm{w}\right)\\ 2\,\mathrm{pH} &= \log\left(\frac{K_\mathrm{b}}{K_\mathrm{a}\,K_\mathrm{w}}\right) \end{align}

It suggests equations' sum, like this:
\begin{align} \ce{HA &<=> A- + H+} & K_\mathrm{a} &= \frac{[\ce{A-}][\ce{H+}]}{[\ce{HA}]}\\ \ce{BOH &<=> B+ + HO-} & K_\mathrm{b} &= \frac{[\ce{HO-}][\ce{B+}]}{[\ce{BOH}]}\\ \ce{A- + B+ + H2O &<=> AH + BOH} & K &= \frac{[\ce{AH}][\ce{BOH}]}{[\ce{A-}][\ce{B+}]} = \frac{K_\mathrm{w}}{K_\mathrm{a}\,K_\mathrm{b}} \end{align}

Finding $\ce{[H+]}$ in relation to $K$

If the salt concentration is $C_\mathrm{s}$, and the mass balance: $$ C_\mathrm{s} = [\ce{AH}] + [\ce{A-}] = [\ce{BOH}] + [\ce{B+}] \tag{Mass balance} $$

Dividing by $C_\mathrm{s}$ we define $\alpha$ and $\beta$ such that $$ \frac{C_\mathrm{s}}{C_\mathrm{s}} = \alpha + (1-\alpha) = \beta + (1-\beta) \tag{Mass balance} $$

And if we admit that $\alpha$, $\beta <<1$ which also implies $\alpha\approx\beta$ (this is possibly a weakness of the answer) then $$\alpha^2 \approx \frac{K_\mathrm{w}}{K_\mathrm{a}\,K_\mathrm{b}},$$ and $$ K_\mathrm{a} = \frac{(1-\alpha)[\ce{H+}]}{\alpha} \approx\frac{[\ce{H+}]}{\alpha}, $$ then \begin{align} (\frac{[\ce{H^+}]}{K_\mathrm{a}})^{2} &\approx \frac{K_\mathrm{w}}{K_\mathrm{a}\,K_\mathrm{b}}\\ [\ce{H^+}]^{2} &\approx \frac{K_\mathrm{w}\,K_\mathrm{a}}{K_\mathrm{b}}, \end{align} and we get the equation.

  • $\begingroup$ As an example, $\ce{pH}$ for ammonium acetate salt is exactly 7 because $pH=\frac{14}{2}$ $\endgroup$
    – user43021
    Commented Jun 3, 2019 at 17:55

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