# How does the pH of solutions of salts in which the cation is acidic and anion basic, depend only on the relation of their constants?

My book is not very explanative about it, I'm trying to get the equation:

$$[\ce{H^+}]=\sqrt{K_\mathrm{w}\times\frac{K_\mathrm{a}}{K_\mathrm{b}}}$$

but I didn't succeed, can someone explain the theory behind and how to get the formula comparing the two ionization constants?

Translation of the actual page of the textbook (Cacace Fondamenti di stechiometria (in Italian)):

Salts which cation has acid features and anions has basic features. A salt like ammonium acetate, $$\ce{NH4Ac}$$, release after the following (complete) dissociation:

$$\ce{NH4Ac -> NH4+ +Ac-}$$

A cation will behave as an acid and an anion will behave as a base. In cases like this one, acid-base equilibrium in solution depends only by the strength of the cation compared to the anion. It can be proved

(but it does not, that's why I'm here)

that in first approximation , the concentration of $$\ce{H+}$$ in those solution is indipendent from the concentration of the salt and it depends only from the comparison between ionization constants of the weak acid and the weak base from which the salt has been formed:

$$[\ce{H^+}] = \sqrt{K_\mathrm{w}\times\frac{K_\mathrm{a}}{K_\mathrm{b}}}$$

From which:

$$\mathrm{pH} = \frac{1}{2}(14 + \mathrm{p}K_\mathrm{a} - \mathrm{p}K_\mathrm{b})$$

It seems to be unclear. I would like to understand the mathematics of the equation that brings the comparison of the two constant became like the one I already wrote. Why is independent from the salt concentration?

• I suspect this is the formula for finding the pH of the solution of the highly-soluble salt of weak acid and weak base that undergoes hydrolysis. If the question gets reopened, I'm going to post an answer with how this formula is derived. – andselisk Jan 16 at 9:18
• @andselisk go ahead :D // This may be related: Calculating pH of diprotic and amphoteric solutions – Martin - マーチン Jan 16 at 13:37
• @Martin-マーチン Thanks for reopening the question:) – andselisk Jan 16 at 15:12

Disclaimer: as we will see later, the formula

$$[\ce{H+}] = \sqrt{\frac{K_\mathrm{w}K_\mathrm{a}}{K_\mathrm{b}}} \label{1}\tag{1}$$

is only applicable to highly soluble salts with equimolar rests of monovalent weak acid and weak base such as $$\ce{NH4OAc}$$, $$\ce{NH4HCO3}$$, $$\ce{NH4HSO3}$$, $$\ce{NH4NO2}$$ and cannot be used for polyvalent electrolytes, poorly soluble salts and when hydrolysis is irreversible (e.g. $$\ce{Al2S3}$$).

Let's derive formula \eqref{1} in for the salt of the general $$\ce{BHA}$$ formula. As we agreed, the salt is highly soluble (also true for your textbook example, ammonium acetate), so the main approximation here is that we assume this salt dissociates completely in solution:

$$\ce{BHA → BH+ + A-} \label{R1}\tag{R1}$$

Each ion participates in hydrolytic equilibrium (\eqref{R2}, \eqref{R3}), plus there is autoprotolysis of water, \eqref{R4}. Subtracting \eqref{R4} from \eqref{R2} and \eqref{R3} we get a brutto equation for hydrolysis \eqref{R5}:

\begin{align} &+\quad&\ce{BH+ + H2O &<=> B + H3O+} \label{R2}\tag{R2}\\ &+\quad&\ce{A- + H2O &<=> HA + HO-} \label{R3}\tag{R3}\\ &-\quad&\ce{2 H2O &<=>[K_\mathrm{w}] H3O+ + HO-} \label{R4}\tag{R4}\\ \hline &&\ce{BH+ + A- &<=>[K_\mathrm{h}] B + HA}, \label{R5}\tag{R5}\\ \end{align}

where $$K_\mathrm{h}$$ is hydrolysis constant. According to the law of mass action, it can be written for \eqref{R5} as such:

$$K_\mathrm{h} = \frac{[\ce{B}][\ce{HA}]}{[\ce{BH+}][\ce{A-}]} \label{2}\tag{2}$$

Now let's multiply both divider and denominator in \eqref{2} by $$[\ce{H3O+}][\ce{HO-}]$$ so that

\begin{align} K_\mathrm{h} &= \left.\frac{[\ce{B}][\ce{HA}]}{[\ce{BH+}][\ce{A-}]}\qquad \right|\cdot \frac{[\ce{H3O+}][\ce{HO-}]}{[\ce{H3O+}][\ce{HO-}]}\\ &= \frac{\color{blue}{[\ce{B}]}\color{red}{[\ce{HA}]}[\ce{H3O+}][\ce{HO-}]}{\color{blue}{[\ce{BH+}]}\color{red}{[\ce{A-}][\ce{H3O+}]}\color{blue}{[\ce{HO-}]}}\\ &= \frac{[\ce{H3O+}][\ce{HO-}]}{\color{red}{K_\mathrm{a}}\color{blue}{K_\mathrm{b}}}\\ &= \frac{K_\mathrm{w}}{K_\mathrm{a}K_\mathrm{b}}\label{3}\tag{3} \end{align}

where $$K_\mathrm{a}$$ and $$K_\mathrm{b}$$ are dissociation constants of the conjugated acid and conjugated base, respectively:

\begin{align} \ce{HA + H2O &<=>[K_\mathrm{a}] A- + H3O+} \label{R6}\tag{R6}\\ \ce{B + H2O &<=>[K_\mathrm{b}] BH+ + HO-} \label{R7}\tag{R7}\\ \end{align}

Since the dissociation of the salt is complete (see \eqref{R1}), its initial concentration $$c_0$$ equals to the concentrations of $$\ce{BH+}$$ and $$\ce{A-}$$ in solution:

$$c(\ce{BH+}) = c(\ce{A-}) = c_0 \label{4}\tag{4}$$

and the corresponding equilibrium concentrations of the ions participating in hydrolysis are

$$[\ce{BH+}] = [\ce{A-}] = (1-α)c_0, \label{5}\tag{5}$$

where $$α$$ is the degree of hydrolysis showing the ratio between the amount of hydrolyzed salt $$n_\mathrm{h}$$ to the total amount of salt $$n_0$$ ($$α = n_\mathrm{h}/n_0$$; $$α\in [0;1]$$). The equilibrium concentrations of the weak acid and base formed during hydrolysis are

$$[\ce{B}] = [\ce{HA}] = αc_0 \label{6}\tag{6}$$

and eq. \eqref{2} for $$K_\mathrm{h}$$ can be rewritten using degree of hydrolysis:

$$K_\mathrm{h} = \frac{[\ce{B}][\ce{HA}]}{[\ce{BH+}][\ce{A-}]} = \frac{α^2c_0^2}{(1-α)^2c_0^2} = \frac{α^2}{(1-α)^2} \label{7}\tag{7}$$

Note that hydrolysis constant doesn't depend on salt's concentration. Since we are interested in $$\ce{pH}$$, it is convenient to express $$[\ce{H3O+}]$$ from $$K_\mathrm{a}$$:

$$K_\mathrm{a} = \frac{[\ce{H3O+}][\ce{A-}]}{[\ce{HA}]};$$

$$[\ce{H3O+}] = K_\mathrm{a}\frac{[\ce{HA}]}{[\ce{A-}]} = K_\mathrm{a}\frac{αc_0}{(1-α)c_0} = K_\mathrm{a}\frac{α}{1-α} \label{8}\tag{8}$$

From \eqref{7} $$α/(1-α) = \sqrt{K_\mathrm{h}}$$ so that \eqref{8} finally becomes the first equation you asked about:

$$[\ce{H3O+}] = K_\mathrm{a}\sqrt{K_\mathrm{h}} = K_\mathrm{a}\sqrt{\frac{K_\mathrm{w}}{K_\mathrm{a}K_\mathrm{b}}} = \sqrt{\frac{K_\mathrm{w}K_\mathrm{a}}{K_\mathrm{b}}} \label{9}\tag{9}$$

An equation for $$\mathrm{pH}$$ is obtained trivially by taking negative decimal logarithm on both parts of \eqref{9}.

• What's the rationale for assuming that $\alpha$ is the same in both cases? – Zhe Jan 16 at 16:23
• @Zhe What do you mean? There is only one case: a single hypothetical salt (see 2nd paragraph) with its own degree of hydrolysis. – andselisk Jan 16 at 16:39
• Sorry, I think I'm not familiar with thinking about it this way. I was basically at the same solution as this except that I did not get to a point where the concentrations could effectively cancel to 1 because that requires an assumption that the relative concentrations of acid/conjugate acid and base/conjugate base are very close to 1. Perhaps that's the first order approximation I needed, but I didn't find a convincing reason why this should just be true. – Zhe Jan 16 at 16:46
• @Zhe Oh, yes, as I mentioned, the salt must be very soluble to assume that. Otherwise you end up with a single equation with 4 variables or something. And if there also were a multi-step hydrolysis, then the math would've really gotten very wild. – andselisk Jan 16 at 16:49
• @andselisk Thank you for your time, it was truly clear. i'll try to make it myself and truly understanding it. Best regards – Gabriele Sortino Jan 17 at 13:28