How to determine the pH of a mixture of two weak acids?

Question

We’ve two solutions:

Solution 1 $\ce{HCOOH}$ its concentration is $c_1=10^{-2}\ \mathrm{mol/l}$ and its volume is $V_1 = 50\ \mathrm{ml}$ and its $\mathrm{pH}_1 = 2.9$.

Solution 2 $\ce{CH_3COOH}$ its concentration is $c_2=10^{-2}\ \mathrm{mol/l}$ and its volume is $V_2 = 50\ \mathrm{ml}$ and its $\mathrm{pH}_2 = 3.4$.

How would be the equation and the ice table, and what is the $\mathrm{pH}$ of the mixture of these two solutions

I used number just to understand how that is work, no other reason.

I tried to do $\mathrm{pH}=\frac{\mathrm{pH}_1+\mathrm{pH}_2}{2}$, but it’s trivial.

Answer 1

Given the following

\begin{align} [\ce{CH3COOH}]&=\pu{1 \cdot 10^{-2}\,mol\,L^{-1}}\\ [\ce{HCOOH}]&=\pu{1 \cdot 10^{-2}\,mol\,L^{-1}}\\ K_{z-\ce{CH3COOH}}&=\pu{1.8 \cdot 10^{-5}\,mol\,L^{-1}}\\ K_{z-\ce{HCOOH}}&=\pu{1.8 \cdot 10^{-4}\,mol\,L^{-1}}\\ [\ce{H3O^{+}}]_1 &= \pu{1.3 \cdot 10^{-3}\,mol\,L^{-1}}\\ [\ce{H3O^{+}}]_2 &= \pu{4.0 \cdot 10^{-4}\,mol\,L^{-1}} \end{align}

Seeing that we could write

$$\ce{HCOOH + CH3COOH + 2H2O-> HCOO^{-} + CH3COO^{-}+2H3O^{+}}$$

We can derive a equilibrium

\begin{align} \frac{[\ce{HCOO^{-}}][\ce{CH3COO^{-}}][\ce{H3O^{+}}]^{2}}{[\ce{HCOOH}][\ce{CH3COOH}]}&=\frac{[\ce{HCOO^{-}}][\ce{H3O^{+}}]}{[\ce{HCOOH}]}\frac{[\ce{CH3COO^{-}}][\ce{H3O^{+}}]}{[\ce{CH3COOH}]}\\ &=K_{z-\ce{HCOOH}}K_{z-\ce{CH3COOH}} \end{align}

I think we still need to know the original concentrations of $\ce{CH3COOH}$ and $\ce{HCOOH}$, we can calculate

$$[\ce{CH3COO^{-}}]=\frac{[\ce{CH3COOH}]}{[\ce{H3O^{+}}]_2}K_{z-\ce{CH3COOH}}= \pu{4.5000 \cdot 10^{-4}\,mol\,L^{-1}}$$

for $[\ce{HCOO^{-}}]$ we get $\pu{1.3846 \cdot 10^{-3}\,mol\,L^{-1}}$ (keeping some significance). So...

\begin{align} n_{\ce{HCOOH}}&=\pu{1.1385 \cdot 10^{-2}\,mol\,L^{-1}} &&\text{(total)}\\ n_{\ce{CH_ 3COOH}}&=\pu{5.225 \cdot 10^{-3}\,mol\,L^{-1}} && \text{(total)} \end{align}

Solving for $y$ (neglecting the already present $\ce{H3O^{+}}$ of water)

\begin{align} K_{z-\ce{HCOOH}}&=\frac{[\ce{HCOO^{-}}][\ce{H3O^{+}}]}{[\ce{HCOOH}]}\\ &=\frac{x(x+y)}{(0.5n_{\ce{HCOOH}}-x)}\\ \end{align} gives $$y = -0.00018 - x + \frac{1.02465 \cdot 10^{-6}}{x},$$ using $y$ in \begin{align} K_{z-\ce{CH3COOH}}&=\frac{[\ce{CH3COO^{-}}][\ce{H3O^{+}}]}{[\ce{CH3COOH}]}\\ &=\frac{y(x+y)}{(0.5n_{\ce{CH3COOH}}-y)} \end{align} gives $$x = 0.000884322 \vee x = 0.00690291$$ (there are more solutions, but these are the only real positive ones). From this we can calculate $$y=0.0000943623 \vee y = -0.00693447.$$ $y$ should be positive, thus $x = 0.000884322$ and $y=0.0000943623$. This gives us a $pH$ of

$$\mathrm{pH}=-\log(0.000884322+0.0000943623)\approx 3.01$$

Which seems to me as correct (it will be slightly more acidic than when adding the amounts of $\ce{H^{+}}$).

Answer 2

Below a more general approach.

Suppose that we have two weak acids $\ce{HA}$ and $\ce{HB}$. The initial concentrations are $C^0_\ce{HA}$ and $C^0_\ce{HB}$, and their constants are ${K_\ce{a}}_\ce{(HA)}$ and ${K_\ce{a}}_\ce{(HB)}$. Suppose yet that volumes, $V_\ce{HA}$ and $V_\ce{HB}$, are additives. So we have:

1. Reactions

$$\ce{HA + H2O <=> H3O+ + A-}\qquad {K_\ce{a}}_\left(\ce{HA}\right)=\frac{\ce{[H3O+][A-]}}{\ce{[HA]}}\tag{1}\label{eq:KAcidHA}$$

$$\ce{HB + H2O <=> H3O+ + B-}\qquad {K_\ce{a}}_\left(\ce{HB}\right)=\frac{\ce{[H3O+][B-]}}{\ce{[HB]}}\tag{2}\label{eq:KAcidHB}$$

$$\ce{2 H2O <=> H3O+ + OH-}\qquad K_\ce{w}=\ce{[H3O+][OH-]}\tag{3}\label{eq:KWater}$$

2. Mass balance

$$C_\ce{HA} = \frac{C^0_\ce{HA}V_\ce{HA}}{V_\ce{HA} + V_\ce{HB}}=\ce{[HA] + [A-]}\tag{4}\label{eq:MassBalanceHA}$$

$$C_\ce{HB} = \frac{C^0_\ce{HB}V_\ce{HB}}{V_\ce{HA} + V_\ce{HB}}=\ce{[HB] + [B-]}\tag{5}\label{eq:MassBalanceHB}$$

3. Change balance

$$\ce{[H3O+] = [OH-] + [A-] + [B-]}\tag{6}\label{eq:ChargeBalance}$$

Replacing ($\ref{eq:KAcidHA}$–$\ref{eq:MassBalanceHB}$) equations on ($\ref{eq:ChargeBalance}$), we have: $$\ce{[H3O+]} = \frac{K_\ce{w}}{\ce{[H3O+]}} + \frac{C_\ce{HA}{K_\ce{a}}_\left(\ce{HA}\right)}{\ce{[H3O+]} + {K_\ce{a}}_\left(\ce{HA}\right)} + \frac{C_\ce{HB}{K_\ce{a}}_\left(\ce{HB}\right)}{\ce{[H3O+]} + {K_\ce{a}}_\left(\ce{HB}\right)}\tag{7}\label{eq:GeneralEquation}$$

or as polynomial

\begin{align} \begin{split} &\ce{[H3O+]}^4\\ +&\ce{[H3O+]}^3\left({K_\ce{a}}_\left(\ce{HA}\right) + {K_\ce{a}}_\left(\ce{HB}\right)\right)\\ +&\ce{[H3O+]}^2\left[{K_\ce{a}}_\ce{(HA)}{{K_\ce{a}}_\ce{(HB)}} -\left(C_\ce{HA}{K_\ce{a}}_\ce{(HA)}+C_\ce{HB}{{K_\ce{a}}_\ce{(HB)}}\right)-K_\ce{w} \right]\\ -&\ce{[H3O+]}\big[\big(C_\ce{HA}+C_\ce{HB}\big){K_\ce{a}}_\ce{(HA)}{{K_\ce{a}}_\ce{(HB)}} + K_\ce{w}\left({K_\ce{a}}_\ce{(HA)}+{{K_\ce{a}}_\ce{(HB)}}\right)\big]\\ -&{K_\ce{a}}_\left(\ce{HA}\right){K_\ce{a}}_\left(\ce{HB}\right)K_\ce{w}\\ =&\ 0 \end{split}\tag{8}\label{eq:GeneralPol} \end{align}

This single equation will exactly solve any equilibrium problem involving the mixture of any two monoprotic acids, in any concentration (as long as they're not much higher than about $\pu{1 mol L-1}$) and any volume. Depending of $K_\ce{a}$ values, we can yet obtain a simpler version.

The ($\ref{eq:GeneralPol}$) equation can simplified considering that ${K_\ce{a}}_\left(\ce{HA}\right){K_\ce{a}}_\left(\ce{HB}\right)K_\ce{w}\ll 1$.

\begin{align} \begin{split} &\ce{[H3O+]}^3\\ +&\ce{[H3O+]}^2\left({K_\ce{a}}_\left(\ce{HA}\right) + {K_\ce{a}}_\left(\ce{HB}\right)\right)\\ +&\ce{[H3O+]}\left[{K_\ce{a}}_\ce{(HA)}{{K_\ce{a}}_\ce{(HB)}} -\left(C_\ce{HA}{K_\ce{a}}_\ce{(HA)}+C_\ce{HB}{{K_\ce{a}}_\ce{(HB)}}\right)-K_\ce{w} \right]\\ -&\big[\big(C_\ce{HA}+C_\ce{HB}\big){K_\ce{a}}_\ce{(HA)}{{K_\ce{a}}_\ce{(HB)}} + K_\ce{w}\left({K_\ce{a}}_\ce{(HA)}+{{K_\ce{a}}_\ce{(HB)}}\right)\big]\\ =&\ 0 \end{split}\tag{9}\label{eq:GeneralPolSimp1} \end{align}

The ($\ref{eq:GeneralPolSimp1}$) equation can simplified considering that ${K_\ce{a}}_\left(\ce{HA}\right){K_\ce{a}}_\left(\ce{HB}\right)\ll 1$ and disregarding the autoionization of water.

\begin{align} \ce{[H3O+]}^2+\ce{[H3O+]}\left({K_\ce{a}}_\ce{(HA)} + {K_\ce{a}}_\ce{(HB)}\right) -\left(C_\ce{HA}{K_\ce{a}}_\ce{(HA)}+C_\ce{HB}{{K_\ce{a}}_\ce{(HB)}}\right) = 0\tag{10}\label{eq:GeneralPolSimp2} \end{align}

The ($\ref{eq:GeneralPolSimp2}$) equation can be solved as usual.

$$\ce{[H3O+]}=\frac{-\left({K_\ce{a}}_\ce{(HA)} + {K_\ce{a}}_\ce{(HB)}\right)+\sqrt{\left({{K_\ce{a}}_\ce{(HA)} + {K_\ce{a}}_\ce{(HB)}}\right)^2+4\left(C_\ce{HA}{K_\ce{a}}_\ce{(HA)}+C_\ce{HB}{{K_\ce{a}}_\ce{(HB)}}\right)}}{2}$$

or using the initial concentrations

$$\ce{[H3O+]}=\frac{-\left({K_\ce{a}}_\ce{(HA)} + {K_\ce{a}}_\ce{(HB)}\right)+\sqrt{\left({{K_\ce{a}}_\ce{(HA)} + {K_\ce{a}}_\ce{(HB)}}\right)^2+4\left(\displaystyle\frac{C^0_\ce{HA}V_\ce{HA}{K_\ce{a}}_\ce{(HA)}}{V_\ce{HA} + V_\ce{HB}}+\frac{C^0_\ce{HB}V_\ce{HB}{K_\ce{a}}_\ce{(HB)}}{V_\ce{HA} + V_\ce{HB}}\right)}}{2}$$

Replacing $C^0_\ce{HA}=\pu{0.01 mol L-1}$, $C^0_\ce{HB}=\pu{0.01 mol L-1}$, $V_\ce{HA}=\pu{0.050 L}$ and $V_\ce{HB}=\pu{0.050 L}$, and using $\text{p}K_\ce{a}=3.77$ for formic acid and $\text{p}K_\ce{a}=4.756$ for acetic acid, we have

$$\ce{pH}=3.06$$

Answer 3

$-\log[\ce{H}]=\text{p(H)}$ you can calculate concentration of $[\ce{H}]$ for each acid, then add the $[\ce{H}]$ for two acids and calculate $\text{p(H)}$ by $-\log[\ce{H}]$.