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.


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 $\ce{H^{+}}$).

  • 1
    $\begingroup$ Checking an online pH calculator, your answer seems correct. You have the right idea of combining both equilibria in a single calculation, though your solution seems a little more complex to read (it could be just me though!). There's a nice,organized and easily extendable way to solve any type of acid-base pH problem, based simply on charge and matter balance equations. If you're curious, check out some of my previous answers to other problems, e.g. here and here. $\endgroup$ – Nicolau Saker Neto May 28 '14 at 0:25
  • $\begingroup$ Another source with the mathematical description of these solutions is given here. $\endgroup$ – Nicolau Saker Neto May 28 '14 at 0:32
  • 1
    $\begingroup$ @NicolauSakerNeto That's good to hear. This is actually more simple than it looks like. You can extend this by solving more equations with more unknown. In this example we used x, y, but if I would be including $H_2O$ we would get x, y, x and an extra equation $K_w=[OH^{-}][H_3O^{+}]$. Thanks for the links, I'll look after it. $\endgroup$ – Jori May 28 '14 at 7:39
  • 1
    $\begingroup$ Also, it seems the links provided do essentially the same thing as me. Setting up equations and then substituting. $\endgroup$ – Jori May 28 '14 at 7:42
  • $\begingroup$ @NicolauSakerNeto Can you give the link of your online pH calculator? $\endgroup$ – Jori Jul 2 '14 at 22:33

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

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

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

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

  1. Mass balance

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

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

  1. 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


  • $\begingroup$ If ($\ref{eq:GeneralPol}$) equation is solved exactly using the matlab or octave the pH value is 3.05. $\endgroup$ – GRSousaJr Sep 16 '19 at 12:18
  • $\begingroup$ @GRSousaJrHow assumption "${K_\ce{a}}_\left(\ce{HA}\right){K_\ce{a}}_\left(\ce{HB}\right)K_\ce{w}\ll 1$" simplifies equation ($8$) to equation ($9$)? How assumption "${K_\ce{a}}_\left(\ce{HA}\right){K_\ce{a}}_\left(\ce{HB}\right)\ll 1$" simplifies equation ($9$) to equation ($10$)? $\endgroup$ – Adnan AL-Amleh Oct 2 '19 at 23:22
  • 1
    $\begingroup$ Look the acetic and propionic acids example. The $\text{p}K_{\ce{a}}$ values are $4.756$ and $4.88$ respectively. $K_{\ce{a}(\ce{Ac})}K_{\ce{a}(\ce{Pro})}K_{\ce{w}} = \pu{2.31E-24}$ and $K_{\ce{a}(\ce{Ac})}K_{\ce{a}(\ce{Pr})} = \pu{2.31E-10}$. The contribution of these values to sum can be disregarding and equations simplified. $\endgroup$ – GRSousaJr Oct 3 '19 at 11:26
  • $\begingroup$ Thanks, I understand, but how the assumption simplifies the quartic equation to the Cubic equation? $\endgroup$ – Adnan AL-Amleh Oct 3 '19 at 21:52
  • 1
    $\begingroup$ Divide the equation by $\ce{[H3O+]}$. $\endgroup$ – GRSousaJr Oct 5 '19 at 1:37

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

  • $\begingroup$ I can't understand what are you trying to do, by the way $pH=-\log [H^+]$ doesn't works for weak acids. $\endgroup$ – Hedwig Apr 27 '14 at 12:33
  • $\begingroup$ pH=−log[H+] work for all acids.you have pH for tow acids then you calculate concentration [H+] by pH=-log[H+] and addition [H+] for tow acids then calculate pH by pH=-log[H+] $\endgroup$ – mansour Apr 28 '14 at 13:38

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