I came across this formula to calculate the resultant [H]+ concentration when two weak acids are mixed (Both acids of equal volumes are mixed together). enter image description here

enter image description hereThe only problem i have with this is that, we are completely ignoring the fact that the concentration of both the acids are going to be halved after they are mixed together since, the volume is doubling. Is there anything i'm missing? The only way, this could be correct is if the concentrations that are given to me in the question are those after the acids have been mixed.

EDIT: This is the question i was talking about enter image description here

This is the solution i was given enter image description here

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    – andselisk
    Apr 24, 2020 at 5:07

2 Answers 2


Let suppose $c, c_1, c_2$ are concentrations of weak acids, and $K_\mathrm{a}, K_\mathrm{a1}, K_\mathrm{a2}$ are their respective acidity(dissociation) constants.

The equations $\ce{[H+]}=\sqrt{K_\mathrm{a} \cdot c}$ for a weak acid, respectively for 2 acids $\ce{[H+]}=\sqrt{K_\mathrm{a1} \cdot c_1 + K_\mathrm{a2} \cdot c_2}$

are approximately valid, if $c, (c_1, c_2) \gg K_\mathrm{a}(, K_\mathrm{a1}, K_\mathrm{a2}) \gg 10^-7 = \sqrt{K_\mathrm{w} }$, what also implies $\ce{[H+]} \gg 10^{-7}$.

Then we can approximate all $\ce{H+}$ comes from the acid dissociation, but OTOH, all acid is in non-dissociated form, leading to $$K_\mathrm{a} = \frac{\ce{[H+][A-]} }{\ce{[HA]}} \simeq \frac{{[H+]}^2 }{ c}$$ and $$\ce{[H+]}=\sqrt{K_\mathrm{a} \cdot c}$$

If, OTOH, $c \ll 10^{-7}=\sqrt{K_\mathrm{w}}$ and $c \ll K_\mathrm{a}$ we can approximate all acid is dissociated.

Then $$K_\mathrm{w} = \ce{[H+][OH-]}$$

$$\ce{[H+]} - \ce{[OH-]} = c_1 + c_2$$

$$\ce{[H+]} - \frac{K_\mathrm{w}}{\ce{[H+]}} = c_1 + c_2$$

$${\ce{[H+]}}^2 - (c_1 + c_2) \cdot \ce{[H+]} - K_\mathrm{w} = 0 $$

$$\ce{[H+]} = \frac{c_1 + c_2 \pm \sqrt {{(c_1 + c_2)}^2 + 4 \cdot K_\mathrm{w} }}{2} $$

$$\ce{[H+]} = \frac{\pu{2.1e-9} \pm \sqrt {{(\pu{2.1e-9})}^2 + \pu{4e-14} }}{2} = \pu{1.0105e-7 mol/L }$$

$$\mathrm{pH} = 6.995$$

It is obvious, that if with full dissociation $\mathrm{pH}$ does not reach even the value $6.99$, the result of partial dissociation cannot be $6.7$


The pH of a mixture of two weak acids can be obtained as: $$\mathrm{pH} = \sqrt{K_\mathrm{\alpha 1}\times c_1 + K_\mathrm{\alpha 2}\times c_2}$$

Ok, that's what we are trying to prove, but the equation is wrong from the start. The right hand side calculates $\ce{[H+]}$, not pH. Note that $c_1$ and $c_2$ are the nominal concentrations of the acids in the mixture.

Consider two weak acids say HOCN and HCOOH

For HOCN: $K_\alpha = \dfrac{\ce{[H+][OCN-]}}{\ce{[HOCN]}}$

For HCOOH: $K_\alpha = \dfrac{\ce{[H+][HCOO-]}}{\ce{HCOOH}}$

Both of the above $K_\alpha$ expressions would be valid for any concentrations of the weak acids and any values of the $K_\alpha$ constants.

Also, $\ce{[H+]}$ in solution will be same and thus:
$$\ce{[H+] = [H+] by HOCN + [H+] by HCOOH}$$ $$= \ce{[OCN-] + [HCOO-]}$$

This assumes that $ \ce{[OCN-] + [HCOO-] \gg [OH-]}$

$$\ce{[H+]} = \dfrac{K_{\alpha,\ \ce{HOCN}}\times \ce{[HOCN]}}{\ce{[H+]}} + \dfrac{K_{\alpha,\ \ce{HCOOH}}\times \ce{[HCOOH]}}{\ce{[H+]}}$$

$$\therefore \ce{[H+]^2} = K_{\alpha,\ \ce{HOCN}}\times \ce{[HOCN]} + K_{\alpha,\ \ce{HCOOH}}\times \ce{[HCOOH]}$$

$$\text{or}\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad$$ $$\ce{[H+]} = \sqrt{ K_{\alpha,\ \ce{HOCN}}\times \ce{[HOCN]} + K_{\alpha,\ \ce{HCOOH}}\times \ce{[HCOOH]}}$$

Now we know that $C_\ce{HOCN} = \ce{[HOCN] + [OCN-]}$. If $\ce{[HOCN] \gg [OCN-]}$ then $C_\ce{HOCN} \approx \ce{[HOCN]}$ and we can substitute $C_\ce{HOCN}$ for $\ce{[HOCN]}$.

Likewise for $C_\ce{HCOOH}$ and $\ce{[HCOOH]}$

Substituting we get the desired result.

$$\therefore \ce{[H+]^*} = \sqrt{K_\mathrm{\alpha,\ \ce{HOCN}}\times C_\ce{HOCN} + K_\mathrm{\alpha,\ \ce{HCOOH}}\times C_\ce{HCOOH}}$$

But the hydrogen ion concentration has been notated as $\ce{[H+]^*}$ to denote that the value is conditional. The condition is that $\ce{[H+]^* \gg 1\times 10^{-7}}$ for the value to be valid.

$\ce{[H+]^* \gg 1\times 10^{-7}}$ is somewhat fuzzy.

  • If $\ce{[H+]^* \gt 1\times 10^{-6}}$ then we can have 1 significant figure in the answer.

  • If $\ce{[H+]^* \gt 1\times 10^{-5}}$ then we can have 2 significant figures in the answer.

  • If $\ce{[H+]^* \gt 1\times 10^{-4}}$ then we can have 3 significant figures in the answer. (3 significant figures for such calculations is more wishful thinking about the chemistry than reality.)

I've been scratching my head on this part, but I believe that $\ce{[H+]^* \gg 1\times 10^{-7}}$ encompasses any other assumptions.


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