$\pu{0.1 mol}$ of sodium formate was added to one litre of $\pu{0.2 M}$ solution of formic acid ($K_\mathrm{a} = 1.8\times10^{-4}$). How much will the $\ce{H+}$ concentration of the acid diminish assuming the salt to be completely dissociated?

I found the concentration of $\ce{H+}$ in formic acid. For calculating $\ce{[H+]}$ in $\ce{HCOONa}$, I am planning to use the equation: $$\mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log\frac{[\text{salt}]}{[\text{acid}]}$$ But I don't know the $\mathrm{p}K_\mathrm{a}$ of $\ce{HCOONa}$. How do I proceed, are there any alternative approaches?

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    $\begingroup$ HCOONa does not have the pKa, nor does it produce [H+]. $\endgroup$ Sep 12 '15 at 8:17
  • $\begingroup$ Oh okay , but then how do i find the final [H+] concentration . There will be an increase in [HCOO-] right so the acid dissociation of HCOOH--[HCOO-] + [H+] will favour backward reaction hence [H+] should decrease $\endgroup$ Sep 12 '15 at 8:22

A rough calculation for the pH of 0.2-ᴍ formic acid gives:

$$ \mathrm{pH} = 1/2 (\mathrm{p}K_\mathrm{a} - \log(0.2)) = 1/2 (3.74 + 0.70) = 2.22 $$

You can check by calculating the equilibrium constant from the concentrations of all the species, and the estimate is pretty good.

The pH of the buffer is 3.44 using the Henderson Hasselbalch expression (see other answer). The change in pH is 1.22, corresponding to a change of a factor of about 17 in the hydronium ion concentration.


$$\ce{HCOONa + HCOOH<=> HCOONa + HCOOH}$$ The constant of this equilibrium equals $1$. The composition of the mixture will not change. \begin{align} \mathrm{pH} &= \mathrm{p}K_\mathrm{a} + \log\frac{\ce{[HCOO-]}}{\ce{[HCOOH]}}\\ \mathrm{pH} &= 3.74 + \log\frac{0.1}{0.2} = 3.44 \end{align}

So, the initial concentration of protons is $[\ce{H^+}]_0 = \pu{10^{-3.74} M}$.

The final concentration of protons is $[\ce{H+}]_\mathrm{f} = \pu{10^{-3.44} M}$.

The decrease in the concentration of $\ce{H^+}$ is $\pu{10^{-3.44} M} - \pu{10^{-3.74} M} = \pu{10^{+0.30} M}$.


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