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$0.1\ \mathrm{mol}$ of sodium formate was added to one litre of 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 ?

My attempt :

Found out the $\ce{[H+]}$in formic acid. Now for calculating $\ce{[H+]}$ in $\ce{HCOONa}$, I am planning to use the equation: $$\mathrm{pH}=\mathrm pK_\mathrm a+\log\frac{[\text{salt}]}{[\text{acid}]}$$ But I don't know the $\mathrm pK_\mathrm a$ of $\ce{HCOONa}$
How do I proceed , any alternative approaches ?

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    $\begingroup$ HCOONa does not have the pKa, nor does it produce [H+]. $\endgroup$ – Ivan Neretin 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 $\endgroup$ – Sujith Sizon Sep 12 '15 at 8:22
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$$\ce{HCOONa + HCOOH<=> HCOONa + HCOOH}$$ The constant of this equilibrium equals $1$.The composition of the mixture will not change. $$\mathrm{pH}= \mathrm{p}ka +\log \frac{\ce{[HC OO^{-}]}}{\ce{[HCOOH]}}$$ $$\mathrm{pH}= 3.74 +\log \frac{{0.1}}{0.2}=3.44$$ So, the initial concentration of $\ce{[H^+]_0= 10^{-3.74}}\mathrm{M}$

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

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

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