0
$\begingroup$

$0.1\ \mathrm{mol}$ 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 ?

$\endgroup$
  • 1
    $\begingroup$ 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
2
$\begingroup$

$$\ce{HCOONa + HCOOH<=> HCOONa + HCOOH}$$ 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}$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.