Question

How many $moles$ of $HCOONa$ must be added to $1L$ of $0.1M$ $HCOOH$ to prepare a buffer solution with a $pH$ of $3.4$?
(Given: $K_a$ for $HCOOH=2\times {10}^{-4}$)

• A. $0.01$
• B. $0.05$
• C. $0.1$
• D. $0.2$

Answer 1

The correct option is B $0.05$

Here the reaction are :

$HCOONA\to HCO{O}^{-}+N{a}^{+}$

$HCOOH\rightleftharpoons HCO{O}^{-}+H^{+}$

Now $Ka=\frac{\left[HCO{O}^{-}\right]\left[H^{+}\right]}{\left[HCOOH\right]}$

Taking log on both sides

$\log Ka=\log HCO{O}^{-}+\log H^{+}-\log HCOOH$

$\Rightarrow pH=pKa+\log \frac{\left[HCO{O}^{-}\right]}{\left[HCOOH\right]}$..........(1)

Since $HCOONa$ is a salt and fully dissociates

$\therefore \left[HCO{O}^{-}\right]\approx \left[HCOONA\right]$

$\therefore 3.4=(-\log (2\times {10}^{-4}\left)\right)+\log \left(\frac{\left[HCO{O}^{-}\right]}{\left[0.1\right]}\right)+1$

$3.4=(4-{\log }_{10}2)+\log \left)\right(HCO{O}^{-})+1$

$\log HCO{O}^{-}=-1.3$

$\left[HCO{O}^{-}\right]={10}^{1.3}=0.05M$

$\therefore$ No of moles of $HCOONa$

required to be added $M1L=0.05mol$