Question

A student prepares a buffer solution which contains $0.05\text{mol}d{m}^{-3}$ $HCOOH$ and $0.250mold{m}^{-3}$ $HCO{O}^{-}N{a}^{+}.$ The $K_a$ of $HCOOH$ is $1.78\times {10}^{-4}.$ Calculate the pH of the solution.
$log1.78=0.250,log5=0.699$

• A. 12.5
• B. 4.45
• C. 6.2
• D. 7.5

Answer 1

The correct option is B 4.45

Given, $\left[HCOOH\right]=0.05\text{mol}d{m}^{-3},\left[HCO{O}^{-}N{a}^{+}\right]=0.250\text{mol}d{m}^{-3},K_a\left(HCOOH\right)=1.78\times {10}^{-4}$
Using the Henderson-Hasselbalch equation for weak bases:
$pH=p{K}_a+log\left[\frac{\left[conjugatebase\right]}{\left[acid\right]}\right]$
$pH=p{K}_a+log\left[\frac{\left[HCO{O}^{-}N{a}^{+}\right]]}{\left[HCOOH\right]}\right]$
$pH=-log{K}_a+log\left[\frac{\left[HCO{O}^{-}N{a}^{+}\right]}{\left[HCOOH\right]}\right]$
$pH=-log{K}_a=+log\left[\frac{\left[HCO{O}^{-}N{a}^{+}\right]}{[HCOOH}\right]$
$pH=-log(1.78\times {10}^{-4})+log\left[\frac{\left[0.250\right]}{\left[0.05\right]}\right]$
$pH=4-log\left(1.78\right)+log\left(5\right)$
$pH=4-0.250+0.699$
$pH=4.45$