Question

How much sodium formate $(HCOONa,68.0069gmo{l}^{-1})$ is to be added to 400 mL of 1.00 M formic acid for a pH=3.50 buffer. Given $K_a=1.77\times {10}^{-4}$
$log1.77=0.248\text{and log}0.56=-2.52$

• A. 20.25 g
• B. 15.23 g
• C. 30.34 g
• D. 5.50 g

Answer 1

The correct option is B 15.23 g

Given, $K_a=1.77\times {10}^{-4},pH=3.50,V=400ml=0.4L,\text{molar mass of formic acid}=68.0069gmo{l}^{-1},\left[HCOOH\right]=1M$
Using the Henderson-Hasselbalch equation for weak acid:
$pH=p{K}_a+log\left[\frac{\left[\text{conjugate base}\right]}{\left[\text{acid}\right]}\right]$
$pH=-log{K}_a+log\left[\frac{\left[HCOONa\right]}{\left[HCOOH\right]}\right]$
$3.50=-log(1.77\times {10}^{-4})+log\left[\frac{\left[HCOONa\right]}{1}\right]$
$3.50=4-log1.77+log\left[HCOONa\right]$
$log\left[HCOONa\right]=3.50-4+0.248+$
$log\left[HCOONa\right]=-0.252$
$\left[HCOONa\right]=0.56$
We know,$\text{Molarity}=\frac{\text{weight of sodium formate}}{\text{molar mass}\times \text{volume}}$
$0.56=\frac{\text{weight}}{68.0069\times 0.4}$
$\text{weight of sodium formate}=15.23g$