Question

The $p{K}_a$ value of acetic acid is 4.7447 at ${25}^{o}C.$ In what ratio of acetic acid and sodium acetate concentration is mixed to obtain a buffer with pH=4?
antilog(-0.7447)=0.18

• A. 0.18:1
• B. 1:0.18
• C. 1.8:1
• D. 1:1.8

Answer 1

The correct option is B 1:0.18

Given, $p{K}_a\left(\text{acetic acid}\right)=4.7447\text{at}{25}^{o}C,pH=4$
Using the Henderson-Hasselbalch equation for weak acid:
$pH=p{K}_a+log\frac{\left[salt\right]}{\left[acid\right]}$
$\Rightarrow 4=4.7447+log\frac{\left[\text{sodium acetate}\right]}{\left[\text{acetic acid}\right]}$
$\Rightarrow -0.7447=log\frac{\left[\text{sodium acetate}\right]}{\left[\text{acetic acid}\right]}$
$\frac{\left[\text{sodium acetate}\right]}{\left[\text{acetic acid}\right]}=0.18$
So, by mixing sodium acetate and acetic acid in the ratio 0.18 : 1 mole per $d{m}^3$.
The buffer of $pH=4$ is obtained.