Question

Consider an acidic buffer solution of acetic acid and sodium acetate at ${25}^{o}C$. What should be the ratio of concentration of sodium acetate and acetic acid so that $pH=4$ is obtained ?

Given :$K_a$ value of acetic acid is $1.8\times {10}^{-5}$ $\text{and}{10}^{(-0.75)}=0.18$

• A. $0.18:1$
• B. $1:0.18$
• C. $1:1.81$
• D. None of the above

Answer 1

The correct option is A $0.18:1$

Given, $K_a\left(\text{acetic acid}\right)=1.8\times {10}^{-5}\text{at}{25}^{o}C,p{K}_a=-{\log }_{10}[1.8\times {10}^{-5}]p{K}_a=5-\log 1.8p{K}_a=5-0.25p{K}_a=4.75\text{Given ,}pH=4$
Formula of $pH$ for an acidic buffer solution is :
$pH=p{K}_a+\log \frac{\left[\text{anion of salt}\right]}{\left[\text{acid}\right]}$
$\Rightarrow 4=4.75+\log \frac{\left[\text{sodium acetate}\right]}{\left[\text{acetic acid}\right]}$
$\Rightarrow -0.75=\log \frac{\left[\text{sodium acetate}\right]}{\left[\text{acetic acid}\right]}\frac{\left[\text{sodium acetate}\right]}{\left[\text{acetic acid}\right]}={10}^{(-0.75)}$
$\frac{\left[\text{sodium acetate}\right]}{\left[\text{acetic acid}\right]}=0.18$
So, by mixing sodium acetate and acetic acid in the ratio $0.18:1$.
The buffer of $pH=4$ is obtained.