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 $pH=5$ is obtained ?

Given :$K_a$ value of acetic acid is $1.8\times {10}^{-5}$ ${\log }_{10}1.8=0.25\text{and}{10}^{\left(0.25\right)}=1.8$

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

Answer 1

The correct option is B $1.8:1$

Given, $K_a\left(\text{acetic acid}\right)=1.8\times {10}^{-5}\text{at}{25}^{o}C,p{K}_a=-{\log }_{10}[1.8\times {10}^{-5}]p{K}_a=5-\log 1.8p{K}_a=5-0.25p{K}_a=4.75\text{Given ,}pH=5$
Formula of pH for an acidic buffer solution is :
$pH=p{K}_a+\log \frac{\left[anionofsalt\right]}{\left[acid\right]}$
$\Rightarrow 5=4.75+\log \frac{\left[\text{sodium acetate}\right]}{\left[\text{acetic acid}\right]}$
$\Rightarrow 0.25=\log \frac{\left[\text{sodium acetate}\right]}{\left[\text{acetic acid}\right]}\frac{\left[\text{sodium acetate}\right]}{\left[\text{acetic acid}\right]}={10}^{\left(0.25\right)}$
$\frac{\left[\text{sodium acetate}\right]}{\left[\text{acetic acid}\right]}=1.8$
So, by mixing sodium acetate and acetic acid in the ratio $1.8:1$ mole per $d{m}^3$.
The buffer of $pH=5$ is obtained.