Question

What amount of sodium propanoate should be added to one litre of an aqueous solution containing $0.02$ mole of propanoic acid $[Ka=1.34\times {10}^{-5}$ at ${25}^{o}C]$ to obtain a buffer solution of pH $4.75$.

• A. $4.52\times {10}^{-2}$ M
• B. $3.52\times {10}^{-2}$ M
• C. $2.52\times {10}^{-2}$ M
• D. $1.52\times {10}^{-2}$ M

Answer 1

The correct option is D $1.52\times {10}^{-2}$ M

Using Handerson equation

$pH=p{K}_a+\log \left(\frac{Moleofsalt}{Moleofacid}\right)$

Number of mole of propanoic acid $=0.020Mole$

$pH=4.75$

$p{K}_a=-\log 1.34\times {10}^{-5}=4.870$

$\left[Acid\right]$ $=0.02mol/L$

$4.75=4.870+\log \left[\frac{salt}{0.02}\right]$

$\left(0.02\right)$ $=$ $4.75-4.870+log\left[salt\right]$

$\left(0.02\right)$ $=$ $-0.12\left[salt\right]$

$\left(0.02\right)$ $=$ anti log $(-0.12)$ $\left[salt\right]$

$\left[salt\right]$ $=$ $0.8869\times 0.02$

$\left[salt\right]$ $=$ $0.17mol/L$

Moles of sodium propionate $=$ conc. $\times$ volume

$=$ $0.017mole.$

$=$ $1.52\times {10}^{-2}M.$

Hence$,$ Option $\left(D\right)$ is correct$.$