Question

How many moles of sodium propionate $\left(C{H}_{3}C{H}_{2}COONa\right)$ should be added to one litre of an aqueous solution containing 0.02 mole of propionic acid$\left(C{H}_{3}C{H}_{2}COOH\right)$ to obtain a buffer solution of $pH=5.87$ ?
Dissociation constant of propionic acid $K_a$ at ${25}^{\circ }C$ is $1.34\times {10}^{-5}$

Take $log\left(1.34\right)=0.13$

• A. 0.1 mol
• B. 0.2 mol
• C. 0.01 mol
• D. 0.02 mol

Answer 1

The correct option is B 0.2 mol

Here for an acidic buffer ,
Sodium propionate is the salt and propionic acid is the acid.
Since volume of the solution is 1 L.
So,
moles = concentration
$\left[C{H}_{3}C{H}_{2}COOH\right]=0.02M$

$p{K}_a=-log\left(K_a\right)p{K}_a=-log(1.34\times {10}^{-5})p{K}_a=(5-0.13)=4.87$

$pH=p{K}_a+log\frac{\left[salt\right]}{\left[acid\right]}pH=p{K}_a+log\left(\frac{\left[C{H}_{3}C{H}_{2}COONa\right]}{\left[C{H}_{3}C{H}_{2}COOH\right]}\right)5.87=4.87+log\left(\frac{\left[C{H}_{3}C{H}_{2}COONa\right]}{0.02}\right)1=log\left(\frac{\left[C{H}_{3}C{H}_{2}COONa\right]}{0.02}\right){10}^1=\frac{\left[C{H}_{3}C{H}_{2}COONa\right]}{0.02}\left[C{H}_{3}C{H}_{2}COONa\right]=0.2M$
$\therefore$ moles of sodium propionate = concentration= 0.2 mol