Question

What amount of sodium propanoate should be added to $1$ liter of an aqueous solution containing $0.02\, mol$ of propanoic acid $\left(K_{ a }=1.0 \times 10^{-5}\right.$ at $\left.25^{\circ} C \right)$ to obtain a buffer solution of $pH \,6 $ ?

• A. 0.1 M
• B. 0.2 M
• C. 0.3 M
• D. 1.3 M

Answer 1

Answer: (B)

Using the expression

$pH =p K_{ a }+\log \frac{[\text { Salt }]}{[\text { Acid }]}$

We get, $6=-\log \left(1.0 \times 10^{-5}\right)+\log \frac{[\text { Salt }]}{[0.02 \,M ]}$

Which gives $6=5+\log \frac{[\text { Salt }]}{[0.02 \,M ]}$

or $\frac{[ Salt ]}{[0.02 M ]}=10$ or $[ {\text{Salt}} ]=0.2 \,M$