Question

$100\text{mL}$ of a buffer solution contains $0.1\text{M}$ each of weak acid $HA$ and salt $NaA$. How many grams of $NaOH$ should be added to the buffer so that its pH becomes 6? $(K_a\text{of HA}={10}^{-5})$

• A. $\text{0.328 g}$
• B. $\text{0.458 g}$
• C. $\text{4.19 g}$
• D. $\text{3.28 g}$

Answer 1

The correct option is A $\text{0.328 g}$

So the given solution will form acidic buffer for which $pH$ is given by
$pH=p{K}_a+log\frac{\left[salt\right]}{\left[acid\right]}$
$6=5+log\frac{\left[salt\right]}{\left[acid\right]}$
$\therefore log\frac{\left[salt\right]}{\left[acid\right]}=1$
$\frac{\left[salt\right]}{\left[acid\right]}=10$
initially 10 mmoles each of salt and acid was present when x mmoles of $NaOH$ is added
$\frac{(10+x)}{(10-x)}=10$
$x=8.18mmoles$
$w_{NaOH}=40\times 8.18\times {10}^{-3}=0.327g$