Question

$100 \,mL$ of a buffer solution contains $1.0\,M$ each of weak acid $HA$ and salt $NaA$. How many gram of $NaOH$ should be added to the buffer so that it $pH$ will be $6$ ?
$\left(K_a\right.$ of $\left.HA =10^{-5}\right)$

• A. $0.328$
• B. 0458
• C. $4.19$
• D. None

Answer 1

Answer: (A)

For acidic buffer, $pH = p K_a+\log \frac{0.1}{0.1}$
$pH = p K_a=-\log \left(10^{-5}\right)=5$
Rule: ABA (In acidic buffer (A), on addition of $S_B( B )$, the concentration of $W_A( A )$ decreases and that of salt increases)
Let $x M$ of $NaOH$ is added
$pH _{ new }=5+\log \left(\frac{0.1+x}{0.1-x}\right)$
$6-5=\log \left(\frac{0.1+x}{0.1-x}\right)$
$\left(\frac{0.1+x}{0.1-x}\right)=\operatorname{Antilog}(1)=10$
Solve for $x$ :
$x=0.082\, M =\frac{0.082}{1000} \times 100$
$=0.0082 \, mol (100 \, mL )^{-1}$
$=0.0082 \times 40 \, g (100\, mL )^{-1}$
$=0.328 \,g$