Question

$100mL$ of a buffer solution contains $1.0M$ 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 $6$ ?
$(K_a$ of $HA=10^{-5})$

• 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 $xM$ of $NaOH$ is added
$p{H}_{new}=5+\log \left(\frac{0.1+x}{0.1-x}\right)$
$6-5=\log \left(\frac{0.1+x}{0.1-x}\right)$
$\left(\frac{0.1+x}{0.1-x}\right)=\mathrm{Antilog}(1)=10$
Solve for $x$ :
$x=0.082M=\frac{0.082}{1000}\times 100$
$=0.0082mol(100mL{)}^{-1}$
$=0.0082\times 40g(100mL{)}^{-1}$
$=0.328g$