Question

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

• A. 4.19
• B. 0.458
• C. 0.328
• D. None

Answer 1

Answer: (C)

For acidic buffer, $pH=p{K}_a+\log \frac{0.1}{0.1}$

$pH=pKa=-\log \left(10^{-5}\right)=5$

Rule: ABA (In acidie 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.

$p{H}_{\text{new }}=5+\log \left(\frac{0.1+x}{0.1-x}\right)$

$6-5=\log \left(\frac{0.1+x}{0.1-x}\right)$

$\left(\frac{0.1+x}{0.1-x}\right)=$ Anti $\log (1)=10$

Solve for $x$ :

$x=0.082M=\frac{0.082}{1000}\times 100$

$=0.0082mol(100mL{)}^{-1}$

$=0.0082\times 40g(100mL{)}^{-1}=0.328g$