100 mL of a buffer solution contains 0.1 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?
(K_a of HA = 10^-5)

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

$\mathrm{pH}={\mathrm{pK}}_{\mathrm{a}}=-\log \left({10}^{-5}\right)=5$

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.

$\begin{array}{l}{\mathrm{pH}}_{\text{new }}=5+\log \left(\frac{0.1+\mathrm{x}}{0.1-\mathrm{x}}\right)\\ 6-5=\log \left(\frac{0.1+\mathrm{x}}{0.1-\mathrm{x}}\right)\\ \left(\frac{0.1+\mathrm{x}}{0.1-\mathrm{x}}\right)=\text{ Antilog }\left(1\right)=10\end{array}$

Solve for x:

$\begin{array}{l}\mathrm{x}=0.082\mathrm{M}=\frac{0.082}{1000}\times 100\\ =0.0082\mathrm{mol}(100\mathrm{mL}{)}^{-1}\\ =0.0082\times 40\mathrm{g}(100\mathrm{mL}{)}^{-1}\\ =0.328\mathrm{g}\end{array}$