How many ml of 0.1M NaOH is added to 60ml of 0.15M $H_{3}P{O}_4$  $(p{K}_{a_1},p{K}_{a_2}andp{K}_{a_3}for{H}_{3}P{O}_{4}are3,8and13)$ such that pH of resulting buffer  solution would be 8.3. (log2 = 0.3)

$$\begin{gathered} pH=p{K}_{a_2}+\log \frac{\left[HP{O}_4^{2-}\right]}{\left[H_{2}P{O}_4^{-}\right]} \\ 8.3=8+\log \frac{\left[HP{O}_4^{2-}\right]}{\left[H_{2}P{O}_4^{-}\right]} \\ \frac{\left[HP{O}_4^{2-}\right]}{\left[H_{2}P{O}_4^{-}\right]}=2 \\ \begin{array}{l}{H}_{3}P{O}_4+NaOH\to Na{H}_{2}P{O}_4\\ 9 x 0\\ 0 x-9 9\end{array} \end{gathered}$$

$\begin{array}{l}Na{H}_{2}P{O}_4+NaOH\to N{a}_{2}HP{O}_4\\ 9 x-9 0\\ 18-x 0 x-9\end{array}$

$$\begin{gathered} \frac{x-9}{18-x}=2 \\ x=15 \\ 15=V\times 0.1,V=150ml \end{gathered}$$