$9.8\times {10}^{-7}g{H}_{3}P{O}_4$  is dissolved in 100mL of a buffer solution of  $pH=5$ (Provides constant concentration of  $H^{+}$) If the equilibrium concentrations of  $H_{3}P{O}_4,H_{2}P{O}_4^{-},HP{O}_4^{2-}$ and $P{O}_4^{-3}$ are  $C_1,C_2,C_3$ and  $C_4$ respectively. Then  $P^{C_1}+P^{C_2}+P^{C_3}+P^{C_4}=$ _________(Round off to nearest integer)
 $(p{C}_i=-\log C_i;wherei=1,2,3,4)$  Given: For   $H_{3}P{O}_4, K_{a_1}={10}^{-3},K_{a_2}={10}^{-8},K_{a_3}={10}^{-12}$

Since $k{a}_1>>k{a}_{2 }$and $k{a}_3$

using the approximation

 $$\begin{gathered} {\left[H_{3}P{O}_4\right]}_0=\left[H_{3}P{O}_4\right]+\left[H_{2}P{O_4}^{-}\right] \\ {\left[H_{3}P{O}_4\right]}_0=Initial concentration of acid \end{gathered}$$