Titration of  $N{a}_{3}P{O}_4$ with  $HCl$ to phenolphthalein end point produces $N{a}_{2}HP{O}_4.$ Subsequent titration to methyl orange end point produces   $Na{H}_{2}P{O}_4.$$\mathbf{\text{'}}\mathbf{x}\mathbf{\text{'}}$  milligrams of  $N{a}_{3}P{O}_4$ and $\mathbf{\text{'}}\mathbf{y}\mathbf{\text{'}}$  milligrams of $N{a}_{2}HP{O}_4$  present in a sample required  $25.00mL,0.12N{H}_{2}S{O}_4$ solution to reach the phenolphthalein end point and an additional $35.00mL$  of the acid to attain the methyl orange end point. Find the value of $(\mathbf{x}+\mathbf{y})$ ____
(At Wts : Na – 23, H – 1, O – 16, P – 31) 
 

 $meqof{H}_{2}S{O}_4$ needed for phenolphthalein end point  $=25\times 0.12=3.0=meqofN{a}_{3}P{O}_4$
Additional $meqof{H}_{2}S{O}_4$  required continuing the titration till methyl orange end point  $=35\times 0.12=4.2$
Out of $4.2meq,3.0meq$  of acid would be consumed for $N{a}_{2}HP{O}_4$  produced from  $N{a}_{3}P{O}_4$ in the previous step of titration.
$\Rightarrow 4.2-3=1.2meqof{H}^{+}$  would be required for neutralizing $N{a}_{2}HP{O}_4$  present in original sample to  $Na{H}_{2}P{O}_4.$
 $$\begin{gathered} \Rightarrow massofN{a}_{3}P{O}_4=3\times {10}^{-3}\times 64=0.492g \\ =492mg \\ \Rightarrow massofN{a}_{3}HP{O}_4=1.2\times {10}^3\times 142=0.1704g \\ =170.4mg \end{gathered}$$