Question

In a gravimetric determination of P, an aqueous solution of dihydrogen phosphate ion $\left[H_{2}P{O}_4\right]$ is treated with a mixture of ammonium and magnesium ions to precipitate magnesium ammonium phosphate. $\left[Mg\right(N{H}_4)P{O}_4.6H_{2}O].$ This is heated and decomposed to magnesium pyrophosphate $\left[M{g}_2{P}_2{O}_7\right],$ which is weighed. A solution of $H_2{P}_4$ yielded $1.054g$ of $M{g}_2{P}_2{O}_7.$ What weight of $Na{H}_{2}P{O}_4$ was present originally?

Answer 1

According to question ,

$H_{2}P{O}^{-}\to Mg\left(N{H}_4\right)P{O}_4.6H_{2}O\to M{g}_2{P}_2{O}_7\to Na{H}_{2}P{O}_4$

$M{g}_2{P}_2{O}_7\to Na{H}_{2}P{O}_4$

1.054g

On applying POAC on $M{g}_2{P}_2{O}_7$ and $Na{H}_{2}P{O}_4$ on P ,

$\frac{2\times wt.M{g}_2{P}_2{O}_7}{mol-mass}=\frac{1\times wt.Na{H}_{2}P{O}_4}{mol-mass}$

$\frac{2\times 1.054}{222}=1\times \frac{x}{120}$

$x=\frac{2\times 120\times 1.054}{222}=1.1395$

weight of $Na{H}_{2}P{O}_4$ was present originally is 1.1395 g.