Question

In a gravimetric determination of $P_1$ an aqueous solution of dihydrogen phosphate ion $H_{2}P{O_4}^{-}$ is treated with the mixture of ammonium and magnesium ions to precipitate magnesium ammonium phosphate, $Mg\left(N{H}_4\right)P{O}_4.6H_{2}O$. This is heated and decomposed to magnesium pyrophosphate, $M{g}_2{P}_2{O}_7$. which is weighted. A solution of $H_{2}P{O_4}^{-}$ yeilded 1.054 g of $M{g}_2{P}_2{O}_7.$ What weight of $Na{H}_{2}P{O}_4$ what present originally?

• A. $1.14g$
• B. $1.62g$
• C. $2.34g$
• D. $1.33g$

Answer 1

Answer: (A)

$1.14g$

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

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

$1.054g$

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

$\frac{2\times wt.gM{g}_2{P}_2{O}_7}{molmass}=\frac{1\times wtofP/Na{H}_{2}P{O}_4}{mol.mass}$

$\begin{array}{c}\frac{2\times 1.054}{222}=1\times \frac{x}{120}\\ x=\frac{2\times 120\times 1.054}{222}\end{array}$

$=1.139\simeq 1.14$