Question

In a gravimetric determination of \[{{{P}}_{{1}}}\] an aqueous solution of dihydrogen phosphate ion ${{{H}}_{{2}}}{{PO}}_{{4}}^{{ - }}$ is treated with the mixture of ammonium and magnesium ions to precipitate magnesium ammonium phosphate, ${{Mg(N}}{{{H}}_{{4}}}{{)P}}{{{O}}_{{4}}}{{.6}}{{{H}}_{{2}}}{{O}}$. This is heated and decomposed to magnesium pyrophosphate, ${{M}}{{{g}}_{{2}}}{{{P}}_{{2}}}{{{O}}_{{7}}}$ which is weighted. A solution of ${{{H}}_{{2}}}{{PO}}_{{4}}^{{ - }}$ yielded 1.054 g of ${{M}}{{{g}}_{{2}}}{{{P}}_{{2}}}{{{O}}_{{7}}}$. What weight of ${{Na}}{{{H}}_{{2}}}{{P}}{{{O}}_{{4}}}$ what present originally?
A.1.14 g
B.1.62 g
C.2.34 g
D.1.33 g

Answer 1

Hint:The method used to determine the concentration or mass of a substance by measuring the change in mass is Gravitational analysis. The amount of an analyte can be determined from the mass of the compound and the change in the concentration or mass. ${{Na}}{{{H}}_{{2}}}{{P}}{{{O}}_{{4}}}$ is also known as monosodium phosphate which is used to detect magnesium ions from salts, used as a food additive.

Complete step by step answer:
The given reactions are:
 ${{{H}}_{{2}}}{{PO}}_{{4}}^{{ - }} \to {{Mg(N}}{{{H}}_{{4}}}{{)P}}{{{O}}_{{4}}}{{.6}}{{{H}}_{{2}}}{{O}}\xrightarrow[{{{heat}}}]{}{{M}}{{{g}}_{{2}}}{{{P}}_{{2}}}{{{O}}_{{7}}}$
${{M}}{{{g}}_{{2}}}{{{P}}_{{2}}}{{{O}}_{{7}}} \to {{Na}}{{{H}}_{{2}}}{{P}}{{{O}}_{{4}}}$
The first reaction shows the formation of magnesium pyrophosphate on heating, whereas in the second reaction sodium dihydrogen phosphate is obtained from magnesium pyrophosphate.
Here the weight of ${{M}}{{{g}}_{{2}}}{{{P}}_{{2}}}{{{O}}_{{7}}}$ (Magnesium pyrophosphate) $ = $1.054 g
Applying, POAC (Principle of Atom Conservation) on ${{M}}{{{g}}_{{2}}}{{{P}}_{{2}}}{{{O}}_{{7}}}$ and ${{Na}}{{{H}}_{{2}}}{{P}}{{{O}}_{{4}}}$ on P atom (since P atom is conserved).
1 mole of ${{Na}}{{{H}}_{{2}}}{{P}}{{{O}}_{{4}}}$ contains 1 mole of P and 1 mole of ${{M}}{{{g}}_{{2}}}{{{P}}_{{2}}}{{{O}}_{{7}}}$ contains 2 moles of P.
$ \Rightarrow {{1}}\,{{ \times }}\,\dfrac{{\left( {{{wt}}{{.}}\,{{of}}\,\,{{Na}}{{{H}}_{{2}}}{{P}}{{{O}}_{{4}}}} \right)}}{{\left( {{{mol}}{{.}}\,{{wt}}\,{{of}}\,{{Na}}{{{H}}_{{2}}}{{P}}{{{O}}_{{4}}}} \right)}}\,{{ = }}\,{{2}}\,{{ \times }}\,\dfrac{{\left( {{{wt}}{{.}}\,{{of}}\,\,{{M}}{{{g}}_{{2}}}{{{P}}_{{2}}}{{{O}}_{{7}}}} \right)}}{{\left( {{{mol}}{{.}}\,{{wt}}\,{{of}}\,{{M}}{{{g}}_{{2}}}{{{P}}_{{2}}}{{{O}}_{{7}}}\,} \right)}}$
$ \Rightarrow \dfrac{{\left( {{{wt}}{{.}}\,{{of}}\,\,{{Na}}{{{H}}_{{2}}}{{P}}{{{O}}_{{4}}}} \right)}}{{{{120}}}}{{ = 2 \times }}\dfrac{{{{1}}{{.054}}}}{{{{222}}}}$
Here, 120 represents the molecular weight of sodium dihydrogen phosphate and 222 represents the molecular weight of magnesium pyrophosphate.
(Now we will put the values given like the weight of ${{M}}{{{g}}_{{2}}}{{{P}}_{{2}}}{{{O}}_{{7}}}$ is 1.054 and also we will calculate the molecular masses of ${{Na}}{{{H}}_{{2}}}{{P}}{{{O}}_{{4}}}$ and ${{M}}{{{g}}_{{2}}}{{{P}}_{{2}}}{{{O}}_{{7}}}$) .
 $ \Rightarrow {{wt}}{{.}}\,{{of}}\,\,{{Na}}{{{H}}_{{2}}}{{P}}{{{O}}_{{4}}}{{ = }}\dfrac{{{{2 \times 1}}{{.054 \times 120}}}}{{{{222}}}}$
Weight of ${{Na}}{{{H}}_{{2}}}{{P}}{{{O}}_{{4}}}$ is $1.139 \ approx 1.14$
(After calculation we get the value for the weight of ${{Na}}{{{H}}_{{2}}}{{P}}{{{O}}_{{4}}}$ is 1.139 g which is approximately 1.14 g)
Hence, the correct option is (A).


Note:
The POAC (Principle of Atom Conservation) states that the total number of atoms of the reactants should be equal to the total number of atoms of the products.
The POAC comes from the Law of Conservation of Mass.
Mass of atoms of an element in reactant = Mass of atoms of an element in the product.
The number of atoms of element of the reactants = Number of atoms of element of the product.
 Moles of atoms of element of the reactants = Moles of atoms of element of the product.