Question

A solution of Na3PO4 that is 0.2000N as a salt is used to precipitate the Mg as MgNH4PO4 from a sample of dolomite containing 14.01% MgCO3. What volume is theoretically required if 2.0000g of sample is dissolved in water and diluted to 250 mL and an aliquot of 50 mL is used?

Answer 1

The ionic reaction of the titration is:

Mg²⁺ + HPO₄^2- + NH₃ $\rightarrow$ MgNH₄PO₄$\downarrow$ .

Here, one mole of HPO₄^2- ions react with one mole of Mg²⁺ ions:

$n(Mg^{+2}) = n(HPO_4^{2-})$ .

It is evident that the number of the moles of HPO₄^2- and of Na₃PO₄ are the same:

$n(HPO_4^{2-}) = n(Na_3PO_4)$

Therefore, the volume of Na₃PO₄ needed for the titration is:

$V(Na_3PO_4) = \frac{n(PO_4^{3-})}{c} = \frac{n(Mg^{2+})}{0.2000}$ .

The number of the moles of Mg²⁺ equals to the number of the moles of MgCO₃ in the aliquot:

$n(Mg^{2+}) = n(MgCO_3)$ .

The number of the moles of MgCO₃ is its mass divided by its molar mass 84.31 g/mol:

$n(MgCO_3) =\frac{m}{M} = \frac{m}{84.31}$ .

The mass of MgCO₃ in the sample of 2.0000 g is:

$m(MgCO_3, sample) = 2.0000\cdot 0.1401 = 0.2802$ g.

The mass of MgCO₃ in 50 mL aliquot of the solution is 250/50=5 times less than in 250 mL solution:

$m(MgCO_3, aliquot) = 0.2802/5 = 0.05604$ g.

Therefore, the number of the moles of MgCO₃ in the aliquot is:

$n(MgCO_3, aliquot) = \frac{0.05604}{84.31} = 0.0006647$ mol.

Finally, the theoretical volume of Na₃PO₄ is:

$V(Na_3PO_4) = \frac{0.0006647}{0.2000} = 0.003323$ L or 3.32 mL.

Answer: 3.32 mL of Na₃PO₄ solution that is 0.2000N is theoretically required if 2.0000g of sampleof dolomite containing 14.01% MgCO₃ is dissolved in water and diluted to 250 mL and an aliquot of 50 mL is used.