Question

In the laboratory, a student adds 24.0 g of ammonium phosphate to a 125 mL volumetric flask and adds water to the mark on the neck of the flask. Calculate the concentration (in mol/L) of ammonium phosphate, the ammonium ion and the phosphate ion in the solution.

Answer 1

a)

Molar mass of (NH4)3PO4,

MM = 3*MM(N) + 12*MM(H) + 1*MM(P) + 4*MM(O)

= 3*14.01 + 12*1.008 + 1*30.97 + 4*16.0

= 149.096 g/mol

mass((NH4)3PO4)= 24.0 g

number of mol of (NH4)3PO4,

n = mass of (NH4)3PO4/molar mass of (NH4)3PO4

=(24.0 g)/(149.096 g/mol)

= 0.161 mol

volume , V = 125 mL

= 0.125 L

Molarity,

M = number of mol / volume in L

= 0.161/0.125

= 1.288 M

This is concentration of (NH4)3PO4

So,

[(NH4)3PO4] = 1.288 M

[NH4+] = 3*[(NH4)3PO4] = 3*1.288 M = 3.864 M

[PO43-] = 1*[(NH4)3PO4] = 1*1.288 M = 1.288 M