Question

The ionic compound, (NH4)2SO4, will fully dissolve in water. If 19.90 g is dissolved in 482.0 mL water, find the following ion concentrations.

The concentration of the NH4+ is:

The concentration of the SO42- is:

Answer 1

Molar mass of (NH4)2SO4 = 2*MM(N) + 8*MM(H) + 1*MM(S) + 4*MM(O)

= 2*14.01 + 8*1.008 + 1*32.07 + 4*16.0

= 132.154 g/mol

mass of (NH4)2SO4 = 19.90 g

we have below equation to be used:

number of mol of (NH4)2SO4,

n = mass of (NH4)2SO4/molar mass of (NH4)2SO4

=(19.9 g)/(132.154 g/mol)

= 0.1506 mol

volume , V = 482.0 mL

= 0.482 L

we have below equation to be used:

Molarity,

M = number of mol / volume in L

= 0.1506/0.482

= 0.3124 M

This is molarity of (NH4)2SO4

So,

[NH4+] = 2*[(NH4)2SO4] = 2*0.3124 M = 0.6248 M

[SO42-] = [(NH4)2SO4] = 0.3124 M

[NH4+] = 0.6248 M

[SO42-] = 0.3124 M