Question

Determine the concentrations of K2SO4, K , and SO42– in a solution prepared by dissolving 2.72 × 10–4 g K2SO4 in 1.00 L of water. Express all three concentrations in molarity. Additionally, express the concentrations of the ionic species in parts per million (ppm). Note: Determine the formal concentration of SO42–. Ignore any reactions with water.

Answer 1

Having these data:

m K2SO4 = 2.72E-4 g;

MW K2SO4 = 174,259 g/mol;

we can calculate moles of potassium sulfate:

mole K2SO4 = m K2SO4 / MW K2SO4

mole K2SO4 = 2.72E-4 g / 174,259 g/mol

mole K2SO4 = 1.56E-6 mol

The dissolution reaction of potassium sulfate in water goes as follows:

K2SO4 (s) ------> 2 K+ (ac) + SO4 2- (ac)

This means that for 1 mole of K2SO4, there are 2 moles of K+ and 1 mole of SO4 2- in dissolution. So, if we have 1.56E-6 mol of K2SO4, then we will have 3.12E-6 mol of K+ and 1.56E-6 mol of SO4 2-

Then we can calculate the concentration of reactant:

[K2SO4] = mole K2SO4 / L water

[K2SO4] = 1.56E-6 mol / 1.00 L = 1.56E-6 M;

the concentration of potassium cation:

[K+] = mole K+ / L water

[K+] = 3.12E-6 mol / 1.00 L = 3.12E-6 M;

and the concentration of sulfate anion:

[SO4 2-] = mole SO4 2- / L water

[SO4 2-] = 1.56E-6 mol / 1.00 L = 1.56E-6 M

Now, having this in mind:

MW K2SO4 = 174,259 g/mol;

MW K+ = 39.0983 g/mol;

MW SO4 2- = 96.06 g/mol;

and asumming 1 L of water equals to 1 kg of water (d H2O = 1.0000 Kg / L at 4 °C), you can calculate concentrations in parts per million (mg/kg):

[K2SO4] = (1.56E-6 mol/L) * (174,259 g/mol) * (1 L / 1 kg) * (1000 mg / 1 g) = 0.271 ppm

[K+] = (3.12E-6 mol/L) * (39.0983 g/mol) * (1 L / 1 kg) * (1000 mg / 1 g) = 0.122 ppm

[SO4 2-] = (1.56E-6 mol/L) * (96.06 g/mol) * (1 L / 1 kg) * (1000 mg / 1 g) = 0.150 ppm