Question

Determine the concentrations of K2SO4, K , and SO42– in a solution prepared by dissolving 1.80 × 10–4 g K2SO4 in 1.50 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

We are given with
1.80 × 10–4 g K2SO4
1.50 L of water

The concentration of K2SO4 is
1.80 × 10–4 g K2SO4 x (1 mole K2SO4 / 174 g K2SO4) x (1 / 1.50 L)
= 6.897 x 10-7 M

Since K2SO4 dissociates into
K2SO4 <=> 2K+  +  SO4-2

The concentration of K+ ions is
2 x 6.897 x 10-7 M = 1.379 x 10-6 M

The concentration of SO4-2 is the same as that of K2SO4
1.379 x 10-6 M

Answer 2

Explanation:

$Molarity=\frac{n}{V(L)}$

n = moles of compound

V =Volume of the solution in Liters

Moles of potassium sulfate ,n= $\frac{1.80\times 10^{-4} g}{174 g/mol}$

Volume of the solution = 1.50 L

Molarity of the solution = $M=\frac{1.80\times 10^{-4} g}{174 g/mol\times 1.50 L}$

Molarity of potassium sulfate solution:

$[K_2SO_4]=M=6.8965\times 10^{-7} mol/L$

1 mole of potassium sulfate gives 2 moles of  potassium ions and 1 mole of sulfate ions.

Molarity of potassium ions in the solution:

$[K^+]=2\times M=2\times 6.8965\times 10^{-7} mol/L=1.3793\times 10^{-6} mol/L$

Molarity of sulfate ions in the solution:

$[SO_4^{2-}]=1\times 6.8965\times 10^{-7} mol/L=6.8965\times 10^{-7} mol/L$

Concentration in ppm:

ppm = Milligram of compo present in 1 liter solution.

ppm = Molarity × Molar mass of compound × 1000

Concentration of potassium ion in ppm:

$1.3793\times 10^{-6} mol/L\times 40 g/mol\times 1000$

$[K^+]=0.05517 ppm$

Concentration of sulfate ions in ppm:

$6.8965\times 10^{-7} mol/L\times 96 g/mol\times 1000$

$[SO_4^{2-}]=0.06620 ppm$