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

Step-by-step explanation:

`Molarity=(n)/(V(L))`

n = moles of compound

V =Volume of the solution in Liters

Moles of potassium sulfate ,n=
`(1.80* 10^(-4) g)/(174 g/mol)`

Volume of the solution = 1.50 L

Molarity of the solution =
`M=(1.80* 10^(-4) g)/(174 g/mol* 1.50 L)`

Molarity of potassium sulfate solution:

`[K_2SO_4]=M=6.8965* 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* M=2* 6.8965* 10^(-7) mol/L=1.3793* 10^(-6) mol/L`

Molarity of sulfate ions in the solution:

`[SO_4^(2-)]=1* 6.8965* 10^(-7) mol/L=6.8965* 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* 10^(-6) mol/L* 40 g/mol* 1000`

`[K^+]=0.05517 ppm`

Concentration of sulfate ions in ppm:

`6.8965* 10^(-7) mol/L* 96 g/mol* 1000`

`[SO_4^(2-)]=0.06620 ppm`

Answer 2

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