Question

Determine the concentrations of K2SO4, K+, and SO42− in a solution prepared by dissolving 2.07 × 10−4 g K2SO4 in 2.50 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

Answer:

[K2SO4] = 4,75x10⁻⁷M ; [K⁺] = 9.50x10⁻⁷M ;  [SO4⁻²] = 4,75x10⁻⁷M

SO4⁻²: 0.045ppm  ;  K⁺: 0.037ppm

[SO4⁻²] = 4,70x10⁻⁷ F

Explanation:

Determine the equation

K2SO4 → 2K⁺  +  SO4⁻²

Each mole of potassium sulfate generates two moles of potassium cation and one mole of sulfate anion

Molar mass K2SO4: 174.26 g/m

Moles of K2SO4: grams / molar mass

2.07x10⁻⁴g / 174.26 g/m = 1.18x10⁻⁶ moles

Molarity: Moles of solute in 1 L of solution

1.18x10⁻⁶ moles / 2.5 L = 4,75x10⁻⁷M (K2SO4)

K⁺ : 4,75x10⁻⁷M . 2 = 9.50x10⁻⁷M

SO4⁻²: 4,75x10⁻⁷ M

1 mol of K2SO4 has 2 moles of K and 1 mol of SO4

1.18x10⁻⁶ moles of K2SO4 has 1.18x10⁻⁶ moles of SO4 and 2.37x10⁻⁶ moles of K.

1.18x10⁻⁶ moles of SO4⁻² are 1.13x10⁻⁴ grams (moles. molar mass)

2.37x10⁻⁶ moles of K are 9.26x10⁻⁵ grams (moles. molar mass)

These grams are in 2.5 L of water, so we need μg/mL to get ppm

2.5 L = 2500 mL

1.13x10⁻⁴ grams SO4⁻² are 113.35 μg (1 μg = 1x10⁶ g)

9.26x10⁻⁵ grams K⁺ are 92.6 μg (1 μg = 1x10⁶ g)

113.35 μg /2500 mL = 0.045ppm

92.6 μg /2500 mL = 0.037ppm

Formal concentration of SO4⁻² :

Formality = Number of formula weight of solute / Volume of solution (L)

(1.13x10⁻⁴ grams / 96.06 g ) / 2.5 L = 4,70x10⁻⁷ F