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
