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.

[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

for step 7 table c on the lab "newtons law of action"   the average time it took the car to travel .25 meters   was 1.92 seconds.

the average time it took the car to travel 0.50 meters was 2.61 seconds

hope this : )