What is the concentration of SO42- in mg/L (ppm) for a solution that is 0.00385 M of K2SO4? Do NOT use scientific notation in your answer.

Answer 1

Hello, Akshay,

0.00385 = 3.85 x 10^-3 M. M is moles/liter.

The chemical equation for dissolution of K2SO4 in water is:

K2SO4 + H2O = SO4^2- + 2K⁺

We only need this to determine how many SO4 are provided by each K2SO4. Obviously, in this case there is only one SO4 per K2SO4. This means that for every mole of K2SO4 we get a mole of SO4. 

Specifically, we are given a 0.00385 M solution of K2SO4. M simply means moles per liter, so state it as 0.00385 moles of K2SO4 per liter. 

For convenience, assume you have 1 liter. That means we have 0.00385 moles of K2SO4 that is dissolved to produce the same number of moles of SO4^2- . (1 mole per mole, as per the balanced equation). 

Next, calculate the mass of the resulting SO4^2- :

The molar mass of SO4^2- is  96.1 g/mole. The mass is therefore:

0.00385 moles SO4^2- * 96.1 g/mole = 0.370 g SO4^2- (with 3 sig figs).

The answer is requested in PPM. In this case that is mg/L.

Since we conveniently assumed that we have 1 Liter to start with, we have:

·     0.370 g SO4^2- per liter, or

·     370 mg/liter, after converting the g to mg by multiplying by 1,000.

·       [Note:  The PPM should be based on mass, but since water’s density is 1 gram / 1 ml , the ppm value is the same in this case as the per volume number.]

I hope this helps,

Bob

Answer 2

K₂SO₄ dissociates to 2K⁺ + SO₄^2-

For each mole of K₂SO₄, you will generate 2 moles K⁺ and 1 mole of SO₄^2-

0.00385 mol/L K₂SO₄ x 1 mol SO₄^2-/mole K₂SO₄ = 0.00385 mole SO₄^2-/L

molar mass SO₄^2- = 96.06 g/mol

0.00385 mol SO₄^2-/L x 96.06 g/mol x 1000 mg/g = 369.8mg/L = 369 ppm (to 3 sig. figs.)