[Hemidol]

Assume that 21.9675 g of (NH4)2SO4 is dissolved in enough water to make 500.0 mL of solution.
(a) What is the molarity of (NH4)2SO4?
(b) What is the molarity of the ammonium cation?
(c) What is the molarity of the sulfate anion?
The molar mass of (NH4)2SO4 is 132.140 g/mol.

Here is my work thus far:

(a) 21.9675g (NH4)2SO4 * (1 mole / 132.140g/mol) = 0.166226524 mole
(0.166226524 mole / 500 mL) * (1000 mL / 1 L) = 0.332488 M

Question relating to b-c, for finding the molarity of a specific cation / anion can I just find the molar mass of that specific cation / anion and do what I have just done here?

Thanks for reading.

[MrTeo]

Yes, because a salt is a strong electrolyte so in a watery solution we will find it fully dissociated in NH₄⁺ and SO₄^2-. Just pay attention to the fact that every mole of (NH₄)₂SO₄ gives use twice the number of moles of cation.

Being more precise you could evaluate the hydrolysis of NH₄⁺ which lowers the concentration of the cation and forms NH₃ and H₃O⁺, but as they don't give you the K_b you don't need to consider this fact.

[Hemidol]

Yes, because a salt is a strong electrolyte so in a watery solution we will find it fully dissociated in NH₄⁺ and SO₄^2-. Just pay attention to the fact that every mole of (NH₄)₂SO₄ gives use twice the number of moles of cation.

Being more precise you could evaluate the hydrolysis of NH₄⁺ which lowers the concentration of the cation and forms NH₃ and H₃O⁺, but as they don't give you the K_b you don't need to consider this fact.

Snack for reply!

Is that because in (NH₄)₂SO₄ there is two mole NH₄⁺ as indicated by the bolded two below?

(NH₄)₂SO₄

Thus I can write:

(b) NH₄⁺ has a molar mass of 15.018g

21.9675g (NH4)2SO4 * (1 mole / 132.140g/mol) * (2 moles NH4 / 1 mole (NH4)2SO4) * (15.018g / 1 mole NH4) = 4.993308839 / 500 mL * (1000 mL / 1 L) = answer

I have a feeling I did that incorrectly.

[plankk]

You don't need to use molar mass of cation or anion. You know the molarity of salt. You also know (for the formula (NH₄)₂SO₄) that for one mole of you salt you receive (after dissolution) two moles of cation and one of anion. On the basis of that easy ratio calculate molarity of both ions.

[Hemidol]

You don't need to use molar mass of cation or anion. You know the molarity of salt. You also know (for the formula (NH₄)₂SO₄) that for one mole of you salt you receive (after dissolution) two moles of cation and one of anion. On the basis of that easy ratio calculate molarity of both ions.

Ah I see. So then I can write:


21.9675g (NH4)2SO4 * (1 mole / 132.140g/mol) * (2 moles NH4 / 1 mole (NH4)2SO4) = (0.33248827 / 500 mL) * (1000 mL / 1 L) = 0.66497654 M NH4+

And then for S:


21.9675g (NH4)2SO4 * (1 mole / 132.140g/mol) * (1 moles SO4 / 1 mole (NH4)2SO4) = (0.33248827 / 500 mL) * (1000 mL / 1 L) = 0.166226524 mole
(0.166226524 mole / 500 mL) * (1000 mL / 1 L) = 0.332488 M SO4

Which doesn't make sense to me because how can SO4 anion have the same Molarity as the entire compound? 

P.S. Sorry for the double post.

[Borek]

Which doesn't make sense to me because how can SO4 anion have the same Molarity as the entire compound?

Thats OK - take a look at formula, it has stoichiometric coeffcient of 1.

(NH₄)₂SO₄ -> (NH₄)₂(SO₄)₁