Concentration problem??

Question

If 87g of K2SO4 (molar mass 174g) is dissolved in enough water to make 250 mL of solution, what are the concentrations of the potassium and the sulfate ions?

(A) [K+] = 0.020M [SO4 (-2)] = 0.020 M
(B) [K+] =1.0 M [SO4 (-2)] = 2.0 M
(C) [K+] = 2.0 M [SO4 (-2)] = 1.0 M
(D) [K+] = 2.0 M [SO4 (-2)] = 2.0 M
(E) [K+] = 4.0 M [SO4 (-2)] = 2.0 M

I found the mols of K2SO4 but what do I do after this? Please, can anyone help?? thank you in advance

Answer 1

Write out the stoich equation

K2SO4 -----> 2 K(+) + SO4 (2-)

You have 87g * 1mol/174g = 0.5 moles of K2SO4
So after dissolution you will have
2* 0.5 moles K(+)
and 0.5 moles of SO4(2-)

All in 250 mL which is 250mL * 1 L/1000mL = 0.250 L
Remember that M is short for moles/litre
So you have 2* 0.5 moles / 0.250 L = 4M K+

Answer 2

one mole of K2SO4 produce 2 moles of K+ ion and 1 mole of SO4 ion. Either choice C or E.
Now number mole of K2SO4 in 250 mL of solution
[87 g /174(g/mole) ] /250 mL * 1000mL / L = 2M

Therefore choice E have to be the one.

Answer 3

dear friend

the formula for the dissociation of K2SO4----> 2K+ + SO4-

according to this formula of law of mass action, one mole of K2SO4 will dissociate into two moles of K+ ions and one mole of SO4 ions so acording to the question above 87g is equivalent to 0.5 mole of K2SO4. Therefore it produces one mole of K+ ions and 0.5 mole of SO4 ions.

thank you

Answer 4

If you know the moles of K2SO4, then each mole of this will ionize to produce 2 moles of K+. So the concentration of K+ will be twice the original concentration of K2SO4