Determine the molarity of sodium cations in solution when 6.21 g of sodium sulflate are mixed with 2.97 g of sodium nitrate in pure water to produce 350.0 mL of solution.......please explain every step. thanks!

Answer 1

moles Na2SO4 = grams/molar mass.
moles Na^+ is twice that.

moles NaNO3 = grams/molar mass.
moles Na^+ is same.

M Na^+ in soln = total moles Na^+/total volume in L.

Answer 2

6.21g/142.042 = 0.0437

2.97g/84.994 = 0.3494

M=moles/L
0.0437+0.3494/350.0= 1.12x10^-03

So I got = 1.12x10^-03
but my book says [Na+]= 0.350M

Where did I go wrong?

Answer 3

6.21g/142.042 = 0.0437
<If you re-read my instructions, you will note that Na^+ in Na2SO4 is TWICE moles Na2SO4; therefore, this should be 0.04372 x 2 = 0.08744

2.97g/84.994 = 0.3494
I suspect this is a typo. It should be 0.03494.

M=moles/L
0.0437+0.3494/350.0= 1.12x10^-03
(0.08744+0.03494/0.350 = 0.34966M which rounds to 0.350M to three s.f. Its a minor point but if you re-read my instructions, they were to use total volume in L (not mL).
So I got = 1.12x10^-03
but my book says [Na+]= 0.350M

Where did I go wrong?

1. You didn't double the Na2SO4.
2. You made a typo or an error in the second part (the NaNO3).
3. You didn't use volume in L.
Considering that you had specific instructions to work the problem, all of this can be laid to "not paying close attention to details." I hope this helps.

Answer 4

Thank you it has!