Question

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</b>

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

M=moles/L
0.0437+0.3494/350.0= 1.12x10^-03
<b>(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).</b>
So I got = 1.12x10^-03
but my book says [Na+]= 0.350M

Where did I go wrong?
<b>
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.</b><i>
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. </i>

Answer 4

Thank you it has!