# chemistry

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Posted by Jenni on Monday, December 13, 2010 at 3:32pm.

Determine the molarity of nitrate ions in a solution prepared by mixing 25.0 mL of 0.50 M Fe(NO3)3 and 35.0 mL of 1.00 M Mg(NO3)2 Please explain fully do not know where to start?

chemistry - DrBob222, Monday, December 13, 2010 at 4:33pm

moles Fe(NO3)3 = M x L = ??
moles NO3^- will be 3x that.

moles Mg(NO3)2 = M x L = ??
moles NO3^- will be 2x that.

Then M NO3^- = total moles NO3^-/total volume in L.

chemistry - Jenni..can you look again i'm still not getting it thx, Monday, December 13, 2010 at 6:24pm
(0.50)(25.0) = 12.5

(1.00)(35.0) = 35

47.5/60 = 0.79M is my answer but my book says

chemistry - DrBob222, Monday, December 13, 2010 at 9:27pm
You didn't follow any of my instructions. I have typed my bolded instructions below each of your lines.
0.50)(25.0) = 12.5

moles Fe(NO3)3 = M x L = ??
moles NO3^- will be 3x that.
M x L = 0.50M x 0.025 L(not 25 mL) = 0.0125 moles Fe(NO3)3
Then nitrate is 3x that or 0.0125*3 = 0.03750 moles nitrate ion

(1.00)(35.0) = 35
moles = M x L = 1.00M x 0.0350 L (not mL) = 0.0350 moles Mg(NO3)2.
Then nitrate is twice that or
2*0.0350 = 0.0700 M in nitrate.

47.5/60 = 0.79M is my answer but my book says
M nitrate = total moles/total volume in L. (0.0375+0.0700)/0.06L = 1.79 M in nitrate.

So it is

chemistry - jenni, Tuesday, December 14, 2010 at 3:43pm
so I times 0.0125 by 3 b/c of the 3 at the end of the bracket?

and I times 0.0350 by 2 b/c of the 2 at the end of the bracket?

but why because when i'm doing my molar mass i do include that last # in the molar mass so why do we times it by that #?

• chemistry -

And you are correct. When calculating molar mass Fe(NO3)3 it is atomic mass Fe + 3*mass NO3^-. And for Mg(NO3)2 it is atomic mass Mg + 2*mass NO3^-. I don't see that we used the molar mass of either of these substances in this problem.
The problem asks for molarity of the nitrate ion. So you determine the moles Fe(NO3)3 by M x L = 0.0125, then you multiply by 3 because there are 3 nitrate ions in 1 molecule of Fe(NO3)3. The total nitrate from the Fe(NO3)3 source then is 0.0375 moles. Next you convert M x L of Mg(NO3)2 to moles = 1.0 x 0.035 = 0.035 moles Mg(NO3)2 and we multiply that by 2 because we want to know the nitrate and not magnesium nitrate. So 0.0350 x 2 = 0.07 moles nitrate from that source. The total nitrate, then, is 0.0375 moles from the Fe(NO3)3 and 0.07 moles from the Mg(NO3)2 for a total of 0.1075 moles. The total volume is 60 mL so M = moles/L = 0.1075/0.06 = 1.79 M in NO3^-.

• chemistry -

thanks!

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