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Topic: Lead(II) nitrate + Sodium Sulfate  (Read 17328 times)

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Offline pimcub2011

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Lead(II) nitrate + Sodium Sulfate
« on: November 09, 2008, 07:55:00 PM »
A sample of 1.05 g of lead (II) nitrate is mixed with 122 mL of 0.101  M sodium sulfate solution.

I got the chemical reaction and found that the limiting reactant is Pb(NO3)2.

I also found the concentration of SO4. (0.0750 M)

How do I go about finding the concentration of Na and NO3?

Offline Arkcon

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Re: Lead(II) nitrate + Sodium Sulfate
« Reply #1 on: November 09, 2008, 07:59:22 PM »
A sample of 1.05 g of lead (II) nitrate is mixed with 122 mL of 0.101  M sodium sulfate solution.

I got the chemical reaction and found that the limiting reactant is Pb(NO3)2.

I also found the concentration of SO4. (0.0750 M)

How do I go about finding the concentration of Na and NO3?

Well, lets start with the balanced equation.  The numbers before each compound aren't just for balancing -- they tell you how many moles of each compound the reaction consumes/produces.
Hey, I'm not judging.  I just like to shoot straight.  I'm a man of science.

Offline pimcub2011

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Re: Lead(II) nitrate + Sodium Sulfate
« Reply #2 on: November 10, 2008, 04:40:38 PM »
Ok, Pb(NO3)2+Na2SO4=PbSO4+2NaNO3, and I know how to set up a mole ratio, but what am I supposed to do first?

Offline Borek

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Re: Lead(II) nitrate + Sodium Sulfate
« Reply #3 on: November 10, 2008, 06:08:33 PM »
To find out concentrations of Na+ and NO3- you don't even need reaction equation, as they are just spectators. How much you put in that much you have in the solution. Just find out moles and final volume, and use molarity definition.
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