GC question

[299678]

what volume of 0.03 M NaOH is needed to titrate 30 ml of 0.1N H3PO4?

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[loveoforganic]

Normality refers to the number of titratable protons. H3PO4 has 3 titrable protons, so one molarity is a third of the normality. Therefore, you have .1/3 M H3PO4.

Can you solve from there?

 

[299678]

That's how I solved the problem but gave me wrong answer. Ok
when I use the equation M1V1 = M2V2, Normality = M x i so 0.1 = M*3, so M = 0.03
so when I plug in the equation,
0.03M x V1 = 30 x 0.03, so V1 = 30mL

However, when I use N1V1 = N2V2,
Normality = molarity on monoprotic, so
Normality of NaOH = 0.03 so plug in
0.03 x V1 = 0.1 x 30, so V1 = 100 mL

the correct answer is 100 mL

any other ideas?

 

[dentalWorks]

I never ever use M1V1 = M2V2 or N1V1 = N2V2 when the number of protons / OH's aren't 1:1....

Here is how you should do this problem:

step 1) determine how many MOLES of H3PO4 you currently have

step 2) determine how many MOLES of NaOH you need
In your first step we established that you will have 0.00297 moles of protons floating around. To titrate that, you will need 0.00297 moles of OH

 

[loveoforganic]

(.033 M H3PO4)(.03 L H3PO4) = .001 mol H3PO4

(.001 mol H3PO4)*(3 mol NaOH / 1 mole H3PO4) = .003 mol NaOH

(.003 mol NaOH) / (.03 M NaOH) = .1 L = 100 mL

 

[299678]

love to see different approaches to solve a problem+++1 . Thank you guys !

 

[UCB05]

I never ever use M1V1 = M2V2 or N1V1 = N2V2 when the number of protons / OH's aren't 1:1....

Normality implies proper ratios for the reaction given. N1V1=N2V2 is fine, since you basically undid, then redid that in the first pic. Rounding error gives the 1ml difference in your answer.