Volume of NaOH needed to neutralize H3PO4

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Homework Statement



what volume of 0.350 mol/L NaOH(aq) is required to completely neutralize 54.5 mL of 1.31 mol/L H3PO4(aq)?

Homework Equations



c1v1 = c2v2

The Attempt at a Solution



You would just solve for v2 but I am not getting the right answer for some reason.
 
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No, you follow my earlier advice - write reaction equation & read the page linked to.
 
so, 3c1v1 = c2v2

then v1 = c2v2/3c1 ?

This would give me 6.80x10^-2 L.
 
Borek said:
68 mL is not a correct volume of NaOH.

Yes, I learned the hard way.

I wrote the balanced equation of H3PO4 + 3 NaOH -> 3 H2O + Na3PO4

3CnaohVnaoh = Ch3po4Vh3po4

Is this wrong?
 
Yes, this is wrong. You got it reversed.

How many moles of NaOH needed to react with one mole of H3PO4? What shoudl be multiplied by three - number fo moles of NaOH, or number of moles of H3PO4?

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If you multiply the H3PO4 by 3, you would have a ratio of 1:1 with NaOH:H3PO4.

So you could write c1v1=3c2v2?
 
If 1 is NaOH and 2 is H3PO4 then yes. But you should be more specific, as what you have wrote is ambiguous.

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