## Find the molecular mass of acid in acid base neutralization

1. The problem statement, all variables and given/known data
1.5000 g of diprotic weak acid H2A was dissolved in 100.00 mL volumetric flask.
25.00 mL aliqouts of this solution was titrated with a monoprotic strong base NaOH (0.08000 M). The titre volume of NaOH was 40.00 mL. Calculate the molecular weight of H2A.

2. Relevant equations
N/A

3. The attempt at a solution
1 mole of H2A contains 2 moles of H+
1 mole of NaOH contains 1 mole of OH-

2CaVa = CbVb
2 (Ca)(0.02500 L) = (0.0800 M)(0.04000L)
Ca = 0.0640 M

#moles of H2A = CaVa
= (0.0640 M)(0.10000 L)
= 0.00640 mol

molecular mass of H2A = mass/# moles
= 1.5000 g / 0.00640 mol
= 234.375 g/mol

My answer is 234.375 but something seems odd about it. Can anyone tell me if I missed something or did something wrong? Thanks!

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 Quote by jessica.so 2CaVa = CbVb
Question is - is that 2 in the correct place?

--
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 I've always been taught that the exponent in CaVa = CbVb equations is the number of moles of H[SUP]+[SUP] ions or OH[SUP]-[SUP] ions which is why I put the 2 infront of the CaVa. Another time I attempted this question but I balanced the neutralization first. H2A + 2 NaOH --> 2 H2O + Na2A CaVa = 2 CbVb (Ca)(0.02500 L) = 2(0.0800 M)(0.04000L) Ca = 0.256 M #moles of H2A = CaVa = (0.256 M)(0.10000 L) = 0.0256 mol molecular mass of H2A = mass/# moles = 1.5000 g / 0.0256 mol = 58.59 g/mol This answer seemed wrong too somehow. One of the questions that came up was that the acid we are neutralizing is a weak acid. 25mL of this weak acid was neutralized by 40mL of strong base. Shouldn't a weak acid take less base to neutralize?

## Find the molecular mass of acid in acid base neutralization

Exponent? I guess you mean stoichiometric coefficient. But you were right the first time, somehow I got it reversed.

Now, the problem is - 0.00640 moles - is it whole 1.5 g sample?

Strength of the acid has nothing to do with amount of base that it needs to be neutralized. It is all in stoichiometry.

 So sorry! Yes, I did mean the coefficient. So...I'm using the first attempt... 2CaVa = CbVb 2(Ca)(0.02500 L) = (0.0800 M)(0.04000L) Ca = 0.0640 M #moles of H2A = CaVa = (0.0640 M)(0.10000 L) = 0.00640 mol molecular mass of H2A = mass/# moles = 1.5000 g / 0.00640 mol = 234.375 g/mol So I found the molarity of the H2A that was neutralized. Because a 25.00 mL sample was taken from the 100.00 mL solution, the molarity stays the same. 1.5000 g made the 100.00 mL solution of H2A so I think the 0.00640 mol is the whole 1.5 g sample.
 Admin have you titrated 100 mL aliquot?
 No, I have not titrated 100mL