Find the molecular mass of acid in acid base neutralization

jessica.so

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^-

2C_aV_a = C_bV_b
2 (C_a)(0.02500 L) = (0.0800 M)(0.04000L)
C_a = 0.0640 M

#moles of H₂A = C_aV_a
= (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!

Borek

Question is - is that 2 in the correct place?

--
www.titrations.info, www.chemistry-quizzes.info, www.ph-meter.info

jessica.so

I've always been taught that the exponent in C_aV_a = C_bV_b 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 C_aV_a.

Another time I attempted this question but I balanced the neutralization first.
H₂A + 2 NaOH --> 2 H₂O + Na₂A

C_aV_a = 2 C_bV_b
(C_a)(0.02500 L) = 2(0.0800 M)(0.04000L)
C_a = 0.256 M

#moles of H₂A = C_aVa
= (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?

Borek

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.

jessica.so

So sorry! Yes, I did mean the coefficient.

So...I'm using the first attempt...

2C_aV_a = C_bV_b
2(C_a)(0.02500 L) = (0.0800 M)(0.04000L)
C_a = 0.0640 M

#moles of H₂A = C_aV_a
= (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.

Borek

have you titrated 100 mL aliquot?

jessica.so

No, I have not titrated 100mL