# Calculating molecular weight of an acid based on titration data

If $\pu{2.600 g}$ of a weak diprotic acid were dissolved in $\pu{100 mL}$ of distilled water and a $\pu{10 mL}$ aliquot of this solution required $\mathrm{21.60\ mL}$ of $\mathrm{0.1000\ M\ NaOH}$ to reach the first endpoint, what are the equivalent and formula weights of $\ce{H2A}$?

I assumed to do this: I calculate the grams of acid in solution which is

$$\mathrm{\frac{2.600\ g}{100\ mL} \cdot 10\ mL = 0.2600\ g}$$

After that I would calculate the moles of NaOH used

$$\mathrm{0.1000\ mol/L \cdot 21.60\ mL \cdot 1\ L/1000\ mL = 0.002160\ mol}$$

Then I use this to find the moles of acid present in solution; this is the part I am unsure about. I am assuming: because at the equivalence point half of the acid has been converted to it's conjugate base and so ($\mathrm{2\ mol\ NaOH = 1\ mol\ H_2A}$).

If this is correct then

$$\mathrm{0.002160\ mol\ NaOH \cdot 2\ mol\ H_2A/mol\ NaOH = 0.004320\ mol\ H_2A}$$

Which I then divide the g/mol

$$\mathrm{\frac{0.2600\ g}{0.00432\ mol} = 60.19\ g/mol}$$

Because the acid is diprotic the equivalence weight would be

$$\mathrm{\frac{60.19\ g/mol}{2} = 30.09\ g/mol}$$

I don't think this is the correct answer because I am trying to do this using real titration data I took in lab and the MW seems to be at $\approx 1/2$ the value it should be. If I am not supposed to double the NaOH moles, is there a reason for it?

• The reason is in the words "...to reach the first endpoint". What is the meaning of these? – Ivan Neretin Sep 29 '15 at 5:34

This being said, your first EP can be described by $$n_A = n_B$$ (and not as you did by assuming $n_A = 2\ n_B$, which also would not be the description of the second EP, which is $2\ n_A = n_B$ instead.)
You have to find the molecular mass of your acid, why the upper equation needs to be expanded to $$\frac{m_A}{M_A} = c_B V_B$$
Now you simply have to solve for $M_A$ and put in your numbers (0.26 g, 0.1 mol/L, 21.6 mL).