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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?

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  • $\begingroup$ The reason is in the words "...to reach the first endpoint". What is the meaning of these? $\endgroup$ – Ivan Neretin Sep 29 '15 at 5:34
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Your error comes from your misunderstanding of what the equivalent points indicate, or as Ivan mystically said:

The reason is in the words "...to reach the first endpoint".


An n-protic acid, has exactly n equivalent points. This means for your titration, that you need one base equivalent to reach the first equivalent point (EP) and two base equivalents to reach the second EP. As you are given the volume that is needed to reach the first EP, everything is quite easy, as the ratio of acid to base is exactly 1:1.

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).

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