In a titration of a diprotic acid, 0.120 g of an unknown solid diprotic acid H2X

Question

In a titration of a diprotic acid, 0.120 g of an unknown solid diprotic acid H2X is dissolved in 100.0 mL water and titrated to the first equivalence point with 15.63 mL of 0.100 M NaOH.  What is the molar mass of the acid?  (the first equivalence point is when one H+ per molecule of H2X has completely reacted with the NaOH)

How many additional mL of the NaOH solution will be required to reach the second equivalence point? (the second equivalence point is when two H+ per molecule of H2X has completely reacted with the NaOH).  I believe the answer is just an additional 15.63, but I'm not sure.  It could maybe be 15.63 * 2 = 31.26

Explanation / Answer

First equivalence point:

H2X + NaOH => NaHX + H2O

Moles of H2X = moles of NaOH = volume x concentration of NaOH

= 15.63/1000 x 0.100 = 0.001563 mol


Molar mass = mass/moles of HX

= 0.120/0.001563

= 76.8 g/mol


Second equivalence point:

NaHX + NaOH => Na2X + H2O

Moles of NaOH = moles of NaHX produced at first equivalence point = moles of H2X = 0.001563 mol


Volume of additional NaOH = moles/concentration of NaOH

= 0.001563/0.100

= 0.01563L = 15.63 mL


(Note: total volume of NaOH added = 2 x 15.63 = 31.26 mL, but additional volume of NaOH for second equivalence point = 15.63 mL since the other 15.63 mL of NaOH was already added for the first equivalence point)

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