Question

A student required 26.42 mL of 0.1013 M NaOH solution to titrate 0.154 grams of a solid triprotic acid to the phenolphthalein end point. Assuming she did her calcilations correctly, what did she report as the molar mass of her acid? (Please show all work as clear as possible) (refer to the balanced molecular equation)

Answer 1

The general reaction of triprotic acid and NaOH is as follows:

3 NaOH + H3A = 3 H2O + 3 NaA

First calculate the moles of NaOH as follows:

Number of moles = molarity * volume in L

= 0.1013 * 26.42 /1000

= 0.00268 moles NaOH

Now calculate the number of moles of triprotic acid as follows:

0.00268 moles NaOH * 1 mole H3A / 3 mole NaOH

= 0.000892 Mole H3A

And we know that, molar mass = amount in g / number of moles

= 0.154 g / 0.000892 Mole H3A

= 172.65 g/ mole

Hence the molar mass of her acid is 172.65 g/ mole