Question

A solution of a theoretical triprotic acid was prepared by dissolving 5.826 g of solid in enough DI water to make 500 Ml solution. 11.75 mL of a .530 M solution was required to titrate 20.00 mL of this acids solution

1) What is the concentration of the acid solution?

2) What is the molar mass of the acid?

Last question is if 14.18 mL of NaOH are required to titrate 15 mL of a 0.6 M oxalic acid solution,

what is the concentration of NaOH ?

Answer 1

The identity of the titrant is not given so i am assuming that NaOH is used to titrate against the acid solution.

The milliequivalents of the acid should be equal to the milliequivalents of the base

n1M1V1 = n2M2V2    for a titration of acid base & n1/n2 = basicity/acidity of the acid/base

3xM x 20 ml = 1x .530 x 11.75 mL

M of the acid solution =0.103 Moles/litre

Molarity of the acid solution = (Moles of the acid /volume of the solution)x1000

                                  0.103 = {(5.826 g/Molar mass)/500} x 1000

                              Molar mass of the acid = 113.12 g/mole

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last question

4.18 mL of NaOH are required to titrate 15 mL of a 0.6 M oxalic acid solution,

n1M1V1 = n2M2V2

4.18 x M X 1 = 2 x .6 X 15

Molarity of NaOH = 4.3 M