Question

# If   14.99 mL of NaOH are required to titrate 15.00 mL of a   0.46 M oxalic acid solution,...

If   14.99 mL of NaOH are required to titrate 15.00 mL of a   0.46 M oxalic acid solution, what is the concentration of the NaOH?

A solution of a theoretical triprotic acid was prepared by dissolving 4.980 g of solid in enough DI water to make 500.0 mL of solution.   10.10 mL of a 0.448 M solution was required to titrate 20.00 mL of this acid's solution.

1. What is the concentration of the acid solution?

2.

What is the molar mass of the acid?

Hint: You need to calculate the total moles in the 500.0 mL solution (the full 500.0 mL was NOT titrated).

2 NaOH + H2C2O4 ----> C2O4Na2 + 2H2O

n1 = 2 mole NaOH , n2 = 1 mole H2C2O4

M1V1/n1 = M2V2/n2

14.99*M1/2 = 0.46*15/1

M1 = molarity of NaOH = 0.921 M

3 NaOH + H3X ----> Na3X + 3H2O

n1 = 3 mole NaOH , n2 = 1 mole H3X

M1V1/n1 = M2V2/n2

10.10*0.448/3 = M2*20/1

M2 = molarity of H3X = 0.0754 M

Molarity = n/V

n = M2*V in L   = 0.0754*0.5 = 0.0377 mole

nO of moles of acid = wt/mwt

0.0377 = 4.98/mwt

Mwt of acid   = 4.98/0.0377 = 132.1 g/mol

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