If 14.99 mL of NaOH are required to titrate 15.00 mL of a 0.46 M oxalic acid sol
ID: 901877 • Letter: I
Question
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).
Explanation / Answer
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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