Question

A diprotic acid (H2X) titrated with a solution of NaOH for 25 ml of the acid 87.42 ml of a 1.95 M solution of NaOH was required to reach the equivalent point, the concentration of the diprotic acid equals
0.14
6.8
3.4
0.29

Answer 1

The concentration of the diprotic acid equals 3.4 M

We'll begin by writing the balanced equation for the reaction.

H₂X + 2NaOH —> Na₂X + 2H₂O

From the balanced equation above,

• The mole ratio of the acid, H₂X (nA) = 1

• The mole ratio of the base, NaOH (nB) = 2

From the question given above, the following data were obtained:

• Volume of acid, H₂X  (Va) = 25 mL

• Volume of base, NaOH (Vb) = 87.42 mL

• Molarity of base, Ca(OH)₂ (Mb) = 1.95 M

• Molarity of acid, H₂X (Ma) =?

MaVa / MbVb = nA/nB

(Ma × 25) / (1.95 × 87.42) = 1/2

(Ma × 25) / 170.469 = 1/2

Cross multiply

Ma × 25 × 2 = 170.469

Ma × 50 = 170.469

Divide both side by 50

Ma = 170.469 / 50

Ma = 3.4 M

Therefore, the concentration of the acid, H₂X is 3.4 M

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