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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