Question:
0.1914 g of an organic acid is dissolved in about 20 ml of water . 25 ml of 0.12 N NaOH is required for the complete neutralization of the acid solution. The equivalent weight of the acid is
Answer:
Given,
Total mass of organic acid = 0.1914 g
No of moles of organic acid = 0.1914/M
60 mL of 0.5 M solution of NaOH
No of moles of NaOH = 0.12 x 25 x 10-3 = 3 x 10-3 mol
No of moles of organic acid = No of moles of NaOH
0.1914/M = 3 x 10-3
M = 0.19143 x 10-3 = 63.8 g
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