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If 50 mL of 0 . 5 M oxalic acid is required to neutralise 25 mL of NaOH solution, the amount of NaOH in 50 mL of given NaOH solution is_______g.

Correct answer

0

Step-by-step solution

50 ml of 0 . 5 M oxalic acid is completely neutralised by 25 ml of NaOH solution. For neutralisation reactions, N 1 V 1 = N 2 V 2 Normality of oxalic acid = Molarity × 2 ⇒ N oxalic acid = 0 . 5 × 2 = 1 N 50 × 1 = 25 × N NaOH For sodium hydroxide, molarity is the same as normality. Molarity of sodium hydroxide = 2 M The number of moles of sodium hydroxide= 50 × 2 × 10 - 3 mol . Hence, the mass of sodium hydroxide = 50 × 2 × 10 - 3 × 40 = 4 g

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