A 100 m l solution of 0.1 N - H C l was titrated with 0.2 N - N a O H solution. The titration was discontinued after adding 30 m l of N a O H solution. The remaining titration was completed by adding 0.25 N - K O H solution. The volume of K

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A 100 m l solution of 0.1 N - H C l was titrated with 0.2 N - N a O H solution. The titration was discontinued after adding 30 m l of N a O H solution. The remaining titration was completed by adding 0.25 N - K O H solution. The volume of K O H required for completing the titration is

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Correct answer: A. 16 m l

In the neutralization of acid and base N × V of both must be equivalent N × V o f H C l = 0.1 × 100 = 10 N × V o f N a O H = 0.2 × 30 = 6 N 1 V 1 ml = N × V NaOH ml + N × V ml KOH 0.1 × 100 = 0.2 × 30 + 0.25 × V 10 = 6 + 0.25 V V = 400 25 V = 16 m l

A 100 m l solution of 0.1 N - H C l was titrated with 0.2 N - N a O H solution. The titration was discontinued after adding 30 m l of N a O H solution. The remaining titration was completed by adding 0.25 N - K O H solution. The volume of K O H required for completing the titration is

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