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JEE Main & Advanced Chemistry Analytical Chemistry Question Bank Volumetric Analysis

  • question_answer
    A 100 ml solution of 0.1 \[N-HCl\] was titrated with 0.2 \[N-NaOH\] solution. The titration was discontinued after adding 30 ml of \[NaOH\] solution. The remaining titration was completed by adding 0.25 \[N-KOH\] solution. The volume of \[KOH\] required for completing the titration is [MP PMT 1997]

    A) 16 ml

    B) 32 ml

    C) 35 ml

    D) 70 ml

    Correct Answer: A

    Solution :

    In the neutralization of acid and base N × V of both must be equivalent N × V of HCl  = 0.1 × 100 = 10 N × V of NaOH = 0.2 × 30   = 6 as to obtain 10 N × V of base                4 N × V of base is required N × V of KOH = 0.25 × 16 = 4 N1V1 = \[\underset{NaOH}{\mathop{N\times V\ }}\,+\ \underset{KOH}{\mathop{N\times V}}\,\] 0.1 × 100 = 0.2 × 30 + 0.25 × V 10 = 6 + 0.25 V \[V=\frac{400}{0.25}\] Þ V = 16 ml


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