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100 ml solution of 0.1 n HCl was titrated with 0.2 n NaOH solutions. 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: [2 M]rn(A)rn70 mlrn(b)rn32 mlrn(c)rn35 mlrn(d)rn16 ml
Accepted Answer
C
Explanation:
Milli equivalents of HCl = 100 * 0.1 = 10 meq
Milli equivalents of NaOH = 30 * 0.2 = 6 meq
Milli equivalents of KOH required = 10 - 6 = 4 meq
Volume of 0.25 N KOH = Volume of KOH in mL/1000 * N
= 4/0.25 * 1000 = 35 mL
Explanation:
Milli equivalents of HCl = 100 * 0.1 = 10 meq
Milli equivalents of NaOH = 30 * 0.2 = 6 meq
Milli equivalents of KOH required = 10 - 6 = 4 meq
Volume of 0.25 N KOH = Volume of KOH in mL/1000 * N
= 4/0.25 * 1000 = 35 mL
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