Question

A 100ml solution of 0.1N HCl was titrated with 0.2N NaOH solution . The titration was discontinued after adding 30ml 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 ___ .

(a) 70 ml. (b) 32 ml. (c)35 ml. (d)16ml

Answer 1

Moles of HCl in 100 ml solution of 0.1 N HCl = (0.1/1000) x 100 = 0.01 moles

Moles of NaOH in 30 mL of 0.2 N NaOH solution = (0.2/1000) x 30 = 0.006 moles
Moles of HCl remaining after titration = 0.01 - 0.006 = 0.004 moles

Hence, we would be needing 0.004 moles of KOH to neutralise the remaining HCl.
Since, 0.25 moles are present in 1000 ml of KOH, then,
0.004 moles are present in
= (1000/0.25) x 0.004 = 16 ml.