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

Question

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 (A) 16 ml (B) 32 ml (C) 35 ml (D) 70 ml

Tardigrade Admin · Accepted Answer

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 N1V1(ml)= undersetNaOHN × V(ml)+ undersetKOHN × V (ml) 0.1× 100 = 0.2× 30+0.25× V 10 = 6+0.25 V V=(400/25) V = 16 ml

$16 m l$

$32 m l$

$35 m l$

$70 m l$

Q. 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 0.25N−KOH solution. The volume of KOH required for completing the titration is

16ml

32ml

35ml

70ml

In the neutralization of acid and base N×V of both must be equivalent N×VofHCl=0.1×100=10 N×VofNaOH=0.2×30=6 N1​V1​(ml)=NaOHN×V​(ml)+KOHN×V(ml)​ 0.1×100=0.2×30+0.25×V 10=6+0.25V V=25400​V=16ml

$16 m l$

$32 m l$

$35 m l$

$70 m l$