Question

A 100 ml solution of 0.1 n hcl was titrated with 0.2n naoh solution.the titration was discontinued after adding 30ml of naoh solution . the remaining the titration was completed by adding 0.25 n koh solution .the volume of koh required for completeting the titration is

Answer 1

Answer: The volume of KOH required for completing the titration is 16 ml.

Explanation:

$\text{no of gram equivalents}={\text{Normality}\times {\text{Volume in L}}$

$\text{no of gram equivalents of} HCl=0.1\times 0.1L=0.01gequivalent$

$HCl+NaOH\rightarrow NaCl+H_2O$

$\text{no of gram equivalents of} NaOH=0.20\times 0.03=0.006gequivalents$

Now gram equivalets left =$(0.01-0.006)=0.004gequivalents$

$0.004=0.25N\times {\text{Volume in L}}$

${\text{Volume in L}}=0.016L=16ml$

Answer 2

Explanation:

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