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InterviewSolution
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A `100 ml` solution of `0.1 N HCl` was titrated with `0.2`? `N NaOH` solution. The titration. The remaining titration war completed by adding `0.25 N KOH` solution. The volume of `KOH` required for completing the titration isA. (a)` 70 ml`B. (b)` 32 ml`C. (c )` 35 ml`D. (d)` 16 ml` |
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Answer» Correct Answer - D Volume m of HCl neutralised by `NaOH` `=("Caustic soda")=V_(1)` `N_(1)V_(1)=N_(2)V_(2), 0.1xxV_(1)=0.2xx60, V_(1)=60 ml` V total (HCl)=100 ml `V_(1)=60 ml` `=40 ml` `40 ml 0.1 N HCl` is now netralised by `KOH (0.25N) rarr (HCl)` `N_(1)V_(1)=N_(2)V_(2) (KOH)` `0.1xx40=0.25xxV_(2), V_(2)=16 ml`. |
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