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`50 cm^(3)` of 0.2 N HCl is titrated against 0.1 N NaOH solution. The titration is discontinued after adding `50 cm^(3)` of NaOH solution. The remaining titration is completed by adding 0.5 N KOH solution. What is the volume of KOH required for completing the titration ?A. `12 cm^(3)`B. `10 cm^(3)`C. `25 cm^(3)`D. `10.5 cm^(3)` |
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Answer» Correct Answer - B `{:("Acid",=,"Base"),(N_(1)V_(1),=,N_(2)V_(2)),(0.2 xx V_(1),=,0.1 xx 50):}` `:. V_(1)=(0.1 xx 50)/(0.2 ) =25cm^(3)` Thus , `25cm^(3)` is the volume of acid neutralized by `50 cm^(3)` of 0.1 N NaOH `:. ` volume of 0.2 N HCl left behind `=50-25=25 cm^(3)` This is neutralized by 0.5 KOH Acid= Base `N_(1)V_(1) = N_(2)V_(2)` `0.2 xx 25 =0.5 xx V_(2) :. V_(2)=(0.2 xx 25)/(0.5 ) =10cm^(3)` `:. `Volume of 0.5 N KOH required `=10cm^(3)` |
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