50 mL of 0.2 N HCl is titrated against 0.1 N NaOH solution. The titration is discontinued after adding 50 mL of NaOH solution. The remaining titration is completed by adding 0.5 N KOH solution. What is the volume of KOH solution required for completing the titration?

50 mL of 0.2 N HCl is titrated against 0.1 N NaOH solution. The titrat

1.	50 mL of 0.2 N HCl is titrated against 0.1 N NaOH solution. The titration is discontinued after adding 50 mL of NaOH solution. The remaining titration is completed by adding 0.5 N KOH solution. What is the volume of KOH solution required for completing the titration?
Answer» Solution :CALCULATION of HCL left after 1st titration : `{:(underset((HCl))(N_(1)xxV_(1))=underset((NaOH))(N_(2)xxV_(2)),,),(0.2xxV_(1)=0.1xx50,or,V_(1)=25mL):}` i.e., 25 mL of 0.2 N HCl has been consumed and therefore 0.2 N HCl left `=50-25=25mL` Calculation of KOH used in 2nd titration : `{:(underset((KOH))(N_(1)xxV_(1))=underset((HCl))(N_(2)xxV_(2)),,),(0.5xxV_(1)=0.2xx25,or,V_(1)=10mL):}` `therefore KOH` solution required for completing the titration = 10 mL.