`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 ?

Explanation

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.