`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

Step I. Calculation of volume of Hcl left after incomplete titration

`(0.2N) xx V_(1)=(0.1 N)xx(50 cm^(3))`
`V_(1)=(("0.1 N")xx("50 cm"^(3)))/(("0.2 N"))="25 cm"^(3)`
Step II. Calculation of volume of KOH for completing the titration

`(0.5 N)xxV_(1)=(0.2 N) xx (25 mL)`
`V_(1)=(("0.2 N")xx("25 mL"))/(("0.5 N"))="10 mL"`.