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50 cm^(3) of 0.2 N HCl is titrated again...

`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)`

Text Solution

Verified by Experts

The correct Answer is:
B
When 0.1 N NaoH is used
`N_(1)V_(1) = N_(2)V_(2)`
(HCl) (NaOH)
` 0.2 xx V = 50 xx 0.1`
`V = (50 xx 0.1)/(0.2) = 25 cm^(3)`
When 0.5 N KOH is used,
`N_(1)V_(1) = N_(3)V_(3)`
(HCl) (KOH)
`0.2 N xx 25 = 0.5 N xx V_(3)`
`V_(3) = (0.2 xx 25)/(0.5) = 10 cm^(3)`
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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?

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Knowledge Check

  • A 100 ml solution of 0.1 N HCl was titrated with 0.2 N NaOH solution. The titration was discontinued after adding 30 ml of NaOH solution. The remaining titration was completed by adding 0.25 N KOH solution. The volume of KOH required for completing the titration is

    A
    70 ml
    B
    32 ml
    C
    35 ml
    D
    16 ml

  • A 100 ml solution of 0.1 N HCl was titrated with 0.2 N NaOH solution. The titration was discontinued after adding 30 ml of NaOH solution. The remaining titration was completed by adding 0.25 N KOH solution. The volume of KOH required for completing the titration is

    A
    16 ml
    B
    32 ml
    C
    35 ml
    D
    70 ml

  • A 100ml solution of 0.1 N HCl was titrated with 0.2N NaOH solution. The titration was discontinued after adding 30ml of NaOH solution. The remaining titration was completed by adding 0.25N KOH solution. The volume of KOH required for completing the titration is

    A
    70ml
    B
    32ml
    C
    35ml
    D
    16ml

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