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Calculate the pH of a solution obtained ...

Calculate the pH of a solution obtained by mixing 50ml of` 0.2`M HCl with `49.9` mL of `0.2`m NaOH solution.

Text Solution

Verified by Experts

The correct Answer is:
3.699
49.9 ml of 0.2 M NaOH will neutralise 49.9 ml of 0.2 M HCl. Hence, HCl left unneutralised = 0.1 ml of 0.2 M . Volume after mixing = 99.9 ml `~~` 100 ml. Hence , applying `N_(1)V_(1)=N_(2)V_(2), 0.1xx0.2=N_(2)xx100 or N_(2)=2xx10^(-4)`. In the final solution, `[HCl]=2xx10^(-4)M`, i.e., `[H_(3)O^(+)]=2xx10^(-4)M`.
Hence, `pH = - log (2xx10^(-4))=4-0.301=3.699`
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Knowledge Check

  • The pH of a solution obtained by mixing 50 ml of 0.4 M HCl with 50 ml of 0.2 M NaOH is

    A
    `-log 2`
    B
    `-log 2 xx 10^(-1)`
    C
    `7.0`
    D
    `2.0`

  • The pH of a solution obtained by mixing 50 ml of 0.4 N HCl and 50 mL of 0.2 N NaOH is :

    A
    13
    B
    12
    C
    `1.0`
    D
    `2.0`

  • The pH of the a solution obtained by mixing 50 ml of 0.4 N HCl and 50 ml of 0.2 N NaOH is

    A
    `-log 2`
    B
    `-log 0.2`
    C
    `1.0`
    D
    `2.0`

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    The pH of a solution obtained by mixing 100mL of 0.1M HCI and 9.9mL of 0.1M NaOH is:

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