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If pK(a) of a weak acid is 5, then pK(b)...

If `pK_(a)` of a weak acid is 5, then `pK_(b)` of the conjugate base will

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`pK_(a)+pK_(b)=14:.pK_(b)=14-pK_(a)=14-5=9`
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Knowledge Check

  • pK_(a) of a weak acid is defined as

    A
    `log_(10)K_(a)`
    B
    `(1)/(log_(10) K_(a))`
    C
    `log_(10). (1)/(K_(a))`
    D
    `-log_(10).(1)/(K_(a))`

  • If K_(a)=10^(-5) for a weak acid, then pK_(b) for its conjugate base would be

    A
    `10^(-10)`
    B
    `9`
    C
    `10^(-9)`
    D
    `5`

  • pK_(a) of a weak acid is defined as :

    A
    `log K_(a)`
    B
    `1/logK_(a)`
    C
    `"log"1/K_(a)`
    D
    `"-log" 1/K_(a)`

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