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If K(h) (hydrolysis constant) for anilin...

If `K_(h)` (hydrolysis constant) for anilinium ion is `2.4xx10^(-5) M`, then `K_(b)` for aniline will be

A

`4.1xx10^(10)`

B

`4.1xx10^(-10)`

C

`2.4xx10^(9)`

D

`2.4xx10^(-19)`

Text Solution

Verified by Experts

The correct Answer is:
B
`K_(h)` for anilinium ion `=2.4xx10^(-5) M`
`K_(h)=(K_(w))/(K_(b))`
`K_(b)=(10^(-14))/(2.4xx10^(-5))=4.1xx10^(-10)`
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Knowledge Check

  • The hydrolysis constant of NaX ( K_(a) of HX is 2 xx 10^(-6) ) is

    A
    `5 xx 10^(-9)`
    B
    `2 xx 10^(-8)`
    C
    `5 xx 10^(-6)`
    D
    `10^(-7)`

  • The Hydrolysis constant of MX is (Given, K_(b) of MOH is 2 xx 10^(-6) and K_(a) of HX is 5 xx 10^(-7) )

    A
    `10^(2)`
    B
    `10^(-2)`
    C
    `10^(-3)`
    D
    `10^(3)`

  • Calculate the hydrolysis constant of the salt containing NO_(2)^(-) . Given the K_(a) " for " HNO_(2) = 4.5 xx 10^(-10)

    A
    `2.22 xx 10^(-5)`
    B
    `2.02 xx 10^(5)`
    C
    `4.33 xx 10^(4)`
    D
    `3.03 xx 10^(-5)`
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