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0.1mol of CH(3)NH(2) (K(b) = 5 xx 10^(-4...

`0.1mol` of `CH_(3)NH_(2) (K_(b) = 5 xx 10^(-4))` is mixed with `0.08mol` of `HCI` and diluted to `1L`. Which statement is correct?

A

The concentration of `H^(o+)` ion is `8 xx 10^(-11)M`.

B

The concentration of `H^(o+)` ion is `8 xx 10^(-5)M`.

C

The `pH` of solution is `9.8`

D

The `pOH` of solution is `10.2`.

Text Solution

Verified by Experts

The correct Answer is:
A, C
`{:(,CH_(3)NH_(2)+,HCIrarr,CH_(3)overset(o+)NH_(3)+,CI^(Theta)),("Initial",0.1,0.08,,),("moles",(0.1-0.08),(0.008-0.008),0.008,-),(,=0.02,=0,,):}`
Since `W_(B)` is left, so it forms basic buffer solution
`[overset(Theta)OH] = K_(b) (["Base"])/(["Salt"]) = 5xx 10^(-4) xx (0.02)/(0.08) = 1.25 xx 10^(-14)`
`[H^(o+)] = (K_(w))/([overset(Theta)OH]) = (10^(-14))/(1.25 xx 10^(-4)) = 8 xx 10^(-11)M`
`pH =- log (8 xx 10^(-11))`
`=- log2^(4) +11`
`=- 0.3 xx 4+11 = 9.8`
`pOH = 14 - 9.8 = 4.2`
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Knowledge Check

  • 0.1 mole of CH_(3)NH_(2)(K_(b) = 5 xx 10^(-4)) is mixed with 0.08 mole of HCl and diluted to one litre. What will be the H^(+) concentration in the solution

    A
    `8 xx 10^(-2)M`
    B
    `8 xx 10^(-11) M`
    C
    `1.6 xx 10^(-11) M`
    D
    `8 xx 10^(-5) M`

  • 0.1 mole of CH_(3)NH_(2) (K_(b) = 5 xx 10^(-4)) is mixed with 0.08 mole for HCl and diluted to one litre. What will be the H^(+) concentration in the solution

    A
    `8 xx 10^(-2)M`
    B
    `8 xx 10^(-11)M`
    C
    `1.6 xx 10^(-11)M`
    D
    `8 xx 10^(-5) M`

  • 0.1 mole of CH_(3)NH_(2) (K_(b)=5xx10^(-4)) is mixed with 0.08 mole of HCl and diluted to one litre. The [H^(+)] in solution is

    A
    `8xx10^(-2) M`
    B
    `8xx10^(-11) M`
    C
    `1.6xx10^(-11) M`
    D
    `8xx10^(-5) M`

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