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Weak acids and bases are not completely ...

Weak acids and bases are not completely ionised when dissolved in polar medium like water
`H rarr H^(+) +A^(-)`
`{:(t_(0),C,O,O),(t_(eq),C-Calpha,Calpha,Calpha):}`
`K_(a)=(C alpha^(2))/(1-alpha)=Calpha^(2),alphasqrt((K_(a))/(C))`
`:. (alpha_(1))/(alpha_(2))=sqrt((Ka_(1))/(Ka_(2))),(alpha_(1))/(alpha_(2))=sqrt((C_(2))/(C_(1)))`
( For two acids at same conc.) (for same acid at diff conc. )
`alpha`and `alpha` are in the ratio 1:2 at same conc. `Ka_(1), =2xx 10^(-4)`, what will be `Ka_(2)`,?

A

`8xx10 ^(-4)`

B

` 2xx10 ^(-4)`

C

` 4xx10 ^(-4)`

D

` 1xx10 ^(-4) `

Text Solution

Verified by Experts

The correct Answer is:
A
` (alpha_1)/(alpha_2) = sqrt((Ka_1)/(Ka_2) )rArr (1)/(2) = sqrt((2xx 10 ^(-4) )/(Ka_2)) , Ka_2 = 8xx 10^(-4) `
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Knowledge Check

  • Weak acids and bases are not completely ionised when dissolved in polar medium like water H rarr H^(+) +A^(-) {:(t_(0),C,O,O),(t_(eq),C-Calpha,Calpha,Calpha):} K_(a)=(Calpha^(2))/(1-aplha)=Calpha^(2),alphasqrt((K_(a))/(C)) :. (alpha_(1))/(alpha_(2))=sqrt((Ka_(1))/(Ka_(2))),(alpha_(1))/(alpha_(2))=sqrt((C_(2))/(C_(1))) ( For two acids at same conc.) (for same acid at diff conc. ) Relative strength of two weak monoprotic acids may be given as

    A
    `([H^(+)]_(1))/([H^(+)]_(2))`
    B
    `(alpha_(1))/(alpha_(2))`
    C
    `sqrt((Ka_(1))/(Ka_(2)))`
    D
    All of the above

  • Weak acids and bases are not completely ionised when dissolved in polar medium like water H rarr H^(+) +A^(-) {:(t_(0),C,O,O),(t_(eq),C-Calpha,Calpha,Calpha):} K_(a)=(Calpha^(2))/(1-alpha)=Calpha^(2),alphasqrt((K_(a))/(C)) :. (alpha_(1))/(alpha_(2))=sqrt((Ka_(1))/(Ka_(2))),(alpha_(1))/(alpha_(2))=sqrt((C_(2))/(C_(1))) ( For two acids at same conc.) (for same acid at diff conc. ) 0.01M CH_(3),COOH is 4.24% ionised. What will be the pereentage ionisation of 0.1 M CH_(3),COOH.

    A
    `1.33%`
    B
    `4.23%`
    C
    `5.24%`
    D
    `0.33%`

  • Weak acids and bases are not completely ionised when dissolved in polar medium like water. {:(HA hArr H^(+) +A^(-) ),(t_0 " "C " " O" "O ),(t_(eq) " " C-Calpha " "Calpha " "Calpha ):} K_a =(Calpha ^(2))/(1-alpha ) ~~Calpha ^(2) ,alpha =sqrt((K_a)/(C)) , therefore (alpha_1)/(alpha_2) =sqrt((Ka_1)/(Ka_2)) , (alpha_1)/(alpha_2) =sqrt((C_2)/(C_1)) (For two acids at same conc.) (for same acid at diff conc.) Relative strength of two weak monoprotic acids may be given as

    A
    ` ([H^(+)])/([H^(+)]_2) `
    B
    `(alpha _1)/(alpha_2) `
    C
    ` sqrt((Ka1)/(Ka_2))`
    D
    All of the above

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