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Determine the hydrolysis constantand deg...

Determine the hydrolysis constantand degree of hydrolysis of 0.01 M solution of ammonium acetate , given that `K_(a) =1.752 xx 10^(-5) , K_(b) =1.74 xx 10^(-5)`

Text Solution

Verified by Experts

The correct Answer is:
`K_(h) =3.28 xx 10^(-5) h = 5.727 xx 10^(-3)`
Hydrolysis constant `(K_(h))` may be calculated as follows :
`K_(h) =(K_(w))/(K_(a) xx K_(b)) =(1.0xx 10^(-14))/((1.752xx10^(-5))xx(1.74xx10^(-5)))=3.28 xx 10^(-5)`
`"Degree of hydrolysis(h)"" "=(K_(h))^(1//2) =(3.28 xx 10^(-5))^(1//2) =5.727 xx 10^(-3)`
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Knowledge Check

  • The percentage hydrolysis of 0.15 M solution of ammonium acetate, K_a for CH_3 COOH is 1.8 x× 10^(-5) and K_b for NH_3 is 1.8 x× 10^(-5)

    A
    0.556
    B
    4.72
    C
    9.38
    D
    5.56

  • What is the pH of an aqueous solution of ammonium acetate (K_a= K_b =1.8 xx 10^(-5))

    A
    `gt 7`
    B
    7
    C
    `lt 7.0`
    D
    Zero

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