The correct relation between hydrolysis constant (Ka) and degree of hydrolysis (h) for the following equilibrium is: CH3COO−+H2Ohydrolysis⇌CH3COOH+OH−
A
h=√Kw⋅CKa
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B
h=√KwKa⋅C
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C
h=√Ka⋅CKw
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D
h=√KaKw⋅C
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Solution
The correct option is Bh=√KwKa⋅C Salt of weak acid and strong base. If C is the concentration and h is the degree of hydrolysis, then at equilibrium : CH3COO−C(1−h)+H2Ohydrolysis⇌CH3COOHCh+OH−Ch
Kh=[CH3COOH][OH−][CH3COO−]=Ch×ChC(1−h)=Ch21−h CH3COOH⇌CH3COO−+H+Ka=[CH3COO−][H+][CH3COOH]H2O⇌H++OH−Kw=[H+][OH−]
We see that Kh=KwKa=Ch21−hh<<<1Then,1−h≈1KwKa=Ch2h=√KwKa⋅C