Question

# The correct relation between hydrolysis constant (Ka) and degree of hydrolysis (h) for the following equilibrium is:  CH3COO−+H2Ohydrolysis⇌CH3COOH+OH−

A
h=KwCKa
B
h=KwKaC
C
h=KaCKw
D
h=KaKwC

Solution

## The correct option is B h=√KwKa⋅CSalt 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⋅CChemistry

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