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sir/ma'am, i wntd to know how does c(1-a)comes in calculating Ka for acids whr c==initial concentration of acid,a=degree of ionisation

sir/ma'am,


i wntd to know how does c(1-a)comes in calculating Ka for acids whr c==initial concentration of acid,a=degree of ionisation

Grade:

3 Answers

apoorva sagarwal
16 Points
8 years ago

 Reactant A is taken with initial concentration "c" .after it under goes the reaction and when equilibrium is attained the products are formed with concentration concentration "ca" each and the reactant concentration becomes "c-ca" = "c(1-a)"

consider the equation

                           A   ---->   B + C 

initially         :      c              0    0

at equilibrium:    c(1-a)        ca   ca

==> Ka =  ca*ca/c(1-a)

           =  ca2/(1-a)

 

Akash Kumar Dutta
98 Points
8 years ago

a=degree of ionisation...that means =the no. of molecules formed ions/total molecules.
now to find the fraction of molecule left we do...(1-a)
reduced concentration left= c(1-a)..note 1-a<1
hence it is clear that the concentration reduces.

Vivek Sharma
16 Points
8 years ago

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