1.

The degree of dissociation (alpha) of CH_(3)COOHsolution having 0.1M concentration and pKa as 6, in presence of strong acid HCl having concentration 0.1M is 10^(-x). Then what is the value of x?

Answer»


Solution :
`{:(t=0,0.1,0,0),(t=t,0.1- 0.1 alpha,0.1 alpha,(0.1 alpha + 0.1)):}`
`HCL rarr underset(0.1)(H^(+)) + underset(0.1)(Cl^(-))`
`K_(a)= (0.1 (1+ alpha) xx alpha xx 0.1)/(0.1 (1 - alpha))`
Since `alpha lt lt 1` therefore
`10^(-6) = (alpha xx 0.1)/(1)`
`alpha= (10^(-6))/(10^(-1))= 10^(-5)`
So the VALUE of x=5


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