Question

calculate the ph of 0.08M solution of hypochlorous acid,HOCL.The ionization constant of the acid is 2.5x 10-5.determine the % dissociation of HOCL?

Answer 1

First we will write the dissociation equation of HOCl 

HOCl(aq) + H20(l) -----H30+ (aq) + ClO- (aq) 

Initial concentration (M) 0.08 M 0 0  

Change to reach 

Equilibrium concentration(M) -x +x +x 

Equilibrium concentration (M) 0.08 -x x x 

Ka = [H3O+][ClO-]/[HOCL] 

Ka= x square/(0.08-x) 

As x <<0.08 then 0.08-x=0.08 

X square/0.08 = 2.5x10 raise to power minus 5 

X square = 2.0x10 raise to power minus 6 

X = 1.41 x 10 raise to power minus 3 

 

[HOCl] dissociated= 1.41x10^ -3 M and [H+] = 1.41x10^ -3 M 

Therefore,% dissociation = {[HOCl] dissociated/ [HOCl]initial] x 100 

= (1.41x10^-3)/0.08 

= 1.76% 

pH = -log[H+] 

pH= -log(1.41x10^-3) 

pH= 2.85

Answer 2

and there is your amswer

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