Calculate the pH of 0.08 M solution of hypochlorous acid , HOCl. The ionization constant of the acid is 2.5xx10^(-5) . Determine the percent dissociation of HOCl.

Calculate the pH of 0.08 M solution of hypochlorous acid , HOCl. The i

1.	Calculate the pH of 0.08 M solution of hypochlorous acid , HOCl. The ionization constant of the acid is 2.5xx10^(-5) . Determine the percent dissociation of HOCl.
Answer» Solution :The ionic equilibrium of weak acid HOCl is as under : `{:("Ionic equili.",HOCl_((aq))+,H_2O_((l)) HARR , H_3O_((aq))^(+), +ClO_((aq))^(-)),("Initial conc.(M)",0.08,-,0.0,0),("Change in reac.(M)",-x,-,+x,+x),("Concentration at equilibrium:",(0.08-x)M,,xM,xM):}` The value of x is very less , 0.08 `gt gt` x So, (0.08-x) M `approx` 0.08 M = 0.08 M `K_a=([H_3O^+][CLO^-])/([HOCl])` `THEREFORE 2.5xx10^(-5)=((x)(x))/0.08` `therefore x^2=0.08xx2.5xx10^(-5)= 2.00xx10^(-6)` `therefore x=(2.0xx10^(-6))^(1/2)` `=1.4142xx10^(-3)` M `pH=-log [H^+]=-log (1.4142xx10^(-3))` =-(0.1505-3.0) =(-2.8495)=2.85 % of dissociation = `"Dissociated [HOCl] x 100"/"Initial [HOCl]"` `=((1.4142xx10^(-3))XX100)/0.08`= 1.768%