[H^+]=1.41xx10^(-3) M in 0.08 M solution of HOCl . Then what is the percentage of dissociation of it ?

[H^+]=1.41xx10^(-3) M in 0.08 M solution of HOCl . Then what is the pe

1.	[H^+]=1.41xx10^(-3) M in 0.08 M solution of HOCl . Then what is the percentage of dissociation of it ?
Answer» Solution :`{:(,HOCl HARR, H^(+) + , OCl^(-)),(,"0.08 M" ,,),("At equili.",(0.08 - C alpha),c alpha,c alpha),(,=0.08 M,=1.41xx10^(-3)M,):}` % of DISSOCIATION = `"dissociated [HOCl]"/"Initial [HOCl]"XX100` `=(1.41xx10^(-3))/0.08xx100`=1.766% OR `c alpha=1.41xx10^(-3)` So, `alpha (1.41xx10^(-3))/c` `=(1.41xx10^(-3))/0.08`=0.017625 % of dissociation = `100 alpha`= 100 x 0.017625 =1.7662% `approx` 1.77%