[qstong0601]

you meassure the ph of a 0.100mol/l HOCl soulution, and find it to be 4.23. what is the Ka for HOCl acid?
this is what i did:
ph=4.23
[H+]=10^-4.23=5.9*10^-5 mol/l
now write the equilibrium reaction and the equilibrium law equation for the ionization of HOCl solution
HOCl<---->               H+       +                 OCl-
Iinitial
0.100                        0                             0
change
-5.9*10^-5            5.9*10^-5                 5.9*10^-5

equilibrium
0.100-5.9*10^-5     5.9*10^-5                 5.9*10^-5
Ka=[H+][OCl-]/[HOCl]

then plug them all in

Ka=(5.9*10^-5)^2/(0.100-5.9*10^-5)
the method that the textbook used though is:
Ka=(5.9*10^-5)^2/0.100

my quesiton is should i use the (0.100-5.9*10^-5) or 0.100 for the concentration of HOCl at equilibrium?
the question given that 0.100mol/L is the orginal concentration of HOCl, aren't we supposed to find out the equilibrium concentration of HOCl, since it is a weak acid that it doesnt react quantitively.

[enahs]

0.1 - 0.000059 = 0.099941 = 0.1 with significant figures.

[Borek]

0.1 - 0.000059 = 0.099941 = 0.1 with significant figures.

In other words: calculate both versions and check how large difference you will get in the final result.

There exist rule of thumb stating that in approximate calculations of acid/base equilibria when you have sum (or difference) of two numbers a and b, if b is less then 5% of a you can assume a+b=a. In most cases this approximation won't change final result, in all cases it greatly simiplifies calculations