We needed details of the acid dissociation behaviour of this acid; that's why I included the #pK_a#. Without this value we could not do the problem. But what does #pK_a# mean? It is measure of the extent of the following, so-called protonolysis reaction:

#HOCl(aq) + H_2O rightleftharpoons H_3O^+ + ""^(-)OCl#

For which we could write the equilibrium equation:

#K_a=([H_3O^+][""^(-)OCl])/([HOCl])#

Now #pK_a=-log_10K_a#, and thus #K_a=10^(-7.53)#.

And now we put numbers into our equation. Clearly, #[H_3O^+]=[""^(-)OCl]=x#, and #[HOCl]=0.100-x#

#K_a=10^(-7.53)=(x^2)/(0.100-x)#

Now this is a quadratic in #x#, which we could solve exactly. Because chemists are simple souls, and this is a weak acid, we could make the approximation that #0.100-x~=0.100#. We have to justify this approx. later.

And thus #x_1=sqrt(10^-7.53xx0.100)=5.43xx10^-5#.

Now we have a first approx. we can recycle this into the first equation:

#x_2=5.43xx10^-5#

And thus #[H_3O^+]=5.43xx10^-5*mol*L^-1#.

#pH=-log_10[H_3O^+]=-log_10(5.43xx10^-5)=4.27#.

This is a weak acid, but #[H_3O^+]# is about 1000 times more concentrated than pure water.