We needed details of the acid dissociation behaviour of this acid; that's why I included the pKa. Without this value we could not do the problem. But what does pKa mean? It is measure of the extent of the following, so-called protonolysis reaction:
HOCl(aq)+H2O⇌H3O++−OCl
For which we could write the equilibrium equation:
Ka=[H3O+][−OCl][HOCl]
Now pKa=−log10Ka, and thus Ka=10−7.53.
And now we put numbers into our equation. Clearly, [H3O+]=[−OCl]=x, and [HOCl]=0.100−x
Ka=10−7.53=x20.100−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 x1=√10−7.53×0.100=5.43×10−5.
Now we have a first approx. we can recycle this into the first equation:
x2=5.43×10−5
And thus [H3O+]=5.43×10−5⋅mol⋅L−1.
pH=−log10[H3O+]=−log10(5.43×10−5)=4.27.
This is a weak acid, but [H3O+] is about 1000 times more concentrated than pure water.
