We know that the formula: $K_a$ $=$ $\frac{[H_3O^+]^2}{c_{HB}-[H_3O^+]}$ is used to calculate $[H_3O^+]$, then the $pH$ of a weak dilute acid solution. The use shows that often, $[H_3O^+] $ is much smaller than $c_ {HB}$: Provided $ [H_3O^+] $ is more than 100 times smaller than $c_{HB}$ , we can neglect $ [H_3O^+] $ in the expression $c_{HB}$ $-$ $[H_3O^+]$ and replace this expression by $c_{HB} $. Let us examine this condition: $[H_3O^+]$ is more than 100 times smaller than $c_{HB}$ $[H_3O^+]$ $\lt $ $\frac{c_{HB}}{100}$ $\frac{(\frac{c_{HB}}{100})^2}{c_{HB}-\frac{c_{HB}}{100}}$ $\gt $ $K_a$ $\frac{c_{HB}}{9900}$ $\gt$ $ K_a$ $\frac{K_a}{c_{HB}}$ $\lt$ $1,01\cdot 10^{-4}$ $\approx$ $ 10^{-4}$
So under this condition may be neglected $[H_3O^+]$ in the expression $c_{HB}$ $-$ $[H_3O^+]$ and it comes: $K_a$ $=$ $\frac{[H_3O^+]^2}{c_{HB}-[H_3O^+]} $ $\approx$ $ \frac{[H_3O^+]^2}{c_{HB}}$ therefore:
$\frac{K_a}{c_{HB}}$ $\lt$ $ 10^{-4}$ $\Rightarrow$ $[H_3O^+]$ $=$ $\sqrt{K_ac_{HB}}$ $pH$ $=$ $\frac{1}{2}pK_a$ $-$ $\frac{1}{2}log(c_{HB})$
Example: Given a weak acid solution $10^{-1}\frac{mol}{L}$ with $K_a$ $=$ $10^{-6}$: As $\frac{K_a}{c_{HB}}$ $=$ $10^{-5}$ $\lt$ $ 10^{-4}$, we have: $pH$ $=$ $\frac{1}{2}6-\frac{1}{2}log10^{-1}$ $=$ $3.5$