Why does the pH of a weak acid not increase by 1 when diluted by a factor of 10?
In a weak acid  $\ce{HB}$ solution, with a nominal concentration $c$, a tiny amount of its molecules are dissociated into $\ce{H^+}$ and $\ce{B^-}$. Let's call this concentration $[\ce{H^+}]$ = $\ce{[B^-]}$ = $x &lt;&lt;c$, so that the following approximation can be made : $c - x$= $c$. The dissociation equilibrium constant $K_a$ of this weak acid $\ce{HB}$ can be approximated by$$K_a\ce{= \frac {[H^+][B^-]}{$c - x$} = \frac{$x$^2}{$c-x$} = \frac{$x$^2}{$c$}}$$ $$\ce{[H^+] = x = \sqrt{K_ac}}$$ $$\ce{$p$H = \frac{1}{2} (log$K$_a - log $c$)}$$ Look at the coefficient $1/2$ before the logarithm. If the concentration $c$ is multiplied by $10$, the log c increases by $1$, but the $p$H changes by $1/2$.