The pH of a weak acid of concentration c is about (pKa minus log c) divided by 2, so 0.1 mol/L ethanoic acid (pKa 4.76) is pH 2.88 with only 1.3 percent of its molecules ionised, and a tenfold dilution moves it half a unit, not a whole one.
A weak acid HA ionises only a little, so [H+] = [A-] = the square root of Ka times c, and taking the negative log gives the half. The less of it there is, the larger the share that ionises, which is why dilution changes the pH by less than for a strong acid at the same concentration: hydrochloric acid is pH 1.00 at 0.1 mol/L, ethanoic acid pH 2.88.
It needs the acid to be weak (under about 5 percent ionised) and the solution not very dilute: below about 10 to the minus 4 mol/L ethanoic acid (a third of it ionised) the shortcut is off by 0.1 or more, and no acid, however dilute, goes above pH 7. For a base the same works on pOH with the pKb.
Estimate with the rule, then check it against the calculator that models it properly.
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