Phosphoric acid, H₃PO₄, has one P=O and three OH groups…

Step 1. The value of n. $\text{H}_{3}\text{PO}_{4}$ has one P=O and no other oxygen atom without hydrogen, so $n = 1$.

Step 2. The estimates. $\text{p}K_{a1} = 8 - 5 \times 1 = 3$. The next two are $3 + 5 = 8$ and $8 + 5 = 13$.

Step 3. Compare. The data are $2.15$, $7.20$ and $12.35$. They are less than one unit below the estimates each time, so the rule is good to within one unit here.

Step 4. The data difference. $7.20 - 2.15 = 5.05$, close to the 5 of the rule.

Step 5. The two chlorine acids. The rule gives $8 - 5 \times 2 = -2$ for $\text{HClO}_{3}$ and $8$ for HOCl. The difference is $10$ pKa units, which is a factor of $10^{10}$ in $K_a$, so 10 powers of ten.