Why is sodium sulfate neutral while its conjugate acid is weak? It is commonly taught in chemistry class that the conjugate base of a weak acid (well, at least if it's not weaker than water) is a weak base, as shown by the relation

$$\mathrm{p}K_\mathrm{a} = 14 - \mathrm{p}K_\mathrm{b}  \iff K_\mathrm{a}=\frac{K_\mathrm{w}}{K_\mathrm{b}}$$

where the $K_\mathrm{a}$ is of the conjugate acid, and $K_\mathrm{b}$ of the base. For weak monoprotic acids this means that their conjugate bases are weak, e.g. various carboxylic acids, phenol and hydrofluoric acid. For polyprotic acids, if the conjugate base's next $K_\mathrm{a}$ is smaller than its $K_\mathrm{b}$, then it would also form an alkaline solution but not fully dissociate, e.g. bicarbonate ion (carbonic acid), bisulfide ion (hydrogen sulfide), ammonia (ammonium ion).

However, according to Wikipedia as well as my teacher, sodium sulfate $\ce{Na2SO4}$ is neutral, but Wikipedia also says the acidic hydrogen on the bisulfate ion has $\mathrm{p}K_\mathrm{a} \approx 2$, so sulfate would have $\mathrm{p}K_\mathrm{a} \approx 12$ and a $1\ \mathrm{M}$ solution of sulfate with a perfectly non-acidic counterion would have $\mathrm{pH} \approx 8$. Is sodium sulfate in fact neutral? If so, what would be the reason? Could it be that $\ce{Na+}$ is slightly acidic? Or is the formula for conjugate basicity inapplicable here?
Given $\text{pK}_{a_1} = -3$ and $\text{pK}_{a_2} = 1.9$ for sulphuric acid, the pH for a 1M solution is indeed expected to be $7.9$, $7.4$ for a 0.1M, and $8.0$ for a saturated solution. So it is slightly basic.