Is there any explicit description of the maximal totally ramified extension of $\mathbb{Q}_p$? It is well known that the maximal unramified extension of $\mathbb{Q}_p$ can be extended by adding the roots of unity of order prime to $p$. Is there any explicit description of the maximal totally ramified extension of $\mathbb{Q}_p$?
 A: A composite of totally ramified extensions need not be totally ramified:
Example 1. (As per LSpice's suggestion) Consider the extensions $\mathbb Q_p(\sqrt{p})$ and $\mathbb Q_p(\sqrt{\varepsilon p})$, where $\varepsilon \in \mathbb Z_p^\times$ is a nonsquare unit (corresponding, for example, to any lift of a generator of $\mathbb F_p^\times$). Then the compositum $\mathbb Q_p(\sqrt{p},\sqrt{\varepsilon p})$ contains the unramified extension $\mathbb Q_p(\sqrt{\varepsilon})$.
Example 2. Consider the extension $\mathbb Q_p \subseteq \mathbb Q_p(\sqrt[n]{p})$, where $n = p^r - 1$. Then the Galois closure of this extension contains all $n$-th roots of unity. But $\mathbb Q_p \subseteq \mathbb Q_p(\zeta_{p^r-1})$ is the unique degree $r$ unramified extension. Thus, we see that the Galois closure of a totally ramified extension need not even be totally ramified.
So there is no such thing as the maximal totally ramified extension. You could in principle still construct some maximal totally ramified extension (i.e. no further extension is totally ramified), but as far as I can tell these fields are not very explicit. By the second example above, it is not a Galois extension of $\mathbb Q_p$.
On the other hand, maximal abelian totally ramified extensions are very well studied: Lubin–Tate theory gives a relatively explicit construction of an abelian field extension $K \to K^{\operatorname{LT}}$ that is linearly disjoint from $K^{\operatorname{ab}}_{\operatorname{nr}}$ such that their compositum equals $K^{\operatorname{ab}}$. Thus, $K^{\operatorname{LT}}$ plays the role of a maximal abelian totally ramified extension in a relatively strong sense. The most explicit case is $K=\mathbb Q_p$; in this case we have $K^{\operatorname{LT}} = \mathbb Q_p(\zeta_{p^\infty})$.
