The situation mentioned in Andreas Blass's answer happens for most complete Heyting algebras that one encounters. In other words, the dual of a frame is usually not a frame even though it is a distributive lattice. For example, the lattice of closed subsets of any non-discrete $T_{1}$-space is not a Heyting algebra.

A bounded distributive lattice $L$ is said to be subfit if whenever $a,b\in L$ and
$a\vee c=1\Rightarrow b\vee c=1$ for all $c$, then $a\leq b$. For example, if $(X,\mathcal{T})$ is a $T_{1}$-space, then the frame $\mathcal{T}$ is subfit. Therefore subfitness is a very weak separation axiom that works particularly well in point-free topology. Therefore, most frames that people study in point-free topology are subfit.

Let $L$ be a complete lattice. Then we say that an element $a\in L$ is linear if whenever $I$ is a set and $b_{i}\in L$ for $i\in I$, then
$$a\wedge\bigvee_{i\in I}b_{i}=\bigvee_{i\in I}(a\wedge b_{i}).$$

We say that an element $a\in L$ is colinear if whenever $I$ is a set and $b_{i}\in L$ for $i\in I$, then

$$a\vee\bigwedge_{i\in I}b_{i}=\bigwedge_{i\in I}(a\vee b_{i}).$$

$\mathbf{Lemma}$ Every complemented element in a complete distributive
lattice is linear and colinear.

$\mathbf{Proof}$ Since the notions of linearity and colinearity are dual, we only need to show that each element on $L$ is linear. Suppose that $L$ is a complete distributive lattice and $a,c\in L$ are complemented elements. Take note that
$$a\vee(a\wedge\bigvee_{i\in I}b_{i})=a=a\vee\bigvee_{i\in I}(a\wedge b_{i}).$$
Furthermore,
$$c\vee(a\wedge\bigvee_{i\in I}b_{i})=(c\vee a)\wedge(c\vee\bigvee_{i\in I}b_{i})=(c\vee\bigvee_{i\in I}b_{i})$$
$$=\bigvee_{i\in I}(b_{i}\vee c)=\bigvee_{i\in I}((b_{i}\vee c)\wedge(a\vee c))=\bigvee_{i\in I}(b_{i}\wedge a)\vee c=(\bigvee_{i\in I}(b_{i}\wedge a))\vee c.$$

Therefore.

$$a\wedge\bigvee_{i\in I}b_{i}=(a\wedge c)\vee(a\wedge\bigvee_{i\in I}b_{i})=
(a\vee(a\wedge\bigvee_{i\in I}b_{i}))\wedge(c\vee(a\wedge\bigvee_{i\in I}b_{i}))$$
$$=(a\vee\bigvee_{i\in I}(a\wedge b_{i}))\wedge(c\vee\bigvee_{i\in I}(a\wedge b_{i}))
=(a\wedge c)\vee\bigvee_{i\in I}(a\wedge b_{i})=\bigvee_{i\in I}(a\wedge b_{i}).$$
$\mathbf{QED}$

$\mathbf{Theorem}$ Let $L$ be a complete distributive fit lattice.
Then an element $a\in L$ is complemented if and only if it is
colinear.

$\mathbf{Proof}$ $\rightarrow$. This direction has been proven already in the above lemma.

$\leftarrow$. Suppose $a\in L$ is colinear. Let $c=\bigwedge\{x\in L|a\vee x=1\}$. Then
$a\vee c=1$ as well by colinearity. Now suppose that $a\wedge c\neq 0$. Then by subfitness, there is some $x\neq 1$ with $1=x\vee(a\wedge c)=(x\vee a)\wedge(x\vee c)$. Therefore $x\vee a=1$, so $x\geq c$, hence $x\geq x\vee c=1>x$, a contradiction. We therefore conclude that $a\wedge c=0$, so $a$ and $c$ are complements.
$\mathbf{QED}$

In particular, if $L$ is a subfit frame and $L^{*}$ is the lattice with the same underlying set as $L$ but with the reverse ordering, then $L^{*}$ is a frame if and only if $L$ is a complete Boolean algebra. Also, if $X$ is a $T_{1}$-space and $\mathcal{C}$ is the lattice of closed sets of $X$, then $\mathcal{C}$ is a frame if and only if $X$ is discrete.

All these facts about frames can be found in Picado and Pultr's new book [1].

- Picado, Jorge, and Aleš Pultr. Frames and Locales: Topology without Points. Basel: Birkhäuser, 2012.