tl;dr: Is there such a thing as a W*-completion of a C*-algebra, and if so, where can I read about it?

I'm wondering about the relationship between (abstract) C*-algebras and W*-algebras. On the one hand, every W*-algebra is a C*-algebra. On the other hand, it seems to me that it should be possible to complete any C*-algebra to a W*-algebra. (Categorially, this would be a reflection.) In the case of commutative algebras, I even think that I know how this works: every commutative C*-algebra is the algebra of continuous functions on some compact Hausdorff space, and we extend this to the W*-algebra of essentially bounded Borel-measurable functions on the space (considered up to equality almost everywhere). [Warning: this is determined in the comments to be wrong.]

So, is this correct? Does it work for noncommutative algebras as well? Is there a good algebro-analytic (without passing through topology) description of this? Is there a good reference, especially online?

Also, in the commutative case, it seems that every state (positive normal linear functional) on a C*-algebra extends uniquely to its W*-completion, so they have the same space of states. Is this correct? Does it extend to the noncommutative case?

Another question is how this relates to concrete algebras (those given as algebras of operators on some Hilbert space). One way to complete a C*-algebra would be to pick a concrete representation and take its weak closure (or double commutant). But I expect that this will depend on the representation chosen. (And my analysis is bad enough that I can't check this for even the commutative case.)

I'd appreciate any help even for the main question, never mind this stuff about states and representations!

notthe enveloping algebra of $C([0,1])$, but a proper quotient thereof. The predual of $C([0,1])^{**}$ is $C([0,1])^*$. Now, $\delta$-measures of points in $[0,1]$ form a norm-discrete subset of continuum cardinality in $C([0,1])^*$, so $C([0,1])^{**}$ does not have separable predual. But the predual of $L^\infty([0,1],\mu)$ is $L^1([0,1],\mu)$, which is norm-separable. $\endgroup$ – Robert Furber Oct 23 '18 at 22:00