I asked this question at math.stackexchange and received no comment.
Let $A$ be a C*-algebra and $\phi$ be a positive linear functional on $A$. Let $\tilde{\phi}$ be its unique $w^*$-continuous extension on $A^{**}$.
Let us put
$$N_{\phi}:=\{a\in A: \phi(a^*a)=0\}~~~,~~~N_{\tilde{\phi}}:=\{x\in A^{**}: \phi(x^*x)=0\}$$
$N_{\phi}$ forms a closed left ideal in $A$ and $N_{\tilde{\phi}}$ forms a $w^*$-closed left ideal in $A^{**}$. It is obvious that $N_{\phi}\subseteq{N_\tilde{\phi}}$.
Q) Assume that $\phi$ is a pure state. I feel strongly (but have no clear proof) that: $N_{\tilde{\phi}}=\overline{N_{\phi}}^{w^*}$.
Remark. The previous assumption does not hold for any positive linear functional. See the following example:
Example 1: Let $\{x_n\}$ be a dense subset in $[0,1]$ and consider $\phi_j:C([0,1])\to \mathbb{C}$ given by
$$\phi_1(f)=\sum \frac{f(x_n)}{2^n}~~~,~~~\phi_2(f)=\int_0^{0.5} f dm $$
where $dm$ is the Lebesgue measure. One may check that $N_{\tilde{\phi_j}}\ne \overline{N_{\phi_j}}^{w^*}$.
It seems that in infinite dimensional commutative $C^*$-algebras the above assertion just holds for pure states however in non-commutative case there are (likely) some non-pure states with this property.
