Let $K$ be a finite extension of $\mathbb{Q}_p$, and let $C=\widehat{\overline{K}}$ be the completion of the algebraic closure of $K$. Let $\mathscr{O}_C$ be the ring of integers in $C$, and let $G_K$ be the absolute Galois group of $K$. I would like to know the following:

What is the annihilator of $H^1_{\mathscr{O}_{C}-mod}(G_K, \mathscr{O}_C(i))$, for $i\in \mathbb{Z}, i\not=0$?

Here $\mathscr{O}_C$-mod is the category of continuous $\mathscr{O}_C$-semilinear representations of $G_K$.

The (proofs of) Tate-Sen theory tell us that this is a torsion module annihilated by some element $\mathscr{O}_K$. I've convinced myself that in principle one could extract such an element from the proofs, though not that the existing proofs would give the best possible result. That said, I'm hoping that someone has already worked this out. In particular, I would like to know:

Is the answer (for fixed $i$) independent of $K?$

**Remarks:**

I think it's not too hard to reduce this to computing $$H^1_{\mathscr{O}_{K_\infty}-mod}(\Gamma, \mathscr{O}_{\widehat{K_\infty}}(i))$$ where $K_\infty$ is a cyclotomic $\mathbb{Z}_p$-extension of $K$ with $\mathbb{Z}_p\simeq \Gamma:=\text{Gal}(K_\infty/K)$. In other words, if $\sigma$ is a topological generator of $\Gamma$, this boils down to computing the annihilator of the cokernel of $$\sigma-\text{Id}: \mathscr{O}_{\widehat{K_\infty}}(i)\to \mathscr{O}_{\widehat{K_\infty}}(i),$$ if I'm not mistaken.

The annihilator of this cokernel is not quite the same as the annihilator of $H^1_{\mathscr{O}_{C}-mod}(G_K, \mathscr{O}_C(i))$, but if I've worked things out correctly, they differ by an amount which is independent of $K$.