How do I calculate the hydronium ion concentration if I know the strontium hydroxide concentration? This is my first question so I would appreciate feedback on how to ask better questions. Forgive me if I don't know how to properly format math/chem. The guide is confusing.

I was taking a practice test on Brønstead–Lowry acids and bases and I came across a question that has totally stumped me:


  Calculate the $\ce{[H3O+]}$ in a $\pu{0.010 M}$ solution of $\ce{Sr(OH)2}$.


Here are my steps:

Write a balanced equation:

$$\ce{Sr(OH)2 &lt;=&gt; Sr^2+ + 2OH-}$$

Initial concentrations: 
$$\ce{[Sr(OH)2]} = \pu{0.010 M}$$

$$\ce{[Sr^2+]} = 0$$
$$\ce{[OH-]} = 0$$

Change in Concentration:
$$\ce{[Sr(OH)2]} = (0.010 - x)~\pu{M}$$

$$\ce{[Sr^2+]} = +x$$
$$\ce{[OH-]} = +2x$$

I am not sure whether $\ce{OH-}$ loses 2 moles to every one of $\ce{Sr(OH)2}$.

Assume $x\ll\pu{0.010 M}$. Now I will set up a $K_\mathrm{a}$ expression:

$$K_\mathrm{a} = \ce{\frac{[H3O][OH]}{[Sr(OH)2]}}$$

$$\frac{K_\mathrm w}{K_\mathrm b} = \frac{x\cdot 2x}{0.010}$$

I don't know what to now.  I feel like I have probably started the problem wrong.
Strontium hydroxide is a strong base, so you can calculate $[\ce{OH-}]$ as $\pu{0.02 M}$, then use

$$K_\mathrm{w} = \ce{[OH-][H3O+]}$$ 

$$1\cdot 10^{-14} = 0.02\cdot [\ce{H3O+}] \quad\to\quad [\ce{H3O+}] = \pu{5e-13 M} \quad\to\quad \mathrm{pH} = 12.3$$