# 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:

$$\ce{Sr(OH)2 <=> Sr^2+ + 2OH-}$$

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.

• As to learning how to format chem, I suggest you learn by opening up the "edit" option for posts containing them, and try imitating it on your posts. – William R. Ebenezer Apr 18 at 18:04

Strontium hydroxide is a strong base, so you can calculate $$[\ce{OH-}]$$ as $$\pu{0.02 M}$$, then use
$$K_\mathrm{w} = \ce{[OH-][H3O+]}$$
$$1\cdot 10^{-14} = 0.02\cdot [\ce{H3O+}] \quad\to\quad [\ce{H3O+}] = \pu{5e-13 M} \quad\to\quad \mathrm{pH} = 12.3$$
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