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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.

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  • $\begingroup$ 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. $\endgroup$ – William R. Ebenezer Apr 18 at 18:04
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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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    $\begingroup$ Welcome to Chemistry.SE! Please note that formulas can be better expressed with \$\ce{ }\$ for chemical formulas/equations, \$ \$ for math term/equations, and \$\pu{ }\$ for units. More information is available in this meta post Also, take a minute to look over the help center and tour page to better understand our guidelines and question policies if you haven't already. $\endgroup$ – A.K. Apr 18 at 19:40

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