Question

Question #d9c3a

Answer 1

Here's what I got.

Explanation:

The idea here is that calcium hydroxide is a strong base, which implies that it dissociates completely in aqueous solution to produce calcium cations and hydroxide anions.

`"Ca"("OH")_ (color(red)(2)(s)) rightleftharpoons "Ca"_ ((aq))^(2+) + color(red)(2)"OH"_ ((aq))^(-)`

Now, keep in mind that calcium hydroxide is not very soluble in water, but the amount that does dissolve dissociates completely.

Calcium hydroxide has a solubility of about `"1.73 g L"^(-1)` at `20^@"C"`.

https://en.wikipedia.org/wiki/Calcium_hydroxide

This is equivalent to

`1.73color(white)(.) color(red)(cancel(color(black)("g")))/"L" * ("1 mole Ca"("OH")_2)/(74.093color(red)(cancel(color(black)("g")))) = "0.0233 mol L"^(-1)`

You can thus say that a saturated solution of calcium hydroxide will be able to dissolve `0.0233` moles of calcium hydroxide for every `"1 L"` of solution.

This implies that all the calcium hydroxide that you dissolve to make your `"0.015-M"` solution will be completely dissociated.

Now, notice that every `1` mole of calcium hydroxide that dissociates produces `color(red)(2)` moles of hydroxide anions.

This means that your solution will have

`["OH"^(-)] = color(red)(2) * ["Ca"("OH")_2]`

which, in your case, is equal to

`["OH"^(-)] = color(red)(2) * "0.015 M" = "0.030 M"`

The `"pOH"` of the solution is defined as

`color(blue)(ul(color(black)("pOH" = - log(["OH"^(-)]))))`

Plug in your value to get

`"pOH" = - log(0.030) = color(darkgreen)(ul(color(black)(1.52)))`

Finally, to get the `"pH"` of the solution, use the fact that an aqueous solution at room temperature has

`color(blue)(ul(color(black)("pH + pOH = 14")))`

You will get

`"pH" = 14 - 1.52 = color(darkgreen)(ul(color(black)(12.48)))`

The answers are rounded to two decimal places, the number of sig figs you have for the concentration of the solution.