Question

Question #753db

Answer 1

I'll show you how to solve parts (c) and (d).

Explanation:

I'll focus on parts (c) and (d), since they're more interesting, and leave parts (a) and (b) to you as practice.

So, for part (c), you know that you're dealing with a solution of potassium hydroxide, a strong base, which you dilute from an initial volume of 10 mL to a final volume of 250 mL.

Use the molarity and initial volume of the solution to find how many moles of potassium hydroxide you're working with

`color(blue)(C = n/V implies n = C * V)`

`n = "0.22 M" * 10.0 * 10^(-3)"L" = "0.0022 moles"`

Since potassium hydroxide is a strong base, it will dissociate completely in aqueous solution to form

`"KOH"_text((aq]) -> "K"_text((aq])^(+) + "OH"_text((aq])^(-)`

Notice that every mole of potassium hydroxide will produce one mole of hydroxide ions. This means that you have

`n_(OH^(-)) = n_(KOH) = "0.0022 moles"`

The molarity of the hydroxide ions after the solution is diluted is

`["OH"^(-)] = "0.0022 moles"/(250 * 10^(-3)"L") = "0.0088 M"`

This means that the `pOH` of the solution will be

`pOH = -log(["OH"^(-)])`

`pOH = -log(0.0088) = color(green)(2.06)`

The pH of the solution will be

`pH_"sol" = 14 - pOH = 14 - 2.06 = color(green)(11.94)`

For part (d), you have to use the molar mass of barium hydroxide, `"Ba"("OH")_2`, to find the number of moles you have in that mass.

`0.50color(red)(cancel(color(black)("g"))) * "1 mole"/(171.34color(red)(cancel(color(black)("g")))) = "0.00292 moles"`

Barium hydroxide will dissociate in aqueous solution to form

`"Ba"("OH")_text(2(aq]) -> "Ba"_text((aq])^(2+) + color(red)(2)"OH"_text((aq])^(-)`

This time, every mole of barium hydroxide will produce `color(red)(2)` moles of hydroxide ions, which means that you have

`n_(OH^(-)) = color(red)(2) * n_(Ba(OH)_2) = 2 * 0.00292 = "0.00584 moles"`

The molarity of the hydroxide ions will thus be

`["OH"^(-)] = "0.00584 moles"/(100 * 10^(-3)"L") = "0.0584 M"`

The solution's `pOH` is

`pOH = -log(0.0584) = color(green)(1.23)`

This means that the pH is

`pH_"sol" = 14 - 1.23 = color(green)(12.77)`