# Question 753db

Sep 10, 2015

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

$\textcolor{b l u e}{C = \frac{n}{V} \implies n = C \cdot V}$

$n = \text{0.22 M" * 10.0 * 10^(-3)"L" = "0.0022 moles}$

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

${\text{KOH"_text((aq]) -> "K"_text((aq])^(+) + "OH}}_{\textrm{\left(a q\right]}}^{-}$

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

${n}_{O {H}^{-}} = {n}_{K O H} = \text{0.0022 moles}$

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 $p O H$ of the solution will be

$p O H = - \log \left(\left[{\text{OH}}^{-}\right]\right)$

$p O H = - \log \left(0.0088\right) = \textcolor{g r e e n}{2.06}$

The pH of the solution will be

$p {H}_{\text{sol}} = 14 - p O H = 14 - 2.06 = \textcolor{g r e e n}{11.94}$

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

${\text{Ba"("OH")_text(2(aq]) -> "Ba"_text((aq])^(2+) + color(red)(2)"OH}}_{\textrm{\left(a q\right]}}^{-}$

This time, every mole of barium hydroxide will produce $\textcolor{red}{2}$ moles of hydroxide ions, which means that you have

${n}_{O {H}^{-}} = \textcolor{red}{2} \cdot {n}_{B a {\left(O H\right)}_{2}} = 2 \cdot 0.00292 = \text{0.00584 moles}$

The molarity of the hydroxide ions will thus be

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

The solution's $p O H$ is

$p O H = - \log \left(0.0584\right) = \textcolor{g r e e n}{1.23}$

This means that the pH is

$p {H}_{\text{sol}} = 14 - 1.23 = \textcolor{g r e e n}{12.77}$