# Question 7db12

Apr 18, 2016

$\text{pH} = 13.13 \to$ for part (a)

#### Explanation:

I'll show you how to solve part (a), and leave part (b) to you as practice.

So, you're mixing nitric acid, ${\text{HNO}}_{3}$, a strong acid, and potassium hydroxide, $\text{KOH}$, a strong base. The balanced chemical equation for this neutralization reaction looks like this

${\text{HNO"_ (3(aq)) + "KOH"_ ((aq)) -> "H"_ 2"O"_ ((l)) + "KNO}}_{3 \left(a q\right)}$

Notice that the two reactants react in a $1 : 1$ mole ratio, which means that a complete neutralization requires equal numbers of moles of nitric acid and potassium hydroxide.

A complete neutralization will leave behind a neutral solution of pH equal to $7$ at room temperature.

If, on the other hand, the neutralization is not complete, the remaining reagent will determine the pH of the solution

• if some of the nitric acid remains unconsumed, then the pH will be lower than $7$
• if some of the potassium hydroxide remains unconsumed, then the pH of the solution will be higher than $7$

Use the molarities and volumes of the two solutions to figure out how many moles of each reactant you have present

$\textcolor{p u r p \le}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{c = {n}_{\text{solute"/V_"solution" implies n_"solute" = c * V_"solution}}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

n_(HNO_3) = "0.0248 mol" color(red)(cancel(color(black)("L"^(-1)))) * overbrace(24.80 * 10^(-3)color(red)(cancel(color(black)("L"))))^(color(blue)("volume in liters"))

${n}_{H N {O}_{3}} = {\text{0.00061504 moles HNO}}_{3}$

and

n_(KOH) = "0.394 mol" color(red)(cancel(color(black)("L"^(-1)))) * overbrace(15.40 * 10^(-3)color(red)(cancel(color(black)("L"))))^(color(blue)("volume in liters"))

${n}_{K O H} = \text{0.0060676 moles KOH}$

Notice that you have about ten times more moles of potassium hydroxide than moles of nitric acid.

This tells you that the moles of acid will be completely consumed way before all the moles of potassium hydroxide get the chance to react, i.e. nitric acid will act as a limiting reagent.

More specifically, the reaction will only consume $0.00061504$ moles of potassium hydroxide. Once the reaction is complete, the solution will contain

${n}_{H N {O}_{3}} = \text{0 moles } \to$ completely consumed

${n}_{K O H} = 0.0060676 - 0.00061504 = \text{0.0054526 moles KOH}$

Potassium hydroxide will dissociate in aqueous solution to form potassium cations, ${\text{K}}^{+}$, and hydroxide anions, ${\text{OH}}^{-}$, in a $1 : 1$ mole ratio

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

This tells you that every mole of potassium hydroxide will produce one mole of hydroxide anions. You will thus have

${n}_{O {H}^{-}} = {\text{0.0054526 moles OH}}^{-}$

The total volume of the solution will be

${V}_{\text{total}} = {V}_{H N {O}_{3}} + {V}_{K O H}$

${V}_{\text{total" = "24.80 mL" + "15.40 mL" = "40.20 mL}}$

The concentration of hydroxide anions will thus be

["OH"^(-)] = "0.0054526 moles"/(40.20 * 10^(-3)"L") = "0.13564 mol L"^(-1)

Use this to calculate the pOH of the solution

color(blue)(|bar(ul(color(white)(a/a)"pOH" = - log(["OH"^(-)])color(white)(a/a)|)))#

You will have

$\text{pOH} = - \log \left(0.13564\right) = 0.87$

As you know, an aqueous solution kept at room temperature has

$\textcolor{p u r p \le}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\text{pH " + " pOH} = 14} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

The pH of the solution will thus be

$\text{pH} = 14 - 0.87 = \textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} 13.13 \textcolor{w h i t e}{\frac{a}{a}} |}}}$