Question 0bc73

Apr 9, 2016

${K}_{b} = 1.8 \cdot {10}^{- 5}$

Explanation:

You're dealing with a neutralization reaction in which $\text{BOH}$, a weak base, reacts with hydrochloric acid, $\text{HCl}$, a strong acid, to produce ${\text{B}}^{+}$, the weak base's conjugate acid, and water.

Your strategy here will be to write the balanced chemical equation for the two reactions between the weak base and the strong acid, and use equation stoichiometry to try and find two relationships between the two expressions of the base dissociation constant, ${K}_{b}$.

Since hydrochloric acid is a strong acid, it will dissociate completely in aqueous solution to form hydronium cations, ${\text{H"_3"O}}^{+}$, and chloride anions, ${\text{Cl}}^{-}$, in a $1 : 1$ mole ratio.

The chloride anions are spectator ions, so I will exclude them from the balanced chemical equations.

Now, let's assume that the $\text{40-mL}$ solution of weak base contains $x$ moles of $\text{BOH}$.

Use the molarity of the hydrochloric acid solution, which will be equivalent to its given normality, to calculate how many moles of strong acid are present in the $\text{5.0-mL}$ sample

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

You will have

n_(H_3O^(+)) = "0.1 mol" color(red)(cancel(color(black)("L"^(-1)))) * overbrace(5.00 * 10^(-3)color(red)(cancel(color(black)("L"))))^(color(purple)("volume in liters"))

$= 5.0 \cdot {10}^{- 4} {\text{moles H"_3"O}}^{+}$

The balanced chemical equation for this reaction looks like this

${\text{ " "BOH"_ ((aq)) + "H"_ 3"O"_ ((aq))^(+) -> "B"_ ((aq))^(+) + 2"H"_ 2"O}}_{\left(l\right)}$

The weak base reacts with the hydronium cations produced by the hydrochloric acid in a $1 : 1$ mole ratio. This means that every mole of weak base that takes part in the reaction consumes one mole of strong acid and produces one mole of conjugate acid.

You know for a fact that the strong acid will be completely consumed by the reaction, since the pH of the solution after the addition of the $\text{5.00 mL}$ sample is well above neutral of acidic levels.

So, if $5.0 \cdot {10}^{- 4}$ moles of acid are consumed by the reaction, you can say that the resulting solution will contain

${n}_{{H}_{3} {O}^{+}} = {\text{0 moles H"_3"O}}^{+} \to$ completely consumed

${n}_{B O H} = \left(x - 5.0 \cdot {10}^{- 4}\right) \textcolor{w h i t e}{a} \text{moles BOH}$

${n}_{{B}^{+}} = 0 + 5.0 \cdot {10}^{- 4} = 5.0 \cdot {10}^{- 4} \textcolor{w h i t e}{a} {\text{moles B}}^{+}$

The total volume of the solution will be

${V}_{\text{total" = "40 mL" + "5.0 mL" = "45 mL}}$

The concentrations of the weak base and conjugate acid will thus be

["BOH"] = (x - 5.0 * 10^(-4)"moles")/(45 * 10^(-3)"L") = (x-5.0 * 10^(-4)) * 10^3/45color(white)(a)"mol L"^(-1)

["B"^(+)] = (5.0 * 10^(-4)"moles")/(45 * 10^(-3)"L") = 1.11 * 10^(-2)color(white)(a)"mol L"^(-1)

Now, the dissociation of the weak base can be described by the following equilibrium reaction

${\text{BOH"_ ((aq)) rightleftharpoons "B"_ ((aq))^(+) + "OH}}_{\left(a q\right)}^{-}$

Use the pH of the solution to find its pOH

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

In this case, you will have

$\text{pOH} = 14 - 10.04 = 3.96$

As you know, the concentration of hydroxide anions can be found using

$\textcolor{p u r p \le}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\left[\text{OH"^(-)] = 10^(-"pOH}\right)} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

This will get you

["OH"^(-)] = 10^(-3.96) = 1.10 * 10^(-4)"M"

By definition, the base dissociation constant will be equal to

${K}_{b} = \left(\left[\text{B"^(+)] * ["OH"^(-)])/(["BOH}\right]\right)$

Plug in these values to get

${K}_{b} = \frac{1.11 \cdot {10}^{- 2} \cdot 1.10 \cdot {10}^{- 4}}{\left(x - 5.0 \cdot {10}^{- 4}\right) \cdot {10}^{3} / 45} = \frac{5.5 \cdot {10}^{- 8}}{x - 5.0 \cdot {10}^{- 4}} \text{ " " } \textcolor{\mathmr{and} a n \ge}{\left(1\right)}$

Now do the same thing for the second hydrochloric acid sample. Keep in mind that you're not adding a separate sample of $\text{20.0 mL}$ of acid.

The total volume of acid added to the weak base solution will be equal to $\text{20.0 mL}$, which implies that you're adding $\text{15.0 mL}$ of acid for this second reaction.

This sample will contain

n_(H_3O^(+)) = "0.1 mol" color(red)(cancel(color(black)("L"^(-1)))) * overbrace(15.0 * 10^(-3)color(red)(cancel(color(black)("L"))))^(color(purple)("volume in liters"))

$= 1.5 \cdot {10}^{- 3} {\text{moles H"_3"O}}^{+}$

Once again, the moles of hydronium ions coming from the acid will be completely consumed by the reaction. This time, the solution already contained $\left(x - 5.0 \cdot {10}^{- 4}\right)$ moles of $\text{BOH}$ and $5.0 \cdot {10}^{- 4}$ moles of ${\text{B}}^{+}$, which means that after the reaction is complete, you will be left with

${n}_{{H}_{3} {O}^{+}} = {\text{0 moles H"_3"O}}^{+} \to$ completely consumed

${n}_{B O H} = x - 5.0 \cdot {10}^{- 4} - 1.5 \cdot {10}^{- 3} = \left(x - 2.0 \cdot {10}^{- 3}\right) \textcolor{w h i t e}{a} \text{moles BOH}$

${n}_{{B}^{+}} = 5.0 \cdot {10}^{- 4} + 1.5 \cdot {10}^{- 3} = 2.0 \cdot {10}^{- 3} \textcolor{w h i t e}{a} {\text{moles B}}^{+}$

The total volume of the solution will now be

${V}_{\text{total" = "45 mL" + "15 mL" = "60 mL}}$

The pOH of the solution, knowing that the pH of the solution is equal to $9.14$, will be

$\text{pOH} = 14 - 9.14 = 4.86$

This means that you have

["OH"^(-)] = 10^(-4.86) = 1.4 * 10^(-5)"M"

The concentrations of the weak base and of its conjugate acid will now be

["BOH"] = (x - 2.0 * 10^(-3)"moles")/(60 * 10^(-3)"L") = (x-2.0 * 10^(-3)) * 10^3/60color(white)(a)"mol L"^(-1)

["B"^(+)] = (2.0 * 10^(-3)"moles")/(60 * 10^(-3)"L") = 3.33 * 10^(-2)color(white)(a)"mol L"^(-1)#

The base dissociation constant will be equal to

${K}_{b} = \frac{3.33 \cdot {10}^{- 2} \cdot 1.4 \cdot {10}^{- 5}}{\left(x - 2.0 \cdot {10}^{- 3}\right) \cdot {10}^{3} / 60} = \frac{2.8 \cdot {10}^{- 8}}{x - 2.0 \cdot {10}^{- 3}} \text{ " " } \textcolor{\mathmr{and} a n \ge}{\left(2\right)}$

You now have two equations with two unknowns, $x$ and ${K}_{b}$. Use these equations to write

$\textcolor{\mathmr{and} a n \ge}{\left(1\right)} = \textcolor{\mathmr{and} a n \ge}{\left(2\right)} \implies \frac{5.5 \cdot \textcolor{red}{\cancel{\textcolor{b l a c k}{{10}^{- 8}}}}}{x - 5.0 \cdot {10}^{- 4}} = \frac{2.8 \cdot \textcolor{red}{\cancel{\textcolor{b l a c k}{{10}^{- 8}}}}}{x - 2.0 \cdot {10}^{- 3}}$

Rearrange to get

$5.5 \cdot \left(x - 2.0 \cdot {10}^{- 3}\right) = 2.8 \cdot \left(x - 5.0 \cdot {10}^{- 4}\right)$

$5.5 x - 2.8 x = 11 \cdot {10}^{- 3} - 1.4 \cdot {10}^{- 3}$

$x = \frac{9.6 \cdot {10}^{- 3}}{2.7} = 3.56 \cdot {10}^{- 3}$

Plug this into equation $\textcolor{\mathmr{and} a n \ge}{\left(1\right)}$ to get the value of ${K}_{b}$

${K}_{b} = \frac{5.5 \cdot {10}^{- 8}}{3.56 \cdot {10}^{- 3} - 5.0 \cdot {10}^{- 4}} = \textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} 1.8 \cdot {10}^{- 5} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

I'll leave the answer rounded to two sig figs.

Jul 4, 2016

This is an alternative to the answer originally provided by Stefan V. This solution utilizes the properties of buffer solutions.

Explanation:

We know that because the pH remains above 7.0, the neutralization is only partial, and therefore we have a buffer solution. For any buffer consisting of a conjugate acid-base pair, we know that
$p H = p {K}_{a} + \log \left[\frac{b a s e}{a c i d}\right]$
or
$p O H = p {K}_{b} + \log \left[\frac{a c i d}{b a s e}\right]$

Here, we'll use the second version, where $\left[b a s e\right] = \left[B O H\right] \mathmr{and} \left[a c i d\right] = \left[{B}^{+}\right]$

Let $x =$ moles of base in the original solution.

After the first addition of $0.005 L \times 0.1 N = 0.0005$ mol HCl,
we have $\left[b a s e\right] = \frac{x - 0.0005}{0.045} M \mathmr{and} \left[a c i d\right] = \frac{0.0005}{0.045} M$
and
$p O H = 14 - 9.14 = 3.96 = p {K}_{b} + \log \left(\frac{0.0005}{x - 0.0005}\right)$

Similarly, after the second addition of $0.002$ mol HCl we have
$\left[b a s e\right] = \frac{x - 0.002}{0.060} M \mathmr{and} \left[a c i d\right] = \frac{0.002}{0.06} M$
and
$p O H = 4.86 = p {K}_{b} + \log \left(\frac{0.002}{x - 0.002}\right)$

We now have 2 equations and two unknowns, so subtracting the first from the second gives

$0.90 = \log \left[\frac{0.002 \left(x - 0.0005\right)}{0.0005 \left(x - 0.002\right)}\right]$
Take the anti-log of both sides and rearrange to give $x = 3.52 \times {10}^{- 3}$ mol

Plugging this back into either of the pOH equations above gives

$p {K}_{b} = 4.74$
or
${K}_{b} = 1.82 \times {10}^{- 5}$