# Question 92142

May 23, 2016

${\Delta}_{\text{pH}} = 0.050$

#### Explanation:

Your buffer solution contains ammonia, ${\text{NH}}_{3}$, a weak base, and ammonium chloride, $\text{NH"_4"Cl}$, the salt of its conjugate base, the ammonium cation, ${\text{NH}}_{4}^{+}$.

The pOH of the buffer can be calculated using the Henderson - Hasselbalch equation

color(blue)(|bar(ul(color(white)(a/a)"pOH" = pK_b + log((["conjugate acid"])/(["weak base"]))color(white)(a/a)|)))

Here you have

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} p {K}_{b} = - \log \left({K}_{b}\right) \textcolor{w h i t e}{\frac{a}{a}} |}}}$

where ${K}_{b}$ is the base dissociation constant.

In your case, the base dissociation constant for ammonia is equal to

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

http://www.bpc.edu/mathscience/chemistry/table_of_weak_bases.html

Use this to calculate the pOH, and subsequently the pH of the buffer before the addition of the strong acid

"pOH" = -log(1.8 * 10^(-5)) + log((0.100color(red)(cancel(color(black)("M"))))/(0.100color(red)(cancel(color(black)("M")))))

$\text{pOH} = 4.74$

The pH of the solution is given by

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

In this case, you would have

$\text{pH} = 14 - 4.74 = 9.26$

So, hydrochloric acid, $\text{HCl}$, is a strong acid that ionizes completely in aqueous solution to form hydronium cations, ${\text{H"_3"O}}^{+}$, and chloride anions, ${\text{Cl}}^{-}$.

Once in solution, the hydronium cation that result from the ionization of the strong acid will react with the ammonia to form ammonium cations and water

${\text{NH"_ (3(aq)) + "H"_ 3"O"_ ((aq))^(+) -> "NH"_ (4(aq))^(+) + "H"_ 2"O}}_{\left(l\right)}$

As you can see, this reaction consumes ammonia and hydronium cations and produces ammonium cations in a $1 : 1$ mole ratio.

Use the molarity and volume of the hydrochloric acid solution to determine how many moles of hydronium cations are delivered to the buffer

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

You will have

n_(H_3O^(+)) = "0.100 mol" color(red)(cancel(color(black)("L"^(-1)))) * overbrace(6.00 * 10^(-3)color(red)(cancel(color(black)("L"))))^(color(blue)("volume in liters"))

${n}_{{H}_{3} {O}^{+}} = {\text{0.00060 moles H"_3"O}}^{+}$

Now, the initial buffer solution contained

n_(NH_3) = "0.100 mol" color(red)(cancel(color(black)("L"^(-1)))) * overbrace(100.0 * 10^(-3)color(red)(cancel(color(black)("L"))))^(color(blue)("volume in liters"))

${n}_{N {H}_{3}} = {\text{0.010 moles NH}}_{3}$

and

n_(NH_4^(+)) = "0.100 mol" color(red)(cancel(color(black)("L"^(-1)))) * overbrace(100.0 * 10^(-3)color(red)(cancel(color(black)("L"))))^(color(blue)("volume in liters"))

${n}_{N {H}_{4}^{+}} = {\text{0.010 moles NH}}_{4}^{+}$

Since you have fewer moles of hydronium cations coming from the strong acid than moles of ammonia, you know for a fact that the hydronium cations will be completely consumed by the reaction.

After the strong acid is added, the solution will contain

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

${n}_{N {H}_{3}} = {\text{0.010 moles" - "0.00060 moles" = "0.0094 moles NH}}_{3}$

${n}_{N {H}_{4}^{+}} = {\text{0.010 moles" + "0.00060 moles" = "0.0106 moles NH}}_{4}^{+}$

The total volume of the solution will now be

${V}_{\text{total" = "100.0 mL" + "6.00 mL" = "106.0 mL}}$

Use this volume to calculate the concentrations of ammonia and nitrate cations after the reaction

["NH"_3] = "0.0094 moles"/(106.0 * 10^(-3)"L") = "0.08868 M"

["NH"_4^(+)] = "0.0106 moles"/(106.0 * 10^(-3)"L") = "0.100 M"

The pOH of the solution will now be

"pOH"_"new" = -log(1.8 * 10^(-5)) + log( (0.100color(red)(cancel(color(black)("M"))))/(0.08868color(red)(cancel(color(black)("M")))))#

$\text{pOH"_"new} = 4.79$

The PH of the solution will be

$\text{pH"_"new} = 14 - 4.79 = 9.21$

The change in pH will thus be

${\Delta}_{\text{pH" = |"pH"_"new" - "pH}} |$

${\Delta}_{\text{pH}} = | 9.21 - 9.26 | = \textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} 0.050 \textcolor{w h i t e}{\frac{a}{a}} |}}}$

So, does this result make sense?

You're adding a strong acid to the original buffer solution, so right from the start you should expect its pH to decrease.

However, because the purpose of a buffer is to resist significant changes in pH upon the addition of strong acids or strong bases, you can see that the change in pH is very, very small.

The pH decreases upon the addition of the acid, but not by a significant amount.