# How do you calculate the change in pH when 3.00 mL of 0.100 M "HCl"(aq) is added to 100.0 mL of a buffer solution that is 0.100 M in "NH"_3(aq) and 0.100 M in "NH"_4"Cl"(aq) ?

Aug 2, 2016

${\Delta}_{\text{pH}} = - 0.026$

#### Explanation:

You're adding hydrochloric acid, $\text{HCl}$, a strong acid, to your buffer, so right from the start you should expect it pH to decrease.

This implies that the change in pH will be negative

${\Delta}_{\text{pH" = "pH"_ "final" - "pH"_ "initial}} < 0$

However, the fact that you're dealing with a buffer solution lets you know that this change will not be significant since the role of a buffer is to resist significant changes in pH that result from the addition of strong acid or strong bases.

So, hydrochloric acid will react with ammonia, ${\text{NH}}_{3}$, a weak base, to produce the ammonium cation, ${\text{NH}}_{4}^{+}$, the ammonia's conjugate acid, and water.

${\text{HCl"_ ((aq)) + "NH"_ (3(aq)) -> "NH"_ (4(aq))^(+) + "Cl}}_{\left(a q\right)}^{-}$

The reaction consumes hydrochloric acid and ammonia in a $1 : 1$ mole ratio. Also, notice that for very mole of hydrochloric acid or ammonia consumed by the reaction, one mole of ammonium cations is produced. Keep this in mind.

Use the molarity of the ammonia solution and the volume of the buffer to calculate how many moles it contains

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

You will have

n_("NH"_3) = "0.100 mol" color(red)(cancel(color(black)("L"^(-1)))) * 100.0 * 10^(-3)color(red)(cancel(color(black)("L")))

$= {\text{0.0100 moles NH}}_{3}$

Do the same for the ammonium cations, ${\text{NH}}_{4}^{+}$

n_("NH"_4^(+)) = "0.100 mol" color(red)(cancel(color(black)("L"^(-1)))) * 100.0 * 10^(-3)color(red)(cancel(color(black)("L")))

$= {\text{0.0100 moles NH}}_{4}^{+}$

Calculate how many moles of hydrochloric acid are being added to the buffer

n_("HCl") = "0.100 mol" color(red)(cancel(color(black)("L"^(-1)))) * 3.00 * 10^(-3)color(red)(cancel(color(black)("L")))

$= \text{0.000300 moles HCl}$

You know that hydrochloric acid and ammonia react in a $1 : 1$ mole ratio, which means that the resulting solution will contain

n_("HCl") = "0 moles HCl" -> completely consumed

n_("NH"_3) = "0.0100 moles" - "0.000300 moles"

$= {\text{0.0097 moles NH}}_{3}$

n_("NH"_4^(+)) = "0.0100 moles" + "0.000300 moles"

$= {\text{0.0103 moles NH}}_{4}^{+}$

The total volume of the buffer will be

${V}_{\text{total" = "100.0 mL" + "3.00 mL" = "103.0 mL}}$

The new concentrations of ammonia and ammonium cations will be

["NH"_3] = "0.0097 moles"/(103.0 * 10^(-3)"L") = "0.094175 M"

["NH"_4^(+)] = "0.0103 moles"/(103.0 * 10^(-3)"L") = "0.100 M"

Notice that the concentration of ammonium cations remained virtually unchanged because the increase in the number of moles was counteracted by the increase in volume.

Now, you can find the change in pH by using the Henderson - Hasselbalch equation, which for a buffer that contains a weak base and its conjugate acid looks like this

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

Notice that the the initial buffer solution had equal concentrations of weak base and conjugate acid. This means that its pOH was equal to

"pOH" = "p"K_b + log( color(red)(cancel(color(black)("0.100 M")))/(color(red)(cancel(color(black)("0.100 M")))))

$\text{pOH" = "p} {K}_{b}$

After the strong acid is added to the buffer, the pOH of the buffer will be

"pOH" = "p"K_b + log( (0.100 color(red)(cancel(color(black)("M"))))/(0.095175color(red)(cancel(color(black)("M")))))

$\text{pOH" = "p} {K}_{b} + 0.026$

You know that an aqueous solution at room temperature has

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

For the initial solution, you have

$\text{pH"_ "initial" = 14 - "p} {K}_{b}$

For the final solution, you have

"pH"_ "final" = 14 - ("p"K_b + 0.026)

$\text{pH"_ "final" = 14 - "p} {K}_{b} - 0.026$

Therefore, you can say that the change in pH is equal to

Delta_ "pH" = color(red)(cancel(color(black)(14))) - color(red)(cancel(color(black)("p"K_b))) - 0.026 - color(red)(cancel(color(black)(14))) + color(red)(cancel(color(black)("p"K_b)))

$\textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{{\Delta}_{\text{pH}} = - 0.026} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

As predicted, the change in pH is negative because the pH of the buffer decreased as a result of the addition of a strong acid.