# Question #d9358

Apr 19, 2015

To solve this problem, you'll need the value of the base dissociation constant, ${K}_{b}$, of ammonia, which is listed as $1.77 \cdot {10}^{- 5}$.

Moreover, you're going to be dealing with a buffer that consists of a weak base, ammonia, and its conjugate acid, ammonium chloride, $N {H}_{4} C l$ (the useful form will be $N {H}_{4}^{+}$), which means that you can use the Henderson-Hasselbalch equation in the form

$p O H = p {K}_{b} + \log \left(\frac{\left[N {H}_{4}^{+}\right]}{\left[N {H}_{3}\right]}\right)$

Calculate the $p {K}_{b}$ by

$p {K}_{b} = - \log \left({K}_{b}\right) = - \log \left(1.77 \cdot {10}^{- 5}\right) = 4.75$

Now you're good to go. in order, the pH of the solution at various points in the titration will be

• Before adding $H C l$

Since ammonia is a weak base, you can expect the pH of the solution to be basic, or greater than 7. An equilibrium will be established that will allow you to calculate the concentration of the hydroxide ion, $O {H}^{-}$

$\text{ } N {H}_{3 \left(a q\right)} + {H}_{2} {O}_{\left(l\right)} r i g h t \le f t h a r p \infty n s N {H}_{4 \left(a q\right)}^{+} + O {H}_{\left(a q\right)}^{-}$
I........0.1..........................................0................0
C.....(-x)..........................................(+x).............(+x)
E.....0.1-x.........................................x.................x

By definition, the acid dissociation constant will be

${K}_{b} = \frac{\left[N {H}_{4}^{+}\right] \cdot \left[O {H}^{-}\right]}{\left[N {H}_{3}\right]} = \frac{x \cdot x}{0.1 - x} = {x}^{2} / \left(0.1 - x\right)$

Because ${K}_{b}$ is so small, you can approximate the term $\left(0.1 - x\right)$ with $0.1$, which will produce

${K}_{b} = {x}^{2} / 0.1 \implies x = 0.00133$

Calculate the solution's pOH by

$p O H = - \log \left(\left[O {H}^{-}\right]\right) = - \log \left(0.00133\right) = 2.88$

As a result, the pH will be

$p {H}_{\text{sol}} = 14 - p O H = 14 - 2.88 = \textcolor{g r e e n}{11.12}$

• After the addition of 10 mL of $H C l$

Now you start adding acid to the solution. Ammonia will react with the strong acid to produce ammonium chloride, which will act as the conjugate acid $\to$ a buffer is formed.

$N {H}_{3 \left(a q\right)} + H C {l}_{\left(a q\right)} \to N {H}_{4} C {l}_{\left(a q\right)}$

The net ionic equation is

$N {H}_{3 \left(a q\right)} + {H}_{\left(a q\right)}^{+} \to N {H}_{4 \left(a q\right)}^{+}$

Calculate the number of moles of ammonia you initially had in solution

$C = \frac{n}{V} \implies n = C \cdot V$

${n}_{\text{ammonia initial" = "0.1 M" * 25 * 10^(-3)"L" = "0.0025 moles}}$

The number of moles of hydrochloric acid added will be

${n}_{\text{HCl" = C * V = "0.1 M" * 10 * 10^(-3)"L" = "0.0010 moles}}$

All the acid $H C l$ will be consumed in the reaction, which means that the number of moles of ammonia will decrease by 0.0010, and 0.0010 moles of $N {H}_{4}^{+}$ will be produced.

The volume of the solution will now be

${V}_{\text{sol" = V_"initial" + V_"HCl" = 25 + 10 = "35 mL}}$

The concentrations of ammonia and ammonium chloride will be

$\left[N {H}_{3}\right] = \text{(0.0025 - 0.0010) moles"/(35 * 10^(-3)"L") = "0.04286 M}$

$\left[N {H}_{4}^{+}\right] = \text{0.0010 moles"/(35 * 10^(-3)"L") = "0.02857 M}$

The pOH of the solution will be

$p O H = p {K}_{b} + \log \left(\frac{\left[N {H}_{4}^{+}\right]}{\left[N {H}_{3}\right]}\right) = 4.75 + \log \left(\left(0.02857 \cancel{\text{M"))/(0.04286cancel("M}}\right)\right)$

$p O H = 4.57 \implies p {H}_{\text{sol}} = 14 - 4.57 = \textcolor{g r e e n}{9.43}$

• After half of the ammonia has been neutralized

When half of the ammonia is neutralized, the number of moles of ammonia that remanin in solution will be equal to the number of moles of ammonium chloride produced.

Since the volume is always the same for both species, this implies that $\left[N {H}_{3}\right] = \left[N {H}_{4}^{+}\right]$.

$p O H = 4.75 + \log \left(1\right) = 4.75$

$p {H}_{\text{sol}} = 14 - 4.75 = \textcolor{g r e e n}{9.25}$

• At the equivalence point

At the equivalence point, all the ammonia has been consumed by the strong acid. This implies that ammonium chloride will be the only species left in solution.

Moreover, the number of moles of ammonium chloride produced will be equal to the initial number of moles of ammonia, 0.0025.

The total volume of the solution will be

${V}_{\text{total" = 2 * V_"initial" = 2 * 25 = "50 mL}}$

The molarity of the ammonium chloride will be (because the ammonia solution and the hydrochloric acid solution had equal molarities)

$C = \frac{n}{V} = \text{0.0025 moles"/(50 * 10^(-3)"L") = "0.05 M}$

The following equilibrium will be established

$\text{ } N {H}_{4 \left(a q\right)} + {H}_{2} {O}_{\left(l\right)} r i g h t \le f t h a r p \infty n s N {H}_{3 \left(a q\right)} + {H}_{3} {O}_{\left(a q\right)}^{+}$
I.....0.05.........................................0.................0
C.....(-x)..........................................(+x)..............(+x)
E....0.05-x......................................x..................x

This time, use the acid dissociation constant, ${K}_{a}$

${K}_{a} = {K}_{w} / {K}_{b} = {10}^{- 14} / \left(1.77 \cdot {10}^{- 5}\right) = 5.65 \cdot {10}^{- 10}$

${K}_{a} = \frac{\left[N {H}_{3}\right] \cdot \left[{H}_{3} {O}^{+}\right]}{\left[N {H}_{4}^{+}\right]} = {x}^{2} / \left(0.05 - x\right) = {x}^{2} / 0.05$

Solving for $x$ will give you the concentration of the hydronium ions, ${H}_{3} {O}^{+}$

$x = 5.3 \cdot {10}^{- 6}$

Therefore, the final pH of the solution will be

$p {H}_{\text{final}} = - \log \left(5.3 \cdot {10}^{- 6}\right) = \textcolor{g r e e n}{5.28}$