# What is the pH at the equivalence point when 35.0 mL of 0.20 M ammonia is titrated by 0.12M hydrochloric acid? K_b for ammonia is 1.8 xx 10^(-5)

Jun 1, 2016

$\text{pH} = 5.19$

#### Explanation:

Ammonia, ${\text{NH}}_{3}$, will react with hydrochloric acid, $\text{HCl}$, to produce aqueous ammonium chloride, $\text{NH"_4"Cl}$, according to the following balanced chemical equation

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

The two reactants react in a $1 : 1$ mole ratio, so right from the start you know that a complete neutralization requires equal numbers of moles of ammonia and of hydrochloric acid.

In other words, you need equal numbers of moles of ammonia and of hydrochloric acid to get to the equivalence point.

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

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

n_(NH_3) = "0.20 mol" color(red)(cancel(color(black)("L"^(-1)))) * overbrace(35.0 * 10^(-3)color(red)(cancel(color(black)("L"))))^(color(blue)("volume in liters"))

${n}_{N {H}_{3}} = {\text{0.0070 moles NH}}_{3}$

This means that the hydrochloric acid solution must also contain $0.0070$ moles of hydrochloric acid. Use this to determine the volume of $\text{0.12 M}$ hydrochloric acid solution needed

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

You will have

V_(HCl) = (0.0070 color(red)(cancel(color(black)("moles"))))/(0.12color(red)(cancel(color(black)("mol")))"L"^(-1)) = "0.0583 L" = "58.3 mL"

The volume of the resulting solution will thus be

${V}_{\text{total}} = {V}_{N {H}_{3}} + {V}_{H C l}$

${V}_{\text{total" = "35.0 mL" + "58.3 mL" = "93.3 mL}}$

Now, notice that the reaction produces aqueous ammonium chloride in $1 : 1$ mole ratios with the two reactants.

If the reaction consumes $0.0070$ moles of ammonia and $0.0070$ moles of hydrochloric acid, it follows that it will produce $0.0070$ moles of ammonium chloride.

Ammonium chloride dissociates completely in aqueous solution to form ammonium cations, ${\text{NH}}_{4}^{+}$, and chloride anion, ${\text{Cl}}^{-}$, in $1 : 1$ mole ratios.

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

The concentration of the ammonium cations in the resulting solution will be

["NH"_4^(+)] = "0.0070 moles"/(93.3 * 10^(-3)"L") = "0.07503 M"

The ammonium cations will hydrolyze to form hydronium cations ,${\text{H"_3"O}}^{+}$. Use an ICE table to find the equilibrium concentration of these cations

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

color(purple)("I")color(white)(aaacolor(black)(0.07503)aaaaaaaaaaaaaacolor(black)(0)aaaaaaaaaaaaacolor(black)(0)
color(purple)("C")color(white)(aaacolor(black)((-x))aaaaaaaaaaaaacolor(black)((+x))aaaaaaaaacolor(black)((+x))
color(purple)("E")color(white)(aacolor(black)(0.07503-x)aaaaaaaaaaacolor(black)(x)aaaaaaaaaaaaacolor(black)(x)

By definition, the acid dissociation constant, ${K}_{a}$, will be

${K}_{a} = \left(\left[{\text{NH"_3] * ["H"_3"O"^(+)])/(["NH}}_{4}^{+}\right]\right)$

In this case, you will have

${K}_{a} = \frac{x \cdot x}{0.07503 - x} = {x}^{2} / \left(0.07503 - x\right)$

Now, you know that for aqueous solutions at room temperature, you have the following relationship between ${K}_{b}$ and ${K}_{a}$

$\textcolor{p u r p \le}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{{K}_{a} \times {K}_{b} = {K}_{W}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

Here

${K}_{W} = {10}^{- 14} \to$ the ionization constant of water

This means that you have

${K}_{a} = {K}_{W} / {K}_{b}$

${K}_{a} = {10}^{- 14} / \left(1.8 \cdot {10}^{- 5}\right) = 5.56 \cdot {10}^{- 10}$

Since ${K}_{a}$ is very, very small compared with the concentration of the ammonium cations, you can use the approximation

$0.07503 - x \approx 0.07503$

This will get you

${K}_{a} = {x}^{2} / 0.07503 = 5.56 \cdot {10}^{- 10}$

Solve for $x$ to find

$x = \sqrt{0.07503 \cdot 5.56 \cdot {10}^{- 10}} = 6.46 \cdot {10}^{- 6}$

Since $x$ represents the equilibrium concentration of the hydronium cations, you will have

["H"_3"O"^(+)] = 6.46 * 10^(-6)"M"

As you know, the pH of the solution is given by

color(blue)(|bar(ul(color(white)(a/a)"pH" = - log(["H"_3"O"^(+)])color(white)(a/a)|)))

This means that you have

$\text{pH} = - \log \left(6.46 \cdot {10}^{- 6}\right) = \textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{5.19} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

Finally, the result makes sense because neutralizing a weak base such as ammonia with a strong acid will result in the formation of the conjugate acid of the base, which will in turn cause the pH to be lower than $7$, i.e. the solution will be acidic.