# Question bdf57

Apr 29, 2016

$\text{pH} = 5.13$

#### Explanation:

The idea here is that the ammonium cation, ${\text{NH}}_{4}^{+}$, acts as a weak acid in aqueous solution, so right from the start you should expect the pH of the solution to be lower than $7$.

Ammonium chloride, $\text{NH"_4"Cl}$, is a soluble ionic compound that dissociates completely in aqueous solution to produce ammonium cations and chloride anions, ${\text{Cl}}^{-}$

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

Since the salt dissociates in a $1 : 1$ mole ratio to form ammonium cations, you can say that

["NH"_4^(+)] = ["NH"_4"Cl" ] = "0.1 mol dm"^(-3)

Now, in order to be able to calculate the pH of the solution, you need to know the acid dissociation constant, ${K}_{a}$, of the ammonium cation, which you'll find listed as

${K}_{a} = 5.6 \cdot {10}^{- 10}$

http://clas.sa.ucsb.edu/staff/Resource%20folder/Chem109ABC/Acid,%20Base%20Strength/Table%20of%20Acids%20w%20Kas%20and%20pKas.pdf

Once in aqueous solution, the ammonium cations will react with water to form ammonia, ${\text{NH}}_{3}$, and hydronium cations, ${\text{H"_3"O}}^{+}$.

Now, because the acid dissociation constant is so small, let us at first include the dissociation of water, given by

$\left[{\text{H"_3"O"^+]["O""H}}^{-}\right] = 1 \cdot {10}^{- 14}$ in molar units, just to be on the safe side. With the autodisdociation included, we then have the following definitions:

$\left[{\text{N""H}}_{3}\right] = x$, the concentration of molecular ammonia produced by the acid-base reaction

$\left[{\text{H"_3"O}}^{+}\right] = x + y$, the concentration of hydronium ions produced by the acid-base reaction plus the dissociation of water.

$\left[{\text{N""H}}_{4}^{+}\right] = 0.1 - x$, the concentration of ammonium ions left over after the acid-base reaction

$\left[{\text{O""H}}^{-}\right] = y$, the concentration of hydroxide ions produced by the water autodissociation.

We now use an ICE table to help you find the equilibrium concentration of the hydronium 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)(aaaaacolor(black)(0.1)aaaaaaaaaaaaaaaaaaaaacolor(black)(0)aaaaaaaacolor(black)(y)
color(purple)("C")color(white)(aaacolor(black)((-x))aaaaaaaaaaaaaaaaacolor(black)((+x))aaaaacolor(black)((+x))
color(purple)("E")color(white)(aaacolor(black)(0.1-x)aaaaaaaaaaaaaaaaaaacolor(black)(x)aaaacolor(black)(x+y)

By definition, the acid dissociation constant will be

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

In your case, this will be equivalent to

${K}_{a} = \frac{x \cdot \left(1 \cdot {10}^{- 7} + x\right)}{0.1 - x} = 5.6 \cdot {10}^{- 10}$

Once again, because ${K}_{a}$ is so small compared with the the initial concentration of the ammonium cations, you can use the approximation

$0.1 - x \approx 0.1$

This will get you

$5.6 \cdot {10}^{- 10} = \frac{x \cdot \left(x + y\right)}{0.1}$

$\textcolor{p u r p \le}{x \cdot \left(x + y\right) = 5.6 \cdot {10}^{- 11}}$

Now we need another equation to get $y$. This comes from the autodissociation equlibrium for water:

$\left[{\text{H"_3"O"^+]["O""H}}^{-}\right] = \textcolor{b r o w n}{\left(x + y\right) \cdot y = 1.0 \cdot {10}^{- 14}}$

We divide the brown equation by the purple one and see that the factor $\left(x + y\right)$ cancels out. So we end up with

$y = 0.000179 x$

And out this back into the purple equation for the ammonium ion hydrolysis:

$\textcolor{p u r p \le}{x \cdot \left(x + 0.00018 x\right) = 1.000179 {x}^{2} = 5.6 \cdot {10}^{- 11}}$

Then take the square root. Remember that $x$, the amount of avid-base reaction, must be positive

$x = \setminus \sqrt{\frac{5.6 \cdot {10}^{- 11}}{1.000179}} = 7.48 \cdot {10}^{- 6} M$

$y = 0.000179 x = 1.35 \cdot {10}^{- 9} M = 0.00134 \cdot {10}^{- 6} M$

Observe that the $y$ value is much less than the ${10}^{- 7}$ M we'd expect for autodissociation in pure water. The weak acid put more hydronium ions into the solution, which gobbled up most of the hydroxide ions and canceled out most of the usual autodissociation. It's an example of Le Chatelier's principle.

The total concentration of hydronium cations present in solution at equilibrium will be

["H"_3"O"^(+)] = overbrace(7.48 * 10^(-6)"M")^(color(purple)("x, coming from NH"_4^(+))) + overbrace(1.34 * 10^(-9)"M")^(color(brown)("y, coming from the self-ionization of water"))

["H"_3"O"^(+)] = 7.48 * 10^(-6)"M"

Here

${\text{M" = "mol dm}}^{- 3}$

As you know, pH is calculated using

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

In your case, you will have

$\text{pH} = - \log \left(7.47 \cdot {10}^{- 6}\right) = \textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} 5.13 \textcolor{w h i t e}{\frac{a}{a}} |}}}$

SIDE NOTE Even with this small dissociation constant, there is so little water dissociation that the concentration of hydronium ion, $x + y$, is almost the same as just the ammonia concentration $x$. This is a good approximation as long as the product of the concentration and the dissociation constant is at least ${10}^{- 12}$.