# What is the "pH" of a "1.5-M" solution of ammonia? The dissociation constant of ammonia at 25.0^@"C" is 1.80 * 10^-5.

Apr 16, 2018

$\text{pH} = 11.72$

#### Explanation:

As you know, ammonia acts as a weak base in aqueous solution, so right from the start, you should expect the $\text{pH}$ of the solution to be $> 7$.

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

The ratio that exists between the equilibrium concentrations of the ammonium cations and of the hydroxide anions and the equilibrium concentration of ammonia is given by the base dissociation constant, ${K}_{b}$.

${K}_{b} = \left(\left[{\text{NH"_4^(+)] * ["OH"^(-)])/(["NH}}_{3}\right]\right)$

Now, ammonia will only partially ionize to produce ammonium cations and hydroxide anions. If you take $x$ $\text{M}$ to be the concentration of ammonia that ionizes, you can say that, at equilibrium, the solution will contain

["NH"_4^(+)] = ["OH"^(-)] = x quad "M"

This happens because every mole of ammonia that ionizes produces $1$ mole of ammonium cations and $1$ mole of hydroxide anions.

So if $x$ $\text{M}$ ionizes, you can expect the solution to contain $x$ $\text{M}$ of the two ions.

The solution will also contain

["NH"_3] = (1.5 - x) quad "M"

When $x$ $\text{M}$ ionizes, the initial concentration of ammonia will decrease by $x$ $\text{M}$.

This means that the expression of the base dissociation constant will now take the form

${K}_{b} = \frac{x \cdot x}{1.5 - x}$

which is equal to

$1.80 \cdot {10}^{- 5} = {x}^{2} / \left(1.5 - x\right)$

Notice that the value of the base dissociation constant is significantly smaller than the initial concentration of the base. This tells you that you can use the approximation

$1.5 - x \approx 1.5$

because the concentration of ammonia that ionizes will be significantly lower than the initial concentration of the base, i.e. the ionization equilibrium will lie to the left.

You now have

$1.80 \cdot {10}^{- 5} = {x}^{2} / 1.5$

Rearrange and solve for $x$ to get

$x = \sqrt{1.5 \cdot 1.80 \cdot {10}^{- 5}} = 0.005196$

Since $x$ $\text{M}$ represents the equilibrium concentration of the hydroxide anions, you can say that

["OH"^(-)] = "0.005196 M"

Now, an aqueous solution at ${25}^{\circ} \text{C}$ has

$\text{pH + pOH = 14}$

Since

"pOH" = - log(["OH"^(-)])

you can say that the $\text{pH}$ of the solution is given by

"pH" = 14 + log(["OH"^(-)])

Plug in your value to find

$\text{pH} = 14 + \log \left(0.005196\right) = \textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{11.72}}}$

The answer is rounded to two decimal places because you have two sig figs for the molarity of the solution.