# Solution of a Weak Base

### In-class Problem Set #1

Unless specifically told otherwise, whenever a problem lists a concentration, that is the value of material added to solution prior to any reactions occurring to achieve equilibrium.  So in the first problem below, 0.155 moles of ammonia were dissolved in 1 liter of solution.  The final concentration of ammonia would be something less than 0.155 moles/liter provided some form of equilibration occurred.

1.  Calculate the pH of a solution that is 0.155 M in ammonia.

The first step in any equilibrium problem is an assessment of the relevant chemical reactions that occur in the solution.  To determine the relevant reactions, one must examine the specie(s) given in the problem and determine which types of reactions might apply.  In particular, we want to consider the possibility of acid-base reactions, solubility of sparingly soluble solids, or formation of water-soluble metal complexes.

When given the name of a compound (e.g., ammonia), it is essential that we know or find out the molecular formula for the compound, and often times we have to look this up in a book or table.  The molecular formula for ammonia is NH3.  Ammonia can be viewed as the building block for a large family of similar compounds called amines in which one or more of the hydrogen atoms are replaced with other functional groups (a functional group is essentially a cluster of atoms - most of these are carbon-containing clusters).  For example, the three compounds below result from replacing the hydrogen atoms of ammonia with methyl (CH3) groups.

CH3NH2          Methyl amine

(CH3)2NH       Dimethyl amine

(CH3)3N          Trimethyl amine

Amines and many other organic, nitrogen-containing compounds constitute one of the major families of bases.  Ammonia is therefore a base.

Bases undergo a very specific reaction with water to produce the hydroxide ion.  The appropriate reaction needed to describe what will happen when ammonia is mixed with water is shown below.

$\ce{NH3 + H2O \leftrightarrow NH4+ + OH-}$

We can describe this reaction by saying that ammonia reacts with water to produce the ammonium cation and hydroxide anion.

Now that we know the reaction that describes the system, we have to ask what K expression is used to represent that particular reaction.  For the reaction of a base, we need an equilibrium constant known as Kb.  The expression for Kb is shown below.

$\mathrm{K_b = \dfrac{[NH_4^+][OH^-]}{[NH_3]}}$

If we examine the tables of equilibrium constants, though, we observe that the table does not list Kb values, but instead only lists Ka values for substances.  A species that is in the reaction that we do find a Ka value for in the table is the ammonium cation.  It is important to note that the species ammonia and ammonium differ by only a hydrogen ion.

$\ce{NH3/NH4+}$

Species that differ from each other by only a hydrogen ion are said to be a conjugate pair.  A conjugate pair always contains a base (ammonia in this case) and an acid (ammonium in this case).  The acid is always the form with the extra hydrogen ion.  The base is the form without the extra hydrogen ion.

The Ka reaction is that of the ammonium ion acting as an acid.

$\ce{NH4+ + H2O \leftrightarrow NH3 + H3O+}$

The equilibrium constant expression for Ka is shown below.

$\mathrm{K_a = \dfrac{[NH_3][H_3O^+]}{[NH_4^+]}}$

Furthermore, the Kb and Ka values for the base and acid form respectively of a conjugate pair have a very specific relationship that is shown below.

$\mathrm{K_b \times K_a = K_w = 1 \times 10^{-14}}$

Remember, Kw is the equilibrium expression that describes the autoprotolysis of water.

$\ce{H2O + H2O \leftrightarrow H3O+ + OH-}$

$\mathrm{K_w = [H_3O^+][OH^-] = 1\times10^{-14}}$

The expression below shows that the result of multiplying Kb times Ka is actually Kw

$\mathrm{K_b \times K_a = \dfrac{[NH_4^+][OH^-]}{[NH_3]} \times \dfrac{[NH_3][H_3O^+]}{[NH_4^+]} = [OH^-][H_3O^+] = K_w}$

Now that the Kb value is known, it is finally possible to solve for the pH of the solution of ammonia.  A useful way to keep track of such problems is to use the reaction as the headings for columns of values that describe the concentrations of species under certain conditions.  The first set of numbers represents the initial concentrations in solution prior to any equilibration.

\begin{align} & &&\ce{NH3}\hspace{25px} +\hspace{25px} \ce{H2O} \hspace{25px}\leftrightarrow &&\ce{NH4+} \hspace{25px} + &&\ce{OH-}\\ &\ce{Initial} &&0.155 &&0 &&10^{-7} \end{align}

We do not need an initial value for water since it’s the solvent.  The hydroxide is given a value of 10-7 M because of the autoprotolysis of the water.  The second set of numbers are expressions for the equilibrium concentrations of the species.  In this case, we want to keep in mind that the value for Kb is small, meaning we do not expect that much product to form.

\begin{align} & &&\ce{NH3}\hspace{25px} +\hspace{25px} \ce{H2O} \hspace{25px}\leftrightarrow &&\ce{NH4+} \hspace{25px} + &&\ce{OH-}\\ &\ce{Initial} &&0.155 &&0 &&10^{-7}\\ &\ce{Equilibrium} &&0.155 - \ce{x} &&\ce{x} &&10^{-7} + \ce{x} \end{align}

If we wanted, these values could now be plugged into the Kb expression and it could be solved using a quadratic.  There may be a way to simplify the problem, though, if we keep in mind that Kb is so small.  In this case, we expect the value of x to be small and we can make two approximations.

The first is that x $$\ll$$ 0.155 so that (0.155 – x) = 0.155

The second is that x $$\gg$$ 10-7 so that (10-7 + x) = x

\begin{align} & &&\ce{NH3}\hspace{25px} +\hspace{25px} \ce{H2O} \hspace{25px}\leftrightarrow &&\ce{NH4+} \hspace{25px} + &&\ce{OH-}\\ &\ce{Initial} &&0.155 &&0 &&10^{-7}\\ &\ce{Equilibrium} &&0.155 - \ce{x} &&\ce{x} &&10^{-7} + \ce{x}\\ &\ce{Approximation} &&0.155 &&\ce{x} &&\ce{x} \end{align}

Now we can plug the approximations in the Kb expression and solve for the value of x.

$\mathrm{K_b = \dfrac{[NH_4^+][OH^-]}{[NH_3]} = \dfrac{(x)(x)}{0.155} = 1.76 \times 10^{-5}}$

$\mathrm{x = [OH^-] = 1.65\times 10^{-3}}$

Before we can use this to calculate the concentration of H3O+ and solve for pH, we first must check the two approximations to make sure they are both valid.

$\dfrac{1.65\times 10^{-3}}{0.155} \times 100 = 1.1\% \hspace{80px} \dfrac{10^{-7}}{1.65\times 10^{-3}} \times 100 = 0.0061\%$

It is worth noting that the assumption that the initial hydroxide or hydronium ion can be ignored is almost always made in these problems.  The only two instances in which this approximation would break down are if:

1. the acid or base is exceptionally weak so that so little dissociation occurs that the initial amount is significant or
2. the acid or base is so dilute that very little dissociation occurs.

Since both approximations are less than 5%, the concentration of H3O+ can be calculated using the Kw expression and the pH can be calculated.

[H3O+] = 6.31$$\times$$10-12

pH = 11.2