How do I calculate the pH of this buffered solution?

Question

Calculate the $\mathrm{pH}$ of a buffered solution containing $\pu{0.5 M}$ ammonia and $\pu{0.5 M}$ ammonium chloride when $\pu{0.15 M}$ $\ce{HCl}$ is added into it. The $\mathrm{p}K_\mathrm{b}$ of ammonia is $4.75$.

This is what I think shoud be going on in the buffer solution:

\begin{align} \ce{NH3(aq) +H2O(l) &<=> NH4+(aq) +OH-(aq)} \\ \ce{NH4Cl(aq) &-> NH4+(aq) +Cl-(aq)} \end{align}

This is where I get stuck thinking that the question told me that I have a buffered solution containing $\pu{0.5 M}$ $\ce{NH3}$ and $\pu{0.5 M}$ $\ce{NH4Cl}$ solution, but is that even possible? How can you make a $\pu{0.5 M}$ ammonia solution? Would not that ammonia exist as ammonium $\ce{NH4+}$ and whether the equations I wrote are correct. I am just so confused!

Sorry for the delay but I'm having exams. Okay, I'll try to do this now by not using the Henderson equation.

Firstly, ammonia will refer to ammonium hydroxide, i.e. $\ce{NH3 -> NH4OH}$:

\begin{array}{cccc} &\ce{&NH4OH &<=> &NH4+ &+ &OH-} \\ &\mathrm{I}: &\pu{0.5 M} & &\pu{0 M} & & \pu{0 M} \\ &\mathrm{E}: &(0.5 - x)\,\pu{M} & &x\,\pu{M} & &x\,\pu{M} \end{array}

Now for the salt:

\begin{array}{cccc} &\ce{&NH4Cl &<=> &NH4+ &+ &Cl-} \\ &\mathrm{I}: &\pu{0.5 M} & &x\,\pu{M} & &\pu{0 M} \\ &\mathrm{E}: &\pu{0 M} & &(0.5 + x)\,\pu{M} & &\pu{0.5 M} \end{array}

$$K_\mathrm{b} = \frac{[\ce{NH4+}][\ce{OH-}]}{[\ce{NH4OH}]} = \frac{(x + 0.5)x}{(0.5 - x)}$$

and since $K_\mathrm{b} = 1.778 \times 10^{-5}$, I can find the value of $x$ to be $1.7778 \times 10^{-5}$.

$$[\ce{H+}][\ce{OH-}] = 10^{-14} \to [\ce{H+}] = \frac{10^{-14}}{[\ce{OH-}]}$$ and $x = [\ce{OH-}]$. So

$$[\ce{H+}] = 5.624 \times 10^{-10}$$

Now I add the $\ce{HCl}$:

\begin{array}{cccc} &\ce{&HCl &<=> &H+ &+ &Cl-} \\ &\mathrm{E}: &\pu{0 M} & &\pu{0.15 M} & &\pu{0.15 M} \end{array}

I will ignore any common ion effect since it will be negligible (I think). The pH will be the total $\ce{H+}$ concentration, so $[\ce{H+}] = 5.624 \times 10^{-10} + 0.15$. This is a pH of about 0.823 which is totally wrong. What am I doing wrong? Also, the volumes of the buffer solution or of the acid solution have not been given.

Okay, so I now know what I was doing wrong (thanks to someone pointing it out). I was adding in $\ce{HCl}$ without thinking that it would actually react with the ammonia in equilibrium in solution (which is just me being either dumb or ignorant).

So I have the situation that:

\begin{array}{cccc} &\ce{&NH4OH &<=> &NH4+ &+ &OH-} \\ &\mathrm{E_\text{(no HCl)}}: &(0.5 - 1.778 \times 10^{-5})\,\pu{M} & &(1.778 \times 10^{-5})\,\pu{M} & &(1.778 \times 10^{-5})\,\pu{M} \\ &\mathrm{E_\text{(HCl)}}: &()\,\pu{M} & &()\,\pu{M} & &()\,\pu{M} \end{array}

Answer 1

You can think of $\ce{NH4OH}$ as a hydrate: $\ce{NH3_{(aq)}}$; no need for complicated reactions interconverting the two.

Concerning the problem you described, at these concentrations all subtleties are negligible, and you basically only need to calculate the nominal concentrations of $\ce{NH4+}$ and $\ce{NH3}$, and plug them into the Henderson-Hasselbach equation.

You start with $\ce{0.5 M NH3}$ and $\ce{0.5 M NH4Cl}$, and add $\ce{HCl}$ to a total nominal concentration of $\ce{0.15 M}$, without altering the total volume (the question is written ambiguously, but that's what it means, most likely). As $\ce{HCl}$ is a very strong acid, it will convert all $\ce{NH3}$ to $\ce{NH4Cl}$ (actually $\ce{NH4+}$, because we are in solution). Therefore your final situation is: $\ce{(0.5-0.15)=0.35 M NH3}$ and $\ce{(0.5+0.15)=0.65 M NH4+}$. This should give you a pH of about 8.98 (lower than the one you would have without the HCl, consistently with the addition of an acid).

So your main error above was that you dissociated $\ce{HCl}$ (correctly), but then just added the resulting $\ce{H+}$ to the total, forgetting that there was a ton of $\ce{NH3}$ around that would react with it.

Another example is when you add a solution of some particular transition metal salt to a solution of a weak acid (could be iron sulfate + hydrogen sulfide for instance). Despite the fact that the salt solution isn't acidic at all, the pH drops considerably compared to the solution of the weak acid alone, simply because the metal and the anion form a very insoluble salt that precipitates, subtracting weak acid anions from the solution and forcing the previously non-dissociated acid to dissociate.

In general, my advice is stick to what you know, don't go looking for complications: if they did not teach you to do very exact/sophisticated pH calculations, they probably don't expect you to pull them out of a magic hat at the exam. And as my stoichiometry teacher used to say, always write down all species you have in solution.

Answer 2

Well, first of all, as you have written before, the main reaction is the following one,

$$\ce{NH3 + H2O <=> NH4+ + OH-}$$

In addition, you know the ammonia constant: $K_\mathrm{b}=10^{-4.75}$

(Note that if the question does not have a mention about equilibrium, it is because you don't need use the equilibrium in order to make the calculations)

So, let's start.

1. Initially you have:

   $$[\ce{NH3}] = [\ce{NH4+}] = \pu{0.5 M}$$

   Then, just taking into account the previous reaction

   $$\ce{NH3 + H2O <=> NH4+ + OH-}$$

   you can get the result easily:

   \begin{align} K_\mathrm{b} &= \frac{[\ce{NH4+}][\ce{OH-}]}{[\ce{NH3}]} \\ [\ce{OH-}] &= K_\mathrm{b} \frac{[\ce{NH3}]}{[\ce{NH4+}]} \\ [\ce{OH-}] &= K_\mathrm{b} = \pu{10^{-4.75} M} \end{align}

   and finally the pH would results as

   $$\mathrm{pH} = 14 - \mathrm{pOH} \overset{\mathrm{pOH} = 4.75}{\to} \mathrm{pH} = 9.25$$

2. You add $\pu{0.15 M}$ $\ce{HCl}$ (acid), where it reacts with, obviously, the ammonia (base) which is found at the equilibrium, so

   \begin{array}{ccccc} & \ce{&NH3 &+ &H2O &<=> &NH4+ &+ &OH-}\\ &i) &0.5 & &0.15 & &0.5 & &/ \\ &f) &0.35 & &/ & &0.65 & &/ \end{array}

   Now, you may observe that both concentrations have changed, so we could obtain, using the same strategy as before we did:

   $$K_\mathrm{b} = \frac{[\ce{NH4+}][\ce{OH-}]}{[\ce{NH3}]} \to [\ce{OH-}] = K_\mathrm{b} \frac{[\ce{NH3}]}{[\ce{NH4+}]} = 10^{-4.75}\frac{0.35}{0.65} \to [\ce{OH-}] = \pu{10^{-5.01} M}$$

   To conclude, the $\mathrm{pH}$ is given by

   $$\mathrm{pH} = 14 - \mathrm{pOH} \overset{\mathrm{pOH}=5.01}{\to} \mathrm{pH} = 8.98$$

   Observe that the buffer turns into more acid as much more acid is introduced in it.

Answer 3

I mainly want to rectify one point that you are confused with and that all the other answers fail to point out explicitly. You say in the beginning that ammonia should react with water according to $(1)$.

$$\ce{NH3 + H2O <=> NH4+ + OH-}\tag{1}$$

Obviously, this is an acid-base reaction. We can deduce where the equilibrium will be just by checking out the $\mathrm pK_\mathrm a$ values of the two Brønsted acidic species, water and ammonium:

\begin{array}{lc}\hline \text{compound} & \mathrm pK_\mathrm a\\ \hline \ce{H2O} & 14\\ \ce{NH4+} & 9.25\\ \hline\end{array}

As we can see, ammonium is more acidic than water by almost five logarithmic units. Therefore, it is safe to assume that in a solution of ammonia in water only a very neglegible amount of ammonia is protonated. Thus, when mixing a $\pu{0.5M}$ nominal solution of ammonia (which is very close to an actual $\pu{0.5M}$ solution) to a $\pu{0.5M}$ solution of ammonium chloride, we can indeed just use the Henderson-Hasselbalch equation to shortcut the way to the $\mathrm{pH}$ value:

\begin{align}\mathrm{pH} &= \mathrm pK_\mathrm a + \lg\frac{[\ce{NH3}]}{[\ce{NH4+}]}\\ &= 9.25 + \lg 1\\ &= 9.25\end{align}

When we then in the second step add $\pu{0.15M}$ of a strong acid, we can simply identify the strongest base in solution and consider $\pu{0.15M}$ of that base additionally protonated. The strongest base around to any appreciable extent is ammonia, so we now have a new concentration of $\pu{0.65M}$ of ammonium and $\pu{0.35M}$ of ammonia. Thus, our new $\mathrm{pH}$ value becomes:

\begin{align}\mathrm{pH} &= \mathrm pK_\mathrm a + \lg\frac{[\ce{NH3}]}{[\ce{NH4+}]}\\[0.5em] &= 9.25 + \lg \frac{\pu{0.35M}}{\pu{0.65M}}\\[0.5em] &= 9.25 + \lg 0.54\\ &= 9.25 - 0.269\\ &= 8.98\end{align}