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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}

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  • $\begingroup$ You are right, consider ammonia as ammonium hydroxide. Use equilibrium calculations and find out hydrogen ion or hydroxide ion concentrations. That should let you find the pH. $\endgroup$ – Pritt says Reinstate Monica Apr 21 '17 at 4:07
  • $\begingroup$ Ammonia does exist as NH4+ , which is essential in a buffer solution as it depends on common ion effect. Use equilibrium concepts to figure out the concentration of the compounds in the solution and apply the Henderson-Hasselbalch equation. Hope this helps. $\endgroup$ – Reya Apr 22 '17 at 19:31
  • $\begingroup$ Okay, but would I assume the initial concentration of ammonia to be of NH3 or NH4+ from the first equation? Because the question says NH3 so assuming the initial concentration of NH3 makes sense to me but at the same time it introduces more variables. So I can't solve it if I assume the initial concentration to be of NH3. $\endgroup$ – Sillysack Buttowski Apr 25 '17 at 12:02
  • $\begingroup$ Don't ever consider it as $\ce{NH3}$. Assume all the ammonia to exist as $\ce{NH4OH}$, although ammonia essentially only contributes to $\ce{[OH^-]}$, since much of the$\ce{[NH4^+]}$ is obtained from the $\ce{NH4Cl}$. $\endgroup$ – Pritt says Reinstate Monica Apr 25 '17 at 12:05
  • $\begingroup$ In order to help you, what is the volume of the buffer and what volume of 0.15 M HCl was added? $\endgroup$ – Bive Apr 28 '17 at 17:15
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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.

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  • $\begingroup$ No, actually I do just fine with acid buffers involving $\ce{HCl}$, it's just these base buffers involving $\ce{NH3}$ that are confusing me. I know that it's easy using the Henderson equation, but I'm just trying really hard to understand what is actually happening in the solution. And the volumes not given is also confusing me. How can I calculate the change in pH of a solution when the volumes of the solutions have not been given. You mentioned something about not altering the total volume? $\endgroup$ – Sillysack Buttowski Apr 29 '17 at 17:38
  • $\begingroup$ The basic concept is that acids will react with bases. $\ce{HCl}$ is a strong acid, $\ce{NH3}$ is a base, the two will react to give $\ce{NH4Cl}$ quantitatively, within the approximate conditions we're applying. I don't really see the difference with adding $\ce{HCl}$ to an acetate/acetic acid buffer: there too you remove acetate anion (a base) by reacting it with $\ce{HCl}$ and turning it to its conjugate acid $\ce{CH3COOH}$. It's just a matter of recognising this general mechanism, I suppose. $\endgroup$ – user6376297 Apr 29 '17 at 18:39
  • $\begingroup$ Oh, and about the volume: yes, I am making an assumption by deciding that 0.15 M HCl means that we add concentrated HCl so that the final nominal concentration of HCl is 0.15 M, but without changing the volume significantly. In this specific case, the total volume would not matter, as it's a buffer solution under the approximate conditions we discussed. However I agree with you, it is confusing when questions/problem are ambiguously formulated. I saw plenty of exam papers that, as far as I am concerned, would constitute sufficient evidence to fire the teacher on the spot. $\endgroup$ – user6376297 Apr 30 '17 at 8:55
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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.

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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}

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