How to Prepare a Buffer Solution?

Question

Problem

I am given the task of preparing three buffer solutions at pH $10$, $9.5$, and $9.0$ . I have available concentrated ammonia and $\pu{3M}$ hydrochloric acid. The buffer capacity desired is $\pu{0.1 M}$.

Attempt at Solving

By adding $\ce{HCl}$ to a solution containing ammonia, it will completely consume the strong acid. Resulting solution will contain ammonia and it's conjugate acid ammonium ion, $\ce{NH4+}$.

$$\ce{NH3 + H+ -> NH4+}$$

Where $$[{\ce{NH4+}]_\mathrm {final}}= [{\ce{HCl}]_\mathrm{initial}}$$ and
$$[{\ce{NH3}]_\mathrm{final}} = [{\ce{NH3}]_\mathrm {initial}} - [{\ce{HCl}]_\mathrm {initial}}$$

I think at this point I should make some assumptions:
$\pu{500 mL}$ buffer to be prepared, $\mathrm pK_\mathrm {a,\ce{NH4+}}$=$9.25$
$[\ce{NH3}] = \pu{2 M}$, ammonia in ethanol solution

From here I think, from the Henderson-Hasselbalch equation.,
$$\mathrm{pH} = \mathrm pK_\mathrm a + \log \left(\frac{[\ce{A^-}]}{[\ce{HA}]}\right)$$
$$\mathrm{pH} = \mathrm pK_\mathrm a + \log \left (\frac{[\ce{NH3]}}{[\ce{NH4+}]}\right)$$
$$\implies \mathrm{pH} = \mathrm pK_\mathrm a + \log \left(\frac{[\ce{NH3}]}{[\ce{NH3}] - [\ce{HCl}]}\right)$$

Making the pH = 10 solution

$$10 = 9.25 + \log\left(\frac{[\ce{NH3}]}{[\ce{NH4+}]}\right)$$

It is pretty visible at this point though that I don't exactly know how to proceed. I have reached one variable but through processes that would not really hold up under scrutiny. The $[\ce{NH3}]$ that I have as my last variable is either $\ce{[NH3]}_\mathrm {initial}$ or $\ce{[NH3]}_\mathrm {final}$ but I have no idea how to determine which it is. I also feel I may have applied the wrong method from the start.

If there is anything I can do to add clarity or stop this post from being thrown out please let me know so I can remedy the situation asap.

Answer 1

You have some of the chemistry wrong.

Looking at your assumptions, concentrated ammonia is typically in aqueous solution, as is concentrated hydrochloric acid. You could have ethanol solutions saturated with $\ce{HCl}$ and $\ce{NH3}$ but that would be unusual.

Concentrated ammonia in aqueous solution is about 18 molar and is usually notated as concentrated ammonium hydroxide.

Since all three buffer solutions are in alkaline solution all the $\ce{HCl}$ will be reacted according to the following reaction:

$\ce{NH3 + HCl -> NH4^+ + Cl^-}$

The $\ce{Cl^-}$ anion is a spectator anion and will not effect the pH.

$\ce{\text{pKa}_{ammonia}= 9.25}$ is sound.

Using the given pKa, at pH = 9.25 then $\ce{[NH3] = [NH4^+]}$. So:

• at pH 9.0 there will be a bit more $\ce{[NH4^+]}$ than $\ce{[NH3]}$.
• at pH 9.5 there will be a bit more $\ce{[NH3]}$ than $\ce{[NH4^+]}$.
• at pH 10.0 there will be a even more $\ce{[NH3]}$ than there was $\ce{[NH4^+]}$ at ph 9.5.

The ratio of $\ce{[NH3]}$ to $\ce{[NH4^+]}$ can be calculated based on the pH using the Henderson-Halbach equation which for the ammonia/ammonium equilibrium is:

pH = 9.25 + log ([NH3]/[NH₄⁺])

---

The problem as stated however seems incomplete.

If you really must make the buffer solution using only concentrated ammonium hydroxide and 3 molar HCl, then there is a unique solution for each buffer.

However if you can add additional water, then there is no unique solution and you'd need to know the buffer capacity for each of the buffers.

---

It seems really odd to have solutions as buffer solutions as strong as 18 molar.

I'll explain further that the equilibrium equation

$\text{K}_a = \dfrac{\ce{[NH3][H^+]}}{\ce{[NH4^+]}}$

should really be written as the activities of the species not the concentrations. Below about 0.1 molar the activity and concentration are pretty equal. But in as concentrated solutions as concentrated as 18 molar the assumption is dicey and some correction would need to be made since the activity of the various species would be less than their actual concentrations.

---

Working towards a solution for 0.10 molar buffers
Edit 1/19/2017, noon

So now the problem is that 0.1 molar buffers are needed for pH values of pH 10, 9.5, and 9.0. (I'll assume that ph 10 is 10.0 so 1 significant figure in concentration.)

Even though we are going to work with pKa of $\ce{[NH4^+]}$, let's nt forget that we are starting with "pure" ammonium hydroxide which ionizes as:

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

In "pure" ammonium hydroxide not much of the ammonia will ionize to ammonium, so we can assume

$\ce{[OH^-] = [NH4^+]}$

and the pH will be 11+. So we can add various amounts of HCl and create the needed buffers.

However there is a consideration. A 0.1 molar buffer means that given 1 liter of solution then:

• If 0.1 moles of a strong acid is added, the the pH drops no more than 1 pH unit.
• If 0.1 moles of a strong base is added the the pH increases no more than 1 pH unit.

Now a buffer would be made "best" at the pKa or pKb value of a chemical so that the buffer capacity was the same for either a strong acid or a strong base.

The buffers at 9.0 and 9.5 are reasonably close to the pKa of ammonium which is 9.25. So these buffers will guard against nearly the same amount of acid as base.

However the buffer at pH 10.0 is a considerable distance from the pKa of ammonium. So it will take much less of a strong base like NaOH to change the pH to 11.0 than a strong acid like HCl to change the pH to 9.0.

So assume that:

• The 0.1 molar buffer capacity means "at least 0.1 molar".
• You're making 1 liter of each of the solutions
• You only need volumes to about 5% for the reagents.

The last assumption is a "gotcha" of sorts. I haven't worked through the solution, but I don't think there is a way to solve all the constraints directly in one pass. I think the "exact" solution would have to be calculated iteratively. No problem for a computer, but it is painful by hand. So overall this is trying to avoid iterating using the quadratic equation.

Answer 2

Here is what you would do in the lab (2 recipes, depending on your needs):

1. You start by diluting the concentrated ammonia to about $\pu{100 m}\mathrm{\small M}$. Then, while measuring the pH, you add sufficient $\ce{HCl}$ to get the pH to $10.0$, and put some of the buffer into a bottle. Then you add some more $\ce{HCl}$ to the remaining buffer until the pH is $9.5$, and put some of it into a second bottle. Finally, you add some more $\ce{HCl}$ to the remaining buffer until the pH is $9.0$, and put that into a third bottle. Now you have the required 3 buffers. This recipe does not give you buffers of defined concentration (and not even of equal concentrations), but your question initially didn't ask for a specific concentration.

2. If you need a specific concentration, take $\pu{100 mL}$ of $\pu{200 m}\mathrm{\small M}$ ammonia, set the pH with concentrated $\ce{HCl}$ and fill up to $\pu{200 mL}$ with water. That will give you a buffer with a concentration of $\pu{100 m}\mathrm{\small M}$ (total concentration of $\ce{NH3}$/$\ce{NH4+}$) at the desired pH.

The reason you do it this way is because buffer solutions are not ideal (by design, you want high concentrations of buffer substances, whereas calculations are accurate for infinitely dilute solutions). Especially if your buffer contains other substances such as high concentrations of $\ce{NaCl}$, it is better to set the pH experimentally rather than do a calculation. Also, it is much faster.