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 3M hydrochloric acid.

Attempt at Solving

By adding HCl to a solution containing ammonia, it will completely consume the strong acid. Resulting solution will contain ammonia and it's conjugate acid ammonium.

NH₃ + H⁺ --> NH₄⁺

Where [NH₄⁺]_final = [HCl]_initial
and...
[NH₃]_final = [NH₃]_initial - [HCl]_initial

I think at this point I should make some assumptions:
500mL buffer to be prepared pKa_ammonia=9.25
Conc NH₃ = 2M Ammonia in ethanol solution

From here I think...
pH = pKa + log ([A^-]/[HA])
pH = pKa + log ([NH3]/[NH₄⁺])
pH = pKa + log ([NH3]/([NH3] - [HCl]))
making the pH 10 solution...
10¹⁰ = 10^9.25 + ( log([NH₃] - log(10¹⁰))

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 [NH3] that I have as my last variable is either [NH3]_initial or [NH3]_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.

Any help at all in general guidance would be great! Thank you everyone.

Additional Info

I am an undergraduate chemistry research student currently in my senior year. I was given this task to be accomplished by next week. I of course will have to provide the calculations and necessary related information though before I will be able to perform any such task.

I am not trying to use this forum to circumnavigate any of that work that I must do. It is however difficult to get time with my research adviser currently so I am just trying to be as prepared as I can be walking into this situation.

I have taken classes were I have been walked through this same process but it definitely seems to be a "if you don't use it, you lose it" sort of deal.

I realize immediately I am new to this forum and am therefore being verbose simply to try and cover all of my bases. I hope that lengthiness in itself isn't more distracting. I have read through the posting rules and the homework questions rules and hope that my post reflects that I have.

I have also searched a great deal looking for related questions but many seem to have fallen victim to going off topic by simply asking a question and not putting any effort into its solution. Once again I hope I have navigated away from this pitfall.

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.

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EDIT FROM COMMENT MADE BY OP - The buffer capacity desired is 0.1M and this was just gained as additional information. – hotmaildotcom1 16 hours ago

Solution

Thank you all for your assistance! Working through it after your guidance this is what I came to. There are more assumptions made that I will explain after the calculations as they will definitely be needed.

I apologize for any sloppy work but it is so much faster for me as a new guy to upload the work rather than type it out.

So after talking with everyone here I reassessed my approach. I will try and address each of my issues in a bullet point.

1. I don't know why it wasn't immediately obvious to me that the NH₃ would need to be diluted to a reasonable concentration. This was a point repeatedly made by the community so thank you all again so much. After I made this assumption I accepted the arbitrary starting concentration of 1.M NH₃. I redid those calculations as I already had a 5mL volumeteric pipette calibrated, so those are the numbers shown.
2. I was struggling with the concept of equilibrium in a significant way. After reading my textbook from analytical, I started to again use ICE tables as can be seen in my work. This greatly simplified the issue for me and I should have done this much sooner.
3. I used the definition of pK_a to solve for pK_a rather than the Henderson-Hasselbalch equation. This is what my textbook used and I was doing well with those examples so I kept with it. It is my understanding that this doesn't change anything.
4. I did not calculate any buffer capacity while doing these equations. I assumed that my overall concentrations of compounds used to make the buffer solution would be significantly larger than the molarity of the reaction being buffered. This proved to be a decent assumption (at least so I believe) as in practice a very large amount of reactant can be added and no noticeable change in pH is observed.

All of these calculations were ran by my PI and he signed off on all of them. Having performed the necessary precautions in order to work with these compounds, I carried out making all of these solutions. They all work like a charm without the need to add any additional acid or base in order to reach the desired pH's. I cannot thank everyone enough for your help and I apologize for taking so long to reply.

Finally I realize that this post is not really organized in a reasonable manner anymore in order to help anyone that might dig it up later. I am hesitant to delete any information in order to make things more legible. I simply did not have enough space in the comments sections in order to show that with everyone's help I was able to achieve a reasonable answer. If everything here checks out after review I will delete the "solutions" section and post it as the answer.

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⁺])

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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.

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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.

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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 100 mM. Then, while measuring the pH, you add sufficient HCl to get the pH to 10.0, and put some of the buffer into a bottle. Then you add some more 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 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 100 mL of 200 mM ammonia, set the pH with concentrated HCl and fill up to 200 mL with water. That will give you a buffer with a concentration of 100 mM (total concentration of NH3/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 NaCl, it is better to set the pH experimentally rather than do a calculation. Also, it is much faster.