Creating a Buffer

  • Thread starter rls623
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Homework Statement



You need to prepare 2000 L of buffer pH 3.9. You have available 0.2 M NaOH solution and 0.2 M Acetic acid (pKa=4.8). How much of each solution do you need?


Homework Equations



I know that I have to use the H-H eqn: pH = pKa + log([HA]/[A-])

The Attempt at a Solution



When I plug the pH and pKa into the eqn, I get that the ratio of [HA]/[A-] = .126

This is where I’m not sure what to do. Is there another equation that I can use and then solve the 2 simultaneously? I feel that that’s probably the case, I just am unsure of how to set it up. Any suggestions would be greatly appreciated.
 

Answers and Replies

  • #2
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typo

It's not 2000 L of solution, but rather 2000mL. Oops.
 
  • #3
symbolipoint
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Use these variable assignments:

K=equilibrium constant for acetic acid, 1.75*10^(-5)
H=Molarity of hydronium
Fs=formality of the sodium salt (acetate)
Fa=formality of acetic acid

V=milliliters volume of the 0.2 M acetic acid to start with
v= 'little v', the milliliters of 0.2 M sodium hydroxide

Let's generalize since we do not yet know exactly how much acetic acid
and NaOH we want. We can adjust for scale later.

Fundamental equilibrium constant equation is
K = (H)(Fs+H)/(Fa-H)

The sodium acetate normality comes from the added sodium hydroxide:
Fs = (0.2)(v)/(V+v)

The formality of the acid is affected by addition of the NaOH:
Fa = (0.2V - 0.2v)/(V+v)

Substitute the F's into the K equational formula:
This will look nasty here because I cannot typeset with what I have here, so
you should do the substitution yourself ---
K = (H)(v(0.2)/(V+v) + H)/((V(0.2)-0.2v)/(V+v) - H)
See, I told you it looks nasty; it will look better when you do it yourself.

Now, you basically use a few algebra steps to clean the equation and obtain
a formula for 'v'. You should obtain something equivalent to this:

v = (0.2KV - HKV - (H^(2))V)/(((H^(2)) + HK +0.2H + 0.2K)

Use this formula for 'v' to find how much acid and how much base to give your
target pH of 3.90. I started with V=100 ml of 0.2M acid, H=10^(-3.9), K=1.75*10^(-5).
You can then change the scale for the 2000 ml size that you want; 100+v is expected
to be too small so you want to use a factor of 2000/(V+v).
 
  • #4
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Thanks for your help. I'm not sure I understand what formality and nornality are though.

I ended up getting an answer to the problem, but I did it differently.

I plugged the pH and pKa values into the H-H eqn to find the ratio of [HAc]/[Ac-]. This gave me that I need 0.125 mol of Ac- for every mol of HAc. I put these values into an IF table as my final values and solved for the number of mols of HAc and OH- I would need initially. I then used these values to determine the ratio of how many liters of each of the solutions I was initially given I would need to come up with the buffer I desired. I then scaled these values back so that my total volume was 2 L.

I ended up with a final answer of 1.8 L of HAc solution and 0.2 L of NaOH.

Does that make sense? I don't know if its right or not, but it seems logical to me.
 
  • #5
symbolipoint
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rls623,

Formality means the number of formula weights of a substance in 1 L of solution. This resembles the definition of Molarity, but they are a bit different from eachother. Formality is representative of how much you put into the solution; Molarity is how much is really present based on behavior when in solution.

If you have for example 0.1 formula units of acetic acid in 1 L of solution, then how is the resulting molarity different? The acid dissociates into hydronium and acetate ions. NOT all of it, just some of it. The FORMALITY is still the same, but the MOLARITY of the acid is now less due to a bit of acetic acid becoming hydronium and acetate.

What would be the molarity of the acid from 0.1 F acetic?
HAc <==> H+ + Ac-
let x=amount of hydronium produced, same as amount of acetate produced;
then [HAc] = molarity of acetic acid = 0.1 - x
 

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