Buffer Solution Calculation Example

In summary: Conjugate Base: Na+...What are the acid and its conjugate base here?Acid: HA...Conjugate Base: Na+...
  • #1
TheShiny
9
0

Homework Statement


Prepare a buffer solution of pH 3.746 from 25 mL of 0.244 M weak acid with pka 3.54

What volume, in mL, of 0.275 M NaOH would need to be added?

I think that I have the answer but I am not sure, could you look over the working to check?
Also, would you do this differently? If so, what would you do?

Homework Equations


Henderson-Hasselbach:
[itex]pH = pka + log_{10} \frac {[A^-]}{[HA]} \Rightarrow [H^+] = ka + \frac{[A^-]}{[HA]} \Rightarrow ka = \frac{[H^+][A^-]}{[HA]}[/itex]

The Attempt at a Solution


Part I
[itex]pH = 3.746 \Rightarrow [H^+] = 10^{-3.746} = 1.79473\times10^{-4}[/itex]

[itex]pka = 3.54 \Rightarrow [H^+] = 10^{-3.54} = 2.88403\times10^{-4}[/itex]

[itex]ka = \frac{[H^+][A^-]}{[HA]} \Rightarrow 2.88403\times10^{-4} = 1.79473\times10^{-4}\times\frac{[A^-]}{[HA]}[/itex]

[itex]\therefore \frac{[A^-]}{[HA]} = 1.60694[/itex]

Pat II
25 mL of 0.244 M acid [itex]\Rightarrow[/itex] 0.0061 M (6.1 mM)

Code:
+---+--------------+--------------+
|   |     Acid     |     Base     |
| I |   0.0061 M   |      0 M     |
| C |     -x M     |     +x M     |
| E | (0.0061-x)M  |      x M     |
+---+--------------+--------------+

[itex]\frac{[A^-]}{[HA]} = \frac{x}{(0.0061 - x)} \therefore \frac{x}{(0.0061 - x)} = 1.60694[/itex]

Solve for [itex]x[/itex]...

[itex]x = 0.00376009[/itex] (is this the moles of acid reacting?)

[itex]\frac{0.00376009}{0.275} = 0.01367305[/itex]

[itex]\therefore[/itex] 13.7 mL NaOH should be added.
 
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  • #2
TheShiny said:
[itex]pH = pka + log_{10} \frac {[A^-]}{[HA]} \Rightarrow [H^+] = ka + \frac{[A^-]}{[HA]} \Rightarrow ka = \frac{[H^+][A^-]}{[HA]}[/itex]

Correct, but completely unnecessary. You could just plug pH and pKa into HH equation, that's what it is for.

[itex]pH = 3.746 \Rightarrow [H^+] = 10^{-3.746} = 1.79473\times10^{-4}[/itex]

Correct, but completely unnecessary.

[itex]pka = 3.54 \Rightarrow [H^+] = 10^{-3.54} = 2.88403\times10^{-4}[/itex]

Wrong, and completely unnecessary. But I assume you meant OK, you just copied wrong.

[itex]\frac{[A^-]}{[HA]} = 1.60694[/itex]

Yep, that's what is important here. As I said above, you could get here much faster.

[itex]\therefore[/itex] 13.7 mL NaOH should be added.

Looks OK to me.

itex tags are for inline formulae, when you post them in their own lines, use just tex.
 
  • #3
TheShiny said:

Homework Statement


Prepare a buffer solution of pH 3.746 from 25 mL of 0.244 M weak acid with pka 3.54

What volume, in mL, of 0.275 M NaOH would need to be added?

I think that I have the answer but I am not sure, could you look over the working to check?
Also, would you do this differently? If so, what would you do?

[snip]

[itex]\therefore[/itex] 13.7 mL NaOH should be added.

This looks correct, though you made your life difficult.

The nice thing about the henderson-Hasselbach equation is that the log part is a ratio of concentrations, but it can just as easily be written as a ratio of moles.

pH = pKa + log [A-]/[HA] = pKa + log ( moles A- / moles HA )

At the half eq. point, moles A- = moles HA, pH = pKa + log (1) ==> pH = pKa

For your buffer, the pH is close to pKa, so you are close to the 1/2 equiv. point. Since pH is a bit larger than pKa, you are a bit past the 1/2 equiv. point, i.e. there is a bit more conj. base than conj. acid. The concentrations of acid and base are pretty close, so the half-eq. point is going to be close to half the original volume of acid -- ca. 12.5 mL.

Finding the exact answer...

3.746 = 3.54 + log (moles A-/moles HA)

moles A- / moles HA = 1.607

moles HA_0 = 25 mL * 0.244 M = 6.10 m mol

x/(6.1 mmol-x) = 1.607 ==> x = 3.76 m mol

vol OH- * 0.275 M = 3.76 m mol

vol OH- = 13.7 mL
 
  • #4
Thank you for your responses.

If I had the question "Prepare a buffer solution of pH 3.924 from 50 mL of 0.236 M solution of a sodium salt of a weak acid, Na+A- where the pka of the weak acid, HA, is 3.956. What volume, in mL, of 0.208 M HCl would need to be added?"

Would I use the same method as with the initial question? What changes, if any, should I make?
 
  • #5
Very similar question and very similar approach, you just need to keep track of what is the acid and what is the conjugate base here.
 
  • #6
Okay, so I've just had a go and I get:

[A-]/[HA] = 0.92896

Base: 50mL * 0.236M = 11.8mM

(11.8-x)/x = 0.92896 => x = 6.11727

6.11727/0.208 = 24.4mL
 
  • #7
And what is the acid here?
 
  • #8
Borek said:
And what is the acid here?

The acid is 0.208 M HCl. I calculated that I should add 24.4mL.
 
  • #9
TheShiny said:
The acid is 0.208 M HCl.

Nope. This is the acid you added, but it is not part of the buffer. Every buffer consist of the conjugate pair - what are the acid and its conjugate base here?
 
  • #10
Acid: HA and C. Base: A- ?
 
  • #11
TheShiny said:
Acid: HA and C. Base: A- ?

Hard to say what you mean. A- typically means "any conjugate base", so it is not specific enough. And there is no such thing as C.

Buffer has ONE acid, not two, so whatever you meant, if you listed two acids you can't be right.
 
  • #12
Borek said:
Hard to say what you mean. A- typically means "any conjugate base", so it is not specific enough. And there is no such thing as C.

Buffer has ONE acid, not two, so whatever you meant, if you listed two acids you can't be right.

I think that the OP meant "C. Base: A-" = "Conjugate Base: A-" I.e. "C. Base" was an abbreviation for "conjugate base."
 
  • #13
Ah, OK. It still doesn't help much as HA and A- are hardly specific (and I suspect OP may be confused about the identity of the acid; that's quite common).
 
  • #14
TheShiny said:
"Prepare a buffer solution of pH 3.924 from 50 mL of 0.236 M solution of a sodium salt of a weak acid, Na+A- where the pka of the weak acid, HA, is 3.956. What volume, in mL, of 0.208 M HCl would need to be added?"
Sorry if I caused confusion, that is the question verbatim.

What I don't fully understand is how the "sodium salt of a weak acid" affects the calculations that I need to do. Because, based on inspection, comparing concentrations as well as pH and pka, I think that the volume to add should be roughly half (25mL) that of the initial solution.

Depending on how I do the sums I arrive at either 24.4mL or 27.3mL and I'm not sure which answer to use.
 
  • #15
OK, so you were right about HA and A-.

Initially solution contains A- only, where does the HA come from?
 
  • #16
That's what I'm not too certain about. Is it the water that the initial solution is diluted with?
 

Related to Buffer Solution Calculation Example

1. What is a pH buffer?

A pH buffer is a solution that resists changes in pH when small amounts of acid or base are added. It is composed of a weak acid and its conjugate base or a weak base and its conjugate acid.

2. How do you calculate the pH of a buffer solution?

The pH of a buffer solution can be calculated using the Henderson-Hasselbalch equation: pH = pKa + log ([conjugate base]/[weak acid]).

3. What is the purpose of a buffer solution?

The purpose of a buffer solution is to maintain a stable pH level in a system. It is commonly used in scientific experiments and in various industries to prevent drastic changes in pH that can negatively impact the desired results.

4. How do you prepare a buffer solution?

A buffer solution can be prepared by mixing a weak acid or base with its conjugate base or acid in a specific ratio. The exact ratio will depend on the desired pH level and the pKa of the weak acid or base.

5. Can you adjust the pH of a buffer solution?

Yes, the pH of a buffer solution can be adjusted by adding small amounts of acid or base. However, the buffer capacity may be affected if too much acid or base is added, leading to a significant change in pH.

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