PH of buffer with addition of conjugate base

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Discussion Overview

The discussion revolves around calculating the pH of a buffer solution consisting of acetic acid and sodium acetate after the addition of sodium acetate. The focus is on the application of the Henderson-Hasselbalch equation and the implications of adding a conjugate base to a buffer system.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • One participant describes the initial setup of the problem, noting the concentrations of acetic acid and sodium acetate and the addition of sodium acetate.
  • Another participant suggests that the approach using the Henderson-Hasselbalch equation is correct but advises checking the math involved.
  • A third participant attempts to apply the equilibrium expression for the acid dissociation constant (Ka) but expresses confusion over the resulting hydrogen ion concentration and its impact on the buffer system.
  • A later reply emphasizes that only the concentration of the conjugate base should be adjusted while the acid concentration remains unchanged, recommending the direct use of the Henderson-Hasselbalch equation.

Areas of Agreement / Disagreement

Participants generally agree on the use of the Henderson-Hasselbalch equation but exhibit uncertainty regarding the calculations and the impact of the added conjugate base on the buffer system. The discussion remains unresolved as participants are still working through the calculations.

Contextual Notes

There are limitations in the discussion regarding the assumptions made about the buffer system's behavior under the addition of the conjugate base, particularly concerning the significance of the small changes in concentrations and the mathematical steps involved in the calculations.

Maharg
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Homework Statement



Consider 1L of a solution that is 0.100 M in acetic acid and 0.100 M in sodium acetate. What will the pH be, when 0.0900 mol of sodium acetate are added?
Ignore changes in volume.

acetic acid pka = 1.75 E-5

Homework Equations



HA --> H+ + A-
pH=pka +log A-/HA


The Attempt at a Solution



I had similar questions that I know how to do with strong acids. The difference in this question is that the added acid/base is a weak one. With the strong base addition I can just add it to A- concentration and subtract it from HA and use Henderson-Hasselbach equation.

This one I tried that but the numbers I know are off. I also tried doing ka = H+*(0.1+.009)/0.1

Solving for H+ and then using that to find out concentrations of A- and HA but the H+ number is far too small to make a difference and I'm fairly sure I can't just add the conjugate bases together.

Can someone please help.
 
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Maharg said:
I also tried doing ka = H+*(0.1+.009)/0.1

0.009? * instead of log? This is a correct approach, just check your math.
 
Borek said:
0.009? * instead of log? This is a correct approach, just check your math.

Oops I made it a mistake, anyways when I did my math I just did 1.75E-5 = H+(0.19)/0.1

I get 9.2E-6

but I have no idea what to do with this. If I do it the same way as a strong base question I would add it to the A- concentration and subtract it from HA, but this value is too small to make a difference to a 0.1 M.
 
You have added conjugate base - so change only base concentration, acid concentration is untouched. Simply plug numbers into HH equation.
 

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