Calculating pH of a Buffer Solution with Added Strong Base

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Discussion Overview

The discussion revolves around calculating the pH of a buffer solution after the addition of a strong base (NaOH) to a mixture of a weak acid (HC3H5O2) and its conjugate base (NaC3H5O2). The focus is on applying the Henderson-Hasselbalch equation and understanding the equilibrium dynamics of the buffer system.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • One participant notes the importance of the equilibrium expression for the weak acid and questions whether the sodium salt's dissociation can be ignored in the calculations.
  • Another participant suggests using the Henderson-Hasselbalch equation, emphasizing that the pH of the buffer depends on the pKa and the ratio of the acid to its conjugate base, and recommends using stoichiometry to find the new amounts after adding NaOH.
  • A participant attempts to calculate the pH using a formula but expresses uncertainty about missing initial concentrations for the conjugate base.
  • Clarification is provided regarding the initial concentration of the conjugate base, indicating that it is indeed present in the solution.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the correct approach to the calculation, with some focusing on the Henderson-Hasselbalch equation while others express confusion about the initial concentrations involved.

Contextual Notes

There is uncertainty regarding the treatment of the sodium salt in the equilibrium calculations and the initial concentrations of the components in the buffer solution.

Who May Find This Useful

This discussion may be useful for students studying buffer solutions, particularly in the context of acid-base chemistry and pH calculations in laboratory settings.

needphyshelp
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Homework Statement


Calculate the pH after 0.030 mol NaOH is added to 1.00 L 0.100 M HC3H5O2 and 0.100 M NaC3H5O2


Homework Equations


Ka=[H+][A-]/[HA]
pH=-log([H+]) or 14-pOH=ph


The Attempt at a Solution


HC3H5O2 <-> H+ + C3H5O2-
NaC3H5O2 <-> Na+ + C3H5O2-

I know that Na+ has no acid/base properties, so I think that this equation is unimportant and I will be focusing on the first equation. I also know Kaeq1=1.3x10-5

I tried to set up an equilibrium equation, but I'm not sure if I should be completely ignoring the sodium equation.

1.3x10-5=(x2)/(.100-.030)
the numerator= [H+][C3H5O2-]

Any input would be appreciated...
 
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Wouldn't that yield
pH=log(1.3e-5)+(.030/.1-.030) = 4.52? This answer is incorrect. Is there an initial concentration for C3H5O2- that I am missing?
 
needphyshelp said:
0.100 M NaC3H5O2

needphyshelp said:
Is there an initial concentration for C3H5O2- that I am missing?

Yes.
 
oh... duh..
thanks :-)
 

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