Calculating pH of a Buffer Solution with Added Strong Base

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To calculate the pH of a buffer solution after adding 0.030 mol of NaOH to 1.00 L of 0.100 M HC3H5O2 and 0.100 M NaC3H5O2, the Henderson-Hasselbalch equation can be utilized, focusing on the ratio of the acid and its conjugate base. The equilibrium constant, Ka, is given as 1.3x10^-5, and the concentration changes after NaOH addition must be accounted for using stoichiometry. The initial concentration of the conjugate base, C3H5O2-, is crucial for accurate calculations. A misunderstanding regarding the initial concentrations led to an incorrect pH calculation of 4.52. Correctly applying the buffer equations will yield the accurate pH value.
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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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