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the titration of 25.00 mL of a 0.0800 M acetic acid solution with 0.0700 M KOH. Assume that the temperature is 25 oC. 

What is the pH when 6.00 mL of the KOH solution have been added? 

Feb 21st, 2015

Anytime that you have a weak acid (here, acetic acid) and a strong base (here, KOH) mixed together, AND the mols of acid > mols of base added, then you can safely use the Henderson-Hasselbalch equation.

                           Henderson-Hasselbalch equation:

                                       pH = pka + log(A-/HA)

where   A- is the conjugate base (ion after dissociation)

             HA is the unassociated acid

First, neutralization occurs:

HA + OH- -> H20 + A-

Mols of OH- added  =  0.006L * 0.0700M = 0.00042 mols

Mols of HA before mixing =  0.025L *  0.0800M = 0.002 mols


After neutralization occurs:

Mols of HA after mixing  =  0.002mols -  0.00042 mols = 0.00158 mols

Mols of A- after mixing  = mols of OH- added =  0.00042 mols

Plug and chug into the Henderson Hasselbalch eqn

pH = -log(1.8*10^-5) + log(0.00042/0.00158)

 pH= 4.17  

Feb 21st, 2015

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