##### Diassociation Constant for a Weak Acid

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What's the pH of a solution that is made up of 0.25 M of sodium acetate and 0.50 M of acetic acid? (Ka= acetic acid 1.8 x 10^-5)

Titration of 100mL of 0.100 acetic acid (Ka=1.8 x10^-5) w/ 0.100 M NaOH. What's pH at equivalence, and half equivalence?

Feb 25th, 2015

QUESTION: What's the pH of a solution that is made up of 0.25 M of sodium acetate and 0.50 M of acetic acid? (Ka= acetic acid 1.8 x 10^-5)

ANSWER: By mixing a weak acid and a salt containing its conjugate base, we’ve created a buffer solution.  Buffer solns are solved by using the Henderson-Hasselbalch eqn:

pH = pka + log (A- / HA)

pH = -log(1.E-5) + log (.25/.50)

= 4.44 (with sigfigs in mind)

Feb 25th, 2015

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Feb 25th, 2015
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Feb 25th, 2015
Dec 5th, 2016
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