CHAPTER 15
Buffered Solutions
Buffers – substances that resist a change in pH when an acid or base is added
- usually contains a weak acid and a salt that contains the conjugate base
of the acid
-
or a weak base and a salt that contains the conjugate acid of the base
ex. HF and NaF or NH3 and NH4Cl
For HF and NaF; Ka for HF is 7.2 x 10-4 => HF(aq) ↔ H+ + F Ka = [H+][F-] / [HF] or solving for [H+]: [H+] = Ka x [HF] / [F-]
Taking the negative log of both sides shows: pH = pKa + log[F-] / [HF];
In general, for any weak acid buffer: pH = pKa + log [base] / [acid]
This is called the Henderson-Hasselbach equation.
TTP 1] F: pH of a 0.45 M HF and 0.25 M NaF buffer solution.
Since Ka = 7.2 x 10-4 => pKa = - log (7.2 x 10-4) = 3.14
pH = pKa + log [F-] / [HF] => 3.14 + log (.25) / (.45) = 3.14 – 0.26 = 2.87
TTP 2] If 0.10 moles of solid NaOH is added to 1.00 L of the buffered solution just
described, find the new pH.
By adding base, the [F-] increases by 0.10, and the [HF] decreases by 0.10, thus making
the [F-] 0.35 M, and the [HF] 0.35 M.
Therefore: pH = pKa + log [F-] / [HF] => 3.14 + log (0.35) / (0.35) = 3.14
TTP 3] If 0.10 moles of pure HCl is added to 1.00 L of the buffered solution from TTP
#1, find the new pH.
Since 0.10 moles of HCl is added, this will cause the [HF] to increase to 0.55 M, and the
[F-] decreases to 0.15 M.
Thus: pH = pKa + log [F-] / [HF] => 3.14 + log (0.15) / (0.55) = 2.58
Buffer capacity: you can only add acid to a buffer until the base in the buffer runs out, or
you can only add as much base to a buffer as there is weak acid in the buffer. If you add
more moles of acid or base than there is in the buffer, you exceed the capacity of the
buffer. You might as well be adding the acid or the base to water.
Since pH = pKa + log [base] / [acid] , the ideal buffer has a high concentration of the
weak acid and the weak base, and their ratio is close to one.
Thus, pH = pKa when [base] = [acid].
Thus one would choose a buffer with a pKa closest to the pH I was trying to buffer.
TTP: If a chemist needs to buffer at pH 4.30, which buffer would the chemist choose?
A] chloroacetic acid (Ka = 1.35 x 10-3)
B] propanoic acid (Ka = 1.3 x 10-5)
C] benzoic acid (Ka = 6.4 x 10-5)
D] hypochlorous acid (Ka = 3.5 x 10-8)
Answer: C – benzoic acid
Titration and pH curves: see the calculations on p. 737 – 740
For a strong acid titrated
by a strong base:
pH
vol base
For a weak acid titrated
by a strong base
pH
vol base
note when the ½ point of the
titration is reached
for a strong acid titrated
by a weak base:
pH
vol base
Note when the ½ point of the
titration is reached
note where the equivalence point is above seven and below seven
For a diprotic acid being titrated
by NaOH
pH
vol NaOH