Section 19.1. Acid-Base Buffer Solutions
Section 19.1. Acid-Base Buffer Solutions
In everyday English, a buffer is something that lessens the impact of an external force.
** An acid-base buffer is a solution that lessens the change in [H3O+] that would result when a strong acid or base is added **
A buffer is a concentrated solution of a weak acid (or base), together with a salt containing the conjugate base (or acid).
i.e., a weak acid and its conjugate base or a weak base and its conjugate acid.
How does a Buffer work? The Common-Ion Effect (example of Le Chatelier's Principle)
= the shift in an equilibrium caused by the addition (or removal) of one of the species participating in the equilibrium.
Example: addition of sodium acetate (CH3COONa or NaAc) to acetic acid (CH3COOH or HAc) solution
CH3COOH (aq) + H2O (l) CH3COO- (aq) + H3O+ (aq)
acetic acid (HAc)
acetate ion (Ac-)
if we add CH3COO-
SHIFT
Net Effect:
[H3O+] decreases pH increases
Also, [OH-] increases pOH decreases
19-1
Example: addition of NH4Cl to NH3 solution. NH3 (aq) + H2O (l) NH4+(aq) + OH- (aq) if we add NH4+ Net Effect: [OH-] decreases, pOH increases [H3O+] increases, pH decreases
19-2
New type of problem to solve! ? solutions with two things dissolved
Example: Calculate [H3O+] in a solution that is 0.10 M in HF and 0.20 M in NaF. Also calculate % ionization. Problem: Use HF (aq) H+ (aq) + F- (aq) ? or F- (aq) + H2O (l) HF (aq) + OH- (aq) ? Answer: Since both include HF, F-, H+, OH- (the last two are related by
[H+][OH-] = 1 x 10-14), we can use either equation !!
HF (aq) [init] 0.10 M [change] -x [equil] (0.10-x)
H+ (aq) + F- (aq)
0
0.20 M
+x
+x
x
(0.20 + x)
Ka =
[H ][F-] [HF]
= 6.8 x 10-4 =
x(0.20 x)
(0.10 - x)
0.20 x 0.10
x = (0.10) (6.8 x 10-4)/0.20 = 3.4 x 10-4 M (x ................
................
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