Buffers
Chem 12A
Mrs. Kay
• Buffers help maintain a constant pH.
• They are able to accept small quantities of
acids and bases without drastically
changing their pH
• A buffer is composed of a weak acid
molecule in equilibrium with its conjugate
base and hydrogen ion (H+).
• CH3COOH < --- > CH3COO- + H+
Acidic buffer solution:
• An acidic buffer solution is simply one
which has a pH less than 7.
• Acidic buffer solutions are commonly
made from a weak acid and one of its salts
- often a sodium salt.
• Example: acetic acid and sodium acetate
CH3COOH and CH3COONa
Alkaline buffer solutions
• An alkaline buffer solution has a pH
greater than 7.
• Alkaline buffer solutions are commonly
made from a weak base and one of its
salts.
• A frequently used example is a mixture of
ammonia solution and ammonium chloride
solution. NH3 and NH4Cl
How do buffers work?
• It has to contain things which will remove
any hydrogen ions or hydroxide ions that
you might add to it - otherwise the pH will
change.
Below is a weak acid, if we added CH3COONa, its
like adding CH3COOCH3COOH < -- > CH3COO- + H+
There will be a shift to the left
• The solution will therefore contain these
important things:
• lots of un-ionised acetic acid;
• lots of acetate ions from the sodium acetate;
• enough hydrogen ions to make the solution
acidic.
Other things (like water and sodium ions) which
are present aren't important to the argument.
• Look to below website for a flash on how
buffers work:
• http://www.mhhe.com/physsci/chemistry/e
ssentialchemistry/flash/buffer12.swf
Buffer capacity and pH
• Buffer capacity is the amount of acid/base
the buffer can neutralize before the pH
begins to drastically change.
• The pH of the buffer depends on the Ka
for the acid and on the relative
concentrations of the acid/base of the
buffer
Calculations
What is the pH of a buffer that is 0.12 M in
lactic acid (HC3H5O3) and 0.10 M of
sodium lactate? For lactic acid,
Ka= 1.4 x 10-4
1. Determine the important species
involved
2. Set up ice table
3. Set up equilibrium expression
4. Plug in concentrations at equilibrium
1. HC3H5O3 <-- > C3H5O3 - + H+
HC3H5O3
0.12 M
C3H5O3 0.10 M
H+
0
-x
+x
+x
0.12 – x
0.10 + x
x
• Ka = 1.4 x 10-4 =[C3H5O3 – ][H+]
[HC3H5O3]
• Because of the small Ka, we expect x to
be small relative to 0.12 and 0.10 M, so
we can simplify our equation to…
• Ka = 1.4 x 10-4 =[0.10 ][x]
[0.12]
Solve for x, x =1.7 x 10-4
• Finally, solve for the pH of the buffer
pH = - log [1.7 x 10-4] = 3.77
Notice, when you put the value of x into the
original equation it makes no difference for
your final answer.
Addition of strong acids/bases to
buffers:
• We assume that the strong acid or base added
is consumed completely by the reaction with the
buffer
• HX < -- > H+ + X• By adding a strong acid, the shift is to the left,
therefore increasing [HX] and decreasing [X-]
• By adding a strong base, the OH- is used up by
HX to produce X-, so [X-] increases and [HX]
decreases.
Calculation
• A buffer is made by adding 0.300 mol of
HC2H3O2 and 0.300 mol of NaC2H3O2 to
1.0 L of water. The pH of the buffer is
4.74. Calculate the pH of this solution if
0.020 mol of NaOH is added (ignore
volume changes)
• HC2H3O2 +H2O  H3O+ +C2H3O2HC2H3O2 H2O
H3O+
C2H3O2
0.300
---
0
0.300
Change -0.020
---
+x
+0.020
Final
---
x
0.320
Initial
0.280
-
• Ka = (x)(0.320) / (0.280) = 1.8 x 10-5 so
x = [H3O+]= (0.280)(1.8 x 10-5 )/0.320
= 1.6 x 10-5
• pH = - log (1.6 x 10-5)= 4.80