Adding an acid or a base to a buffer solution is the same as carrying out a titration. We discuss the pH changes in a buffer solutions using titration of weak acid by a strong base. |
The blood is a natural buffer, and so are other body fluids and plant fluids due to mixtures of weak acids and bases present in them.
On this page, we explore the reasons why the pH of buffer solutions resists to change.
Buffer solutions are required for many chemical experiments. They are also useful to standardize pH meters. Thus, there are many suppliers of buffer solutions.
There are also computer programs available to help design and make buffer solutions on the internet. For example:
We illustrate the titration of a weak acid by a strong base using the following examples.
During the titration process and before the equivalent point is reached, some acid has been neutralized by the strong base, and the solution contains a weak acid and its salt. The solution acts as a buffer.
Example 1.
Solution
Let HA represent the weak acid, and assume x M of it is ionized.
Then, the ionization and equilibrium concentration is
HA = H^{+} + A^{-} C_{a}-x x x x^{2} K_{a} = ------ C_{a}-x x^{2} + K_{a}x - C_{a}K_{a} = 0 -K_{a} + (K_{a}^{2} + 4 C_{a}K_{a})^{1/2} x = --------------------- 2 pH = -log(x)
Discussion
The method has been fully discussed in
Weak acids and bases equilibrium.
Symbols are used here, but approximations may be applied to numerical
problems.
Example 2.
Solution
After mixing, the concentrations C_{a} and C_{s}
of the acid HA and its salt NaA respectively are
HA = H^{+} + A^{-} C_{a}-x x x NaA = Na^{+} + A^{-} C_{s} C_{s} Common ion [A^{-}] = x+C_{s} x(x+C_{s}) K_{a} = -------- C_{a}-x x^{2} + (K_{a}+C_{s})x - C_{a} K_{a} = 0 -(K_{a}+C_{s}) + ((K_{a}+C_{s})^{2} + 4 C_{a} K_{a})^{1/2} x = ----------------------------- 2 pH = -log(x)
Discussion
The formulas for x and the pH derived above can be used to estimate
the pH of any buffer solution, regardless how little salt or acid is used
compared to their counter part.
When the ratio C_{a} / C_{s} is between 0.1 and 10, the Henderson-Hasselbalch equition is a convenient formula to use.
[H^{+}] [A^{-}] K_{a} = ---------- [HA] [A^{-}] pK_{a} = pH - log (----) [HA]The Henderson-Hasselbalch equation is
[A^{-}] pH = pK_{a} + log (----) [HA] [C_{s}] + x = pK_{a} + log (----) [C_{a}] - xBecause when x is insignificant in comparision to C_{s}, [A^{-}] = C_{s} and [HA] = C_{a}
Example 3
Base
added[H^{+}] pH See
0.0 mL 0.00316 2.500 A.
0.1 9.07e-4 3.042 B.
1.0 9.07e-5 4.042 C.
5.0 1.0e-5 5.000 D.
10.0 10^{-11.349} 11.349 E.
Solution
Note the sharp increase in pH when 0.1 mL (3 drops) of basic solution is added to the solution.
HA = H+ + A- C_{a}-x x x [A-] = x + 0.0099 (= C_{s}) x (x + 0.0099) K_{a} = -------------- = 1e-5 0.98 - x x^{2} + 0.0099 x = 9.8e-6 - 1e5 x x^{2} + (0.0099 + 1e5) x - 9.8e-6 = 0 x = (-0.0099 + (0.0099^{2} + 4*0.98*1e-5)^{1/2}) / 2 = 0.000907 pH = 3.042
Note that using the Henderson-Hasselbalch equation will not yield the correct solution. Do you know why?
HA = H+ + A- C_{a}-x x x+0.0901 x (x + 0.0901) K_{a} = -------------- = 1e-5 0.818 - x x = (-0.0901 + (0.0901^{2} + 4*0.818*1e-5)^{1/2}) / 2 = 0.0000907 (any approximation to be made?) pH = 4.042Using the Henderson Hasselbalch equation yield
0.0901(=[A^{-}]) pH = pK_{a} + log (------) = 5 - 0.958 = 4.042 (same result) 0.818(=[HA])
HA = H+ + A- C_{a}-x x x+0.3333 x (x + 0.333) K_{a} = -------------- = 1.0e-5 0.333 - x x ^{2} + (0.333-1e-5)x - 0.333*1e-5 = 0 x = (-0.333 + (0.333^{2} + 4*3.33e-6)^{1/2}) / 2 = 0.0000010 pH = 5.000
Note: Using the Henderson Hasselbalch equation yield the same result
0.333(=[A^{-}] pH = pK_{a} + log (----) = 5 + 0.000 = 5.000 0.333(=[HA])
A^{-} + H_{2}O = HA + OH^{-} C_{s}-x x x Equilibrium concentrations [HA] [OH^{-}] [H^{+}] K_{b} = ---------- ---- [A^{-}] [H^{+}] K_{w} 1e-14 x ^{2} = --- = ----- = 1e-9 = ------ K_{a} 1e-5 C_{s}-x x = (0.500*1.0e-9)^{1/2} = 2.26e-5 pOH = -log x = 2.651; pH = 14 - 2.651 = 11.349
Note: The calculation here illustrate the hydration or hydrolysis of the basic salt NaA.
Discussion..
Sketch the titration curve based on the estimates given in this example,
and notice the points made along the way.
Hint..
Learn it by doing! Perform all the calculations outlined in Example 3,
and then you will be an expert for solving buffer and hydration problems.
Hint..
The following relationship is handy to use, but make sure you know why
K_{b} = K_{w} / K_{a} = ?
What is the pH at the equivalence point when a 0.10 M HF solution is
titrated by a 0.10 M NaOH solution?
Hint..
A simple consideration leads to the formulation:
[Na^{+}] = [F^{-}] = 1.0 mL * 0.10 M / 11.0 mL
Answer[F^{-}]=0.0333 M
Answer pH = pKa = 3.17
Hint..
What is the pH of the equivalence point during the titration of 0.10 M
HF using 0.10 M NaOH solution? (Another way of asking the same question.)