1. 16.2 - Aqueous Ionic Equilibrium - Buffers
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Jake Halpern

2. What is a Buffer?
A buffer is a solution that resists change in pH when an acid or base is added to it. Buffers neutralize small amounts of added acid or base to maintain a nearly constant pH level.
A buffer solution can contain either:
Significant amounts of a weak acid and its conjugate base
Significant amounts of a weak base and its conjugate acid

3. Buffers cont.
For example, one buffer found in blood is carbonic acid (H2C03) and its conjugate base, the bicarbonate ion (HCO3-). If a base was added to the blood, the weak acid (H2C03) would react with it, neutralizing it. Likewise, if acid was added to the blood, the conjugate base (HCO3-) would react with it, neutralizing the solution.
However, it’s important to note that a weak acid or base by itself is not a buffer. Even though they partially ionize in water to form their conjugate bases or acids respectively, there is not a sufficient amount of these present to qualify the solution as a buffer.
A solution must contain significant amounts of both a weak acid and its conjugate base (or vice versa) to be considered a buffer.

4. Consider a buffer solution that was made by dissolving acetic acid (HC2H3O2) and sodium acetate (NaC2H3O2) in water.
If a strong base, such as NaOH, was added, which component of the solution would neutralize the base?
The acetic acid would neutralize the NaOH NaOH(aq) + HC2H3O2(aq) <-> H2O(l) + NaC2H3O2(aq)
If a strong acid, such as HCl, was added, which component of the solution would neutralize the acid?
The sodium acetate would neutralize the HCl HCl(aq) + NaC2H3O2(aq) <-> HC2H3O2(aq) + NaCl(aq)
As long as the amount of added acid or base is less than the amount of weak acid or base in the buffer, the added components are neutralized and the pH change is small.

5. Quick Question:
Which solution is a buffer?
A solution that is 0.1M in HNO2 and 0.1M in HCl
A solution that is 0.1M in HNO3 and 0.1M in NaNO3
(c) A solution that is 0.1M in HNO2 and 0.1M in NaCl
(d) A solution that is 0.1M in HNO2 and 0.1M in NaNO2
The correct answer is (d)
NaNO2 is the conjugate base to HNO2, which is a weak acid.
( (b) is incorrect because HNO3 is a strong acid! )

6. Calculating the pH of a Buffer Solution
How do we calculate the pH of a solution that contains an acid and its conjugate base? Consider a solution that contains HC2H3O2 and NaC2H3O2, each at an initial concentration of 0.1 M.
HC2H3O2(aq) + H2O(l)-> H3O+(aq) + C2H3O2-(aq)
The presence of the C2H3O2- already in the solution suppresses the ionization of the HC2H3O2 due to Le Châtlier’s principle (shifts equilibrium left, towards the reactants). In other words, the presence of a conjugate base already in the solution causes the acid to ionize even less than it normally would, resulting in a less acidic solution (higher pH). This is known as the common ion effect.

7. Calculating the pH of a Buffer Solution cont.
To find the pH of a buffer solution containing common ions, set up and solve an equilibrium problem (ICE table) with the initial concentrations of the acid and its conjugate base.
Practice Question #1:
Calculate the pH of a buffer solution that is M in HC2H3O2 and M in NaC2H3O2.
Ka=1.8x10⁻⁵

8. Practice Question:
First write out the balanced chemical equation for the ionization of the acid.
HC2H3O2(aq) + NaC2H3O2(aq) -> H3O+(aq) + C2H3O2-(aq)
[H3O+][C2H3O2-] (x)(0.1+x) x(0.1)
Ka= → 1.8x10⁻⁵ = → 1.8x10⁻⁵= (x is small)
[HC2H3O2] (0.1-x) (0.1)
x=1.8x10⁻⁵ → [H3O+] = x = 1.8x10⁻⁵ → pH=-log[H3O+]
=-log(1.8x10⁻⁵)
= 4.74
[HC2H3O2]
[H3O+]
[C2H3O2-]
Initial
0.100
≈0.00
Change -x +x
Equilibrium
0.100-x x
0.100+x

9. The Henderson-Hasselbalch Equation
Although you can use an equilibrium approach (like the last question) to solve for the pH of a buffer solution, the Henderson-Hasselbalch equation can make the calculation easier.
[A-]
pH = pKa + log On ref. table
[HA] remember, pKa= -log(Ka)
This equation allows us to quickly calculate the pH of a buffer solution from the initial concentrations of the components as long as x is small.
Practice Question #2:
Calculate the pH of a buffer solution that is M in benzoic acid (HC7H5O2) and M in sodium benzoate (NaC7H5O2) using the Henderson-Hasselbalch equation. Ka= 6.5x10⁻⁵

10. Practice Question 2: [A-] pH = pKa + log [HA] (0.150)
= -log(6.5x10⁻⁵) + log
(0.050) =
pH = 4.66
If you want to confirm that x is small, use the calculated pH to find the [H3O+]. The approximation is valid if the [H3O+] is less than 5% (0.050) of the initial concentration of the acid.

11. Calculating pH Changes in a Buffer Solution
Although buffers resist pH changes, the addition of an acid or base does change the pH slightly. Calculating the pH change requires breaking up the problem into two parts.
The Stoichiometric calculation in which we calculate how the addition changes the relative amounts of acid and conjugate base.
The Equilibrium Calculation in which we calculate the pH based on the new amounts of acid and conjugate base.

12. Calculating pH Changes in a Buffer Solution pt. 1
Consider a 1.0 L buffer solution that is 0.1 M in a generic acid HA and 0.1 M in its conjugate base A-. Let’s calculate the pH of the solution after an addition of moles of a strong acid (H⁺). Assume the change in volume from adding the acid is negligible.
As the added acid is neutralized, it converts a stoichiometric amount of the base into its conjugate acid through the neutralization reaction.
H⁺(aq) A⁻(aq) > HA(aq)
Added acid weak base in buffer
Neutralizing moles of the strong acid (H⁺) requires moles of the weak base (A⁻). As a result, the amount of A⁻ decreases by moles and the amount of HA increases by moles. We can visualize these changes in a table like this: H⁺ A⁻ HA
Before Addition
≈0.00 mol
0.100 mol
Addition
mol
After Addition
0.075 mol
0.125 mol
(This isn’t an ICE table!)

13. Calculating pH Changes in a Buffer Solution pt. 2
At this point we’ve seen that adding a small amount of acid to a buffer is equivalent to changing the initial concentrations of the acid and conjugate base in the buffer.
In this case, since the volume is 1.0L, [HA] increased from 0.1 M to M and [A-] decreased from 0.1 M to M. Knowing these new initial concentrations, we can calculate the pH in the same way we calculate the pH of any buffer: either using the equilibrium approach or by using the Henderson-Hasselbalch equation.

14. Practice Question #3:
A 1.0 L buffer solution contains mol of HC2H3O2 and mol NaC2H3O2. Calculate the pH after adding mol of solid NaOH to the buffer. Ignore small changes in volume that may occur upon addition of the base. The Ka for HC2H3O2 is 1.8x10⁻⁵.
(Use either the equilibrium approach or the Henderson-Hasselbalch Equation)

15. A 1. 0 L buffer solution contains 0. 100 mol of HC2H3O2 and 0
A 1.0 L buffer solution contains mol of HC2H3O2 and mol NaC2H3O2. Calculate the pH after adding mol of solid NaOH to the buffer. Ignore small changes in volume that may occur upon addition of the base.The Ka for HC2H3O2 is 1.8x10⁻⁵.
OH⁻(aq) + HC2H3O2(aq) -> H2O(l) + C2H3O2⁻(aq)
If you’re comfortable with it, I’d use the Henderson-Hasselbalch approach to solve this problem. Remember, x must be small to use this equation!
[base] (0.110)
pH = pKa + log pH = -log(1.8x10⁻⁵) + log pH =
[acid] (0.090) pH = 4.83
OH⁻
HC2H3O2
C2H3O2⁻
Before Addition
≈0.00 mol
0.100 mol
Addition
0.010 mol
After Addition
≈0.00
0.090 mol
0.110 mol

16. Buffers containing a Base and Its Conjugate Acid
So far we’ve only seen examples of buffers composed of an acid and its conjugate base (where the conjugate base is an ion), but remember that buffers can also be composed of a base and its conjugate acid (where the conjugate acid is an ion).
For example, a solution with significant amounts of both NH3 and NH4Cl will act as a buffer. The NH3 acts as a weak base, neutralizing small amounts of added acid, and the NH4+ acts as the conjugate acid, neutralizing small amounts of added base.

17. Buffers containing a Base and Its Conjugate Acid cont.
We can calculate the pH of a solution like this the same way that we calculate the pH of a buffer solution containing an acid and its conjugate base, only first we must find Ka or pKa for the conjugate acid of the weak base.
Ka x Kb = Kw or pKa + pKb = 14
Practice Question #4:
Use the Henderson-Hasselbalch equation to calculate the pH of a buffer solution that is 0.50 M in NH3 and 0.20 M in NH4Cl. For NH3, pKb = 4.75.
pKa + pKb = 14
pKa = 14 - pKb
pKa =
pKa = 9.25
[base] (0.50)
pH = pKa + log pH = log
[acid] (0.20)
pH = 9.65

18. Thanks for Listening
Any Questions?
As always, these slides will be on the website if you want to reference them.