Catalyst

A ProblemI have two beakersBeaker 1 – 1 L of pure

waterBeaker 2 – 1 L of human

bloodI pour 5 mL of NaOH in

the pure water and the pH goes from 7 up to 13.2

I pour 5 mL of NaOH into the blood and it goes from a pH of 7.2 to 7.3

A ProblemI have the same two

beakersBeaker 1 – 1 L of pure

waterBeaker 2 – 1 L of human

bloodI pour 5 mL of HCl in the

pure water and the pH goes from 7 up to 2.2

I pour 5 mL of NaOH into the blood and it goes from a pH of 7.2 to 7.1

Justify – TPSWhy does the pure water show such a drastic

change in pH, but the human blood exhibits such a small change in pH?

Lecture 7.7 – Buffers and Weak Acid Titrations

Today’s Learning TargetsLT 7.18 – I can explain what a buffer is and

how a buffer can be created in the laboratory. LT 7.19 – I can calculate the concentration of

compounds in a buffer solution following the addition of a strong acid/base. Furthermore, I can calculate the pH of this solution.

LT 7.20 – I can interpret the titration curve for weak acid/strong base titration and I can calculate the initial pH, half equivalence point, equivalence point and unknown concentrations from titration data.

BuffersA buffer is any solution that resists changes in

pHTwo components of a buffer:

1. A component that neutralizes an acid2. A component that neutralizes a base

A weak acid or a weak base are capable of creating a buffer because they have both of these components, but a strong acid or base cannot create a buffer.

Weak Acids/Bases Make Excellent Buffers

A weak acid or base make an excellent buffer because they have a component that can react with an acid and a component that can react with a base.

CH3COOH + H2O CH3COO- + H+

⇌

Can react with a base!

Can react with an acid!

Therefore, reacting each component produces a part of the equilibrium and little change in pH is observed!

How to Make a BufferThe best buffers have close to equal

concentrations of the conjugate acid/base pair.We can think about the Ka expression for a

weak acid

Therefore, pH is determined by the ratio of conjugate acid/base pair and the value of Ka.

As long as the change in ratio of [HA]/[A-] is small, the change in pH will be small.

€

Ka =[H +][A−]

[HA]

[H +] =Ka[HA]

[A−]

Class ExampleYou have a buffer system of methylamine

(CH3NH2) that has a Kb of 5.0 x 10-4 and a conjugate acid (CH3NH3

+) whose Ka is 2.0 x 10-

11. You have 40 mL of buffer with a concentration of 0.50 M. You add 10 mL of 0.100 M HCl to this buffer. Calculate the pH before adding the HCl and the pH after adding the HCl.

Table TalkYou have a solution of acetic acid (CH3COOH)

that was created by adding 0.300 moles of CH3COOH and 0.300 moles of CH3COO- to enough water to make a 1.000 L solution. You add in 5.0 mL of 4.0 M NaOH. Calculate the pH before adding in the NaOH and the pH after adding in the NaOH (Ka for the solution is 1.8 x 10-5)

Adding Strong Acid/Base to BufferAdding base to a buffer

shows minimal changeAdding base to a neutral

solution causes huge changes

Henderson – Hasselbalch EquationWhen we have a buffer, we do not need to use

ICE tables to determine the pH of the solution.We can use the Henderson-Hasselbalch

equation in order to solve for the pH

€

[H +] =Ka[HA]

[A−]

−log[H +] = −log Ka[HA]

[A−]

⎛

⎝ ⎜

⎞

⎠ ⎟

pH = logKa + −log[HA]

[A−]

€

pH = pKa − log[HA]

[A−]

Class ExampleCalculate the pH of a buffer that is 0.12 M

lactic acid and 0.10 M sodium lactate. The Ka for lactic acid is 1.4 x 10-4

Table TalkCalculate the pH of a buffer composed of 0.12

M benzoic acid and 0.20 M sodium benzoate. The Ka for the solution is 6.4 x 10-5

When Buffers Stop WorkingThe pH at which any buffer works most

effectively is when pH = pKa

This is known at the ½ equivalence pointBuffers usually have a useable range within ±1

pH unit of the pKa

Pushing It to the Next Level!Green = “Cake Walk Level” Buffer ProblemsYellow = “Heating Up Level” Buffer ProblemsRed = “Expert Chemist Level” Buffer ProblemsEach correct answer yields the following

points:Green = 1 pointYellow = 5 pointsRed = 10 points

Titrating a Weak Acid/Base with a Strong Acid/Base When we titrate a weak acid/base with a strong acid/base, the titration curve looks different than the ones previously studied. A strong acid/base titration goes to completion the instant we add the solution.A weak acid/base titration reacts and than reestablishes equilibrium

Buffer Region

Creating a Titration Curve 4 important points to plot:

1. Initial pH – Equilibrium Problem2. Half – Equivalence Point – Buffer Problem3. Equivalence Point – Weak Base Problem4. Excess Strong Acid/Base – strong acid/base pH

problem

Class Example You create a solution of 100.0 mL of 0.100 M HN3 where Ka = 1.9 x 10-5. You titrate the weak acid with 0.100 M NaOH. Create a titration curve for this titration.

Table Talk You have 50.0 mL of 0.100 M CH3COOH (Ka = 1.8 x 10-5) with 0.100 M NaOH. Create a titration curve for this titration using the 4 points.

Collaborative PosterCreate a poster with your group members

solving the problem that you were given.You must explain and justify your answers

using the following vocabulary/phrases:BufferConjugate acid/base pairDominant species in solution

The poster should be explained in such a way that another AP Chemistry student could learn about titration curves from your poster alone.

Exit Ticket1. You create a solution of 50.0 mL of 0.100 M HN3

where Ka = 1.9 x 10-5. You titrate the weak acid with 0.100 M NaOH. Create a titration curve for this titration.

2. You have a solution of acetic acid (CH3COOH) that was created by adding 0.300 moles of CH3COOH and 0.300 moles of CH3COO- to enough water to make a 1.000 L solution. You add in 5.0 mL of 4.0 M NaOH. Calculate the pH before adding in the NaOH and the pH after adding in the NaOH (Ka for the solution is 1.8 x 10-5)

Closing TimeClosing TimeRead: 17.1, 17.2, and 17.3Homework: