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Catalyst. A Problem. I have two beakers Beaker 1 – 1 L of pure water Beaker 2 – 1 L of human blood I 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 Problem. I have the same two beakers
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A Problem • I have two beakers • Beaker 1 – 1 L of pure water • Beaker 2 – 1 L of human blood • I 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 Problem • I have the same two beakers • Beaker 1 – 1 L of pure water • Beaker 2 – 1 L of human blood • I 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 – TPS • Why does the pure water show such a drastic change in pH, but the human blood exhibits such a small change in pH?
Today’s Learning Targets • LT 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.
Buffers • A buffer is any solution that resists changes in pH • Two components of a buffer: • A component that neutralizes an acid • 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 Buffer • The 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.
Class Example • You 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 Talk • 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)
Adding Strong Acid/Base to Buffer • Adding base to a buffer shows minimal change • Adding base to a neutral solution causes huge changes
Henderson – Hasselbalch Equation • When 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
Class Example • Calculate 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 Talk • Calculate 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 Working • The pH at which any buffer works most effectively is when pH = pKa • This is known at the ½ equivalence point • Buffers usually have a useable range within ±1 pH unit of the pKa
Pushing It to the Next Level! • Green = “Cake Walk Level” Buffer Problems • Yellow = “Heating Up Level” Buffer Problems • Red = “Expert Chemist Level” Buffer Problems • Each correct answer yields the following points: • Green = 1 point • Yellow = 5 points • Red = 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
Creating a Titration Curve • 4 important points to plot: • Initial pH – Equilibrium Problem • Half – Equivalence Point – Buffer Problem • Equivalence Point – Weak Base Problem • 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 Poster • Create 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: • Buffer • Conjugate acid/base pair • Dominant 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 Ticket • 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. • 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 Time • Read: 17.1, 17.2, and 17.3 • Homework: