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Catalyst

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- LT 7.21 – I can explain how to write the equilibrium constants for polyprotic acids and identify its equivalence points.

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- Polyprotic acids are acids have more than one H that can be considered acidic.
H2SO3 H+ + HSO3- Ka1 = 1.7 x 10-2

HSO3- H+ + SO32- Ka2 = 6.4 x 10-8

- Each removal of a proton has it’s own Ka value.
- It is always easier to remove the first proton from a polyprotic than it is is to remove the second proton.
- As long as Ka1 > Ka2 by a factor of at least 103, then we only need to consider first Ka when calculating values like pH.

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- The solubility of CO2 in water at 25 0C is 0.0037 M. the common practice is to assume that all the dissolved CO2 is in the form of carbonic acid (H2CO3), which is produced in the reaction:
CO2 + H2O H2CO3

- What is the pH of a 0.0037 M solution of H2CO3. Ka1 = 4.3 x 10-7 and Ka2 = 5.6 x 10-11

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- Calculate the pH of a solution of oxalic acid (H2C2O4). Ka1 = 5.9 x 10-2 and Ka2 = 6.5 x 10-5

- When titrating a polyprotic acid with a strong base, the reaction proceeds in multiple steps.
H3PO4 H+ + H2PO4- Ka1

H2PO4- H+ + HPO42-Ka2

HPO42-H+ + PO43- Ka3

- All of the first species must complete react before the second proton is removed.
- There will be the same number of equivalence points as there are acidic protons

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- How many acidic protons does the following polyprotic acid have according to the graph below?

- How many acidic protons does the following polyprotic acid have according to the graph below?

- You have 45 minutes to work on the drills, integrated practice, and FRQ practice.
- Options:
- Drill – Weak Acid/Base Equilibrium
- Drill – Buffers
- Drill – Weak Acid/Base Titrations
- Integrated Practice – Acids and Bases
- FRQ Practice – FRQ #1: Equilibrium

- Exit Slip in last 10 minutes of class

- Read: 16.6 and 17.3
- Homework: