1. Acids and Bases

2. 2003 AP
#1 A-E

3. Write the Equilibrium, Kb, for the reaction represented

4. Writing an equilibrium expression for 1a
Using the law of mass action given the chemical equilibrium equation
Concentration Products over Reactants raised to their stoichiometric coefficients excluding pure liquids and solids!

5. Writing an equilibrium expression for 1a
In this case, equilibrium expression consists of the products of the concentrations of Conjugate acid of Aniline and Hydroxide Ion over the concentration of Aniline.

6. A Sample of aniline is dissolved in water to produce 25mL of a 0. 10M
A Sample of aniline is dissolved in water to produce 25mL of a 0.10M. The pH of the solution is Calculate the Equilibrium Constant

7. Calculate the Kb for the reaction
Use the equilibrium expression above to determine the Kb.
The concentration of aniline is 0.1M
The Hydroxide and Conjugate acid concentration can both be determined after calculation of the pOH ( ).
Take the pOH and raise it to the 10^(-pOH)

8. Calculate the Kb for the reaction
This will yield the concentration for OH- and Aniline Conjugate because they are produce in the same proportion 1:1.
Multiply these concentrations and divide them by the initial concentration of Aniline.

9. Calculate Kb Assume the initial concentration of [OH-] is negligible.
The reason x is not subtracted from the initial concentration of aniline is because x is so small that it is considered negligible.

10. The solution prepared in part b is titrated with a 0. 10M
The solution prepared in part b is titrated with a 0.10M. Calculate the pH of the solution when 5.0mL of the acid has been added.

11. Calculate pH after 5mL of HCl is added
Set up a net chemical equation for the reaction of Aniline and H+ (Strong Acid)
Since the initial concentrations of OH- and conjugate acid are small they are not taken into account.
Multiply the Molarity of Aniline by the volume in liters (same with the HCl)
From here you have the moles of both Aniline and HCl

12. Calculate pH after 5mL of HCl is added
The H+ will combine with Aniline to form a conjugate acid and goes to completion.
Subtract the number of moles of H+ from the moles of Aniline. This gives a new number of moles of Aniline.
Since all of the H+ moles are consumed, it is equal to the number of moles of the Conjugate acid.

13. Calculate pH after 5mL of HCl is added
To determine the pH, we will use a variation of the Henderson-Hassalbauch. Below
From here we take the pKb calculated above and the mole ratio of the acid over the base.
The pOH can be determined
To get the pOH we subtract the pOH from 14 to get the pH.

14. Calculate the pH at the equivalence point.

15. pH at the equivalence point
We already know the moles of the Aniline.
The moles of Aniline must equal the moles of HCl for the equivalence point to be reached. (Ratio 1:1)
From here we have the only Aniline Conjugate Acid moles.
We must determine the concentration of conjugate acid and the Ka of the conjugate

16. pH at the equivalence point
25mL initially of aniline
To determine the volume of HCl take the moles of HCl and divide it by the concentration which yields the volume required.
Convert all volumes to liters and add the initial volume of aniline with the volume of HCl added.
Take the moles of conjugate acid and divide it by this new volume

17. pH at the equivalence point
To determine the Ka dived the Kb into (1.0 x 10^-14).

18. pH at the equivalence point
Write the Chemical reaction for the behavior of Aniline conjugate in water and make an equilibrium expression.
x is not subtracted from the initial concentration because it is considered negligible.
Multiply the concentration of Conjugate aniline concentration by the Ka. Then take the square root of the product.

19. pH at the equivalence point
Then take –log (x) of the answer
This will give you the pH at the equivalence point
pH = 2.97

20. Which of the following indicators listed is most suitable for this titration.

21. Selecting an Indicator
Based on the calculations above the pH at the end point is 2.97
Erythrosine is optimal because based on the color change in an acidic pH
It is a weak based titration with a strong acid.