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Acids and Bases

Acids and Bases. 2003 AP. #1 A-E. Write the Equilibrium, Kb, for the reaction represented. Writing an equilibrium expression for 1a. Using the law of mass action given the chemical equilibrium equation

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Acids and Bases

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  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. The pH of the solution is 8.82. 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 (14-8.82). • 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. 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.

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