Acids and BasesAcids and Bases

2003 AP 2003 AP

#1 A-E#1 A-E

Write the Equilibrium, Write the Equilibrium, Kb, for the reaction Kb, for the reaction

representedrepresented

Writing an equilibrium Writing an equilibrium expression for 1aexpression for 1a

Using the law of mass action given the Using the law of mass action given the chemical equilibrium equationchemical equilibrium equation

Concentration Products over Reactants Concentration Products over Reactants raised to their stoichiometric coefficients raised to their stoichiometric coefficients excluding pure liquids and solids!excluding pure liquids and solids!

Writing an equilibrium Writing an equilibrium expression for 1aexpression for 1a

In this case, equilibrium expression In this case, equilibrium expression consists of the products of the consists of the products of the concentrations of Conjugate acid of concentrations of Conjugate acid of Aniline and Hydroxide Ion over the Aniline and Hydroxide Ion over the concentration of Aniline. concentration of Aniline.

A Sample of aniline is A Sample of aniline is dissolved in water to dissolved in water to produce 25mL of a produce 25mL of a

0.10M. The pH of the 0.10M. The pH of the solution is 8.82. solution is 8.82.

Calculate the Calculate the Equilibrium Constant Equilibrium Constant

Calculate the Kb for the Calculate the Kb for the reactionreaction

Use the equilibrium Use the equilibrium expression above to expression above to determine the Kb. determine the Kb.

The concentration of The concentration of aniline is 0.1M aniline is 0.1M

The Hydroxide and The Hydroxide and Conjugate acid Conjugate acid concentration can both be concentration can both be determined after determined after calculation of the pOH (14-calculation of the pOH (14-8.82).8.82).

Take the pOH and raise it Take the pOH and raise it to the 10^(-pOH) to the 10^(-pOH)

Calculate the Kb for the Calculate the Kb for the reactionreaction

This will yield the This will yield the concentration for concentration for OH- and Aniline OH- and Aniline Conjugate because Conjugate because they are produce in they are produce in the same proportion the same proportion 1:1. 1:1.

Multiply these Multiply these concentrations and concentrations and divide them by the divide them by the initial concentration initial concentration of Aniline.of Aniline.

Calculate Kb Calculate Kb Assume the initial Assume the initial

concentration of concentration of [OH-] is negligible. [OH-] is negligible.

The reason x is not The reason x is not subtracted from the subtracted from the initial concentration initial concentration of aniline is because of aniline is because x is so small that it x is so small that it is considered is considered negligible. negligible.

The solution prepared in The solution prepared in part b is titrated with a part b is titrated with a 0.10M. Calculate the pH 0.10M. Calculate the pH

of the solution when of the solution when 5.0mL of the acid has 5.0mL of the acid has

been added. been added.

Calculate pH after 5mL of HCl Calculate pH after 5mL of HCl is added is added

Set up a net chemical equation Set up a net chemical equation for the reaction of Aniline and for the reaction of Aniline and H+ (Strong Acid)H+ (Strong Acid)

Since the initial concentrations Since the initial concentrations of OH- and conjugate acid are of OH- and conjugate acid are small they are not taken into small they are not taken into account. account.

Multiply the Molarity of Aniline Multiply the Molarity of Aniline by the volume in liters (same by the volume in liters (same with the HCl) with the HCl)

From here you have the moles From here you have the moles of both Aniline and HCl of both Aniline and HCl

Calculate pH after 5mL of HCl Calculate pH after 5mL of HCl is addedis added

The H+ will combine The H+ will combine with Aniline to form a with Aniline to form a conjugate acid and goes conjugate acid and goes to completion.to completion.

Subtract the number of Subtract the number of moles of H+ from the moles of H+ from the moles of Aniline. This moles of Aniline. This gives a new number of gives a new number of moles of Aniline. moles of Aniline.

Since all of the H+ Since all of the H+ moles are consumed, it moles are consumed, it is equal to the number is equal to the number of moles of the of moles of the Conjugate acid. Conjugate acid.

Calculate pH after 5mL of HCl Calculate pH after 5mL of HCl is addedis added

To determine the pH, we To determine the pH, we will use a variation of the will use a variation of the Henderson-Hassalbauch. Henderson-Hassalbauch. Below Below

From here we take the pKb From here we take the pKb calculated above and the calculated above and the mole ratio of the acid over mole ratio of the acid over the base. the base.

The pOH can be The pOH can be determined determined

To get the pOH we To get the pOH we subtract the pOH from 14 subtract the pOH from 14 to get the pH. to get the pH.

Calculate the pH at Calculate the pH at the equivalence the equivalence

point. point.

pH at the equivalence point pH at the equivalence point We already know the We already know the

moles of the Aniline. moles of the Aniline. The moles of Aniline The moles of Aniline

must equal the moles must equal the moles of HCl for the of HCl for the equivalence point to equivalence point to be reached. (Ratio 1:1)be reached. (Ratio 1:1)

From here we have From here we have the only Aniline the only Aniline Conjugate Acid moles. Conjugate Acid moles.

We must determine We must determine the concentration of the concentration of conjugate acid and the conjugate acid and the Ka of the conjugate Ka of the conjugate

pH at the equivalence pointpH at the equivalence point 25mL initially of aniline25mL initially of aniline To determine the volume of HCl take the moles of To determine the volume of HCl take the moles of

HCl and divide it by the concentration which HCl and divide it by the concentration which yields the volume required. yields the volume required.

Convert all volumes to liters and add the initial Convert all volumes to liters and add the initial volume of aniline with the volume of HCl added. volume of aniline with the volume of HCl added.

Take the moles of conjugate acid and divide it by Take the moles of conjugate acid and divide it by this new volume this new volume

pH at the equivalence pointpH at the equivalence point

To determine the Ka dived the Kb into (1.0 To determine the Ka dived the Kb into (1.0 x 10^-14).x 10^-14).

pH at the equivalence pointpH at the equivalence point Write the Chemical Write the Chemical

reaction for the behavior reaction for the behavior of Aniline conjugate in of Aniline conjugate in water and make an water and make an equilibrium expression.equilibrium expression.

x is not subtracted from x is not subtracted from the initial concentration the initial concentration because it is considered because it is considered negligible. negligible.

Multiply the Multiply the concentration of concentration of Conjugate aniline Conjugate aniline concentration by the Ka. concentration by the Ka. Then take the square Then take the square root of the product. root of the product.

pH at the equivalence pointpH at the equivalence point

Then take –log (x) Then take –log (x) of the answerof the answer

This will give you This will give you the pH at the the pH at the equivalence pointequivalence point

pH = 2.97pH = 2.97

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

Selecting an IndicatorSelecting an Indicator

Based on the calculations above the Based on the calculations above the pH at the end point is 2.97pH at the end point is 2.97

Erythrosine is optimal because based Erythrosine is optimal because based on the color change in an acidic pHon the color change in an acidic pH

It is a weak based titration with a It is a weak based titration with a strong acid. strong acid.