Acid-Base Titration



Acid-Base Titration Curves

NAME:________________________________________ PERIOD:___________

Prelab

Show Calculations.

1. For the titration of 50.0 ml of 0.100M acetic acid with 0.200M NaOH, using a Ka of

1.76 x 10-5, calculate the pH:

a. Initial (0 ml of NaOH added):

b. At the half equivalence point:

c. At the equivalence point:

d. 15 ml beyond the equivalence point:

2. What is the difference between the endpoint in a titration and the equivalence point?

3. Does the use of a weak acid indicator have much effect on the accuracy of the endpoint of a titration? Why or why not?

Acid-Base Titration Curves

Three important types of acid/base reactions will be investigated in this lab. The change in pH when a strong acid is titrated with a strong base, when a weak acid is titrated with a strong base, and when a weak base is titrated with a strong acid. Each will have a characteristic shape when the titration curve is plotted, that is, the pH is plotted on the y- axis and the volume (mL) of added titrant (the solution in the buret) is plotted on the x-axis. A titration curve allows one to experimentally determine the Ka for a weak acid or the Kb for a weak base. The pH will be measured using a pH probe attached to a CBL and graphing calculator.

In the titration of a strong acid with a strong base, hydrochloric acid, HCl, will be reacted with sodium hydroxide, NaOH. The concentration of the NaOH solution will be given and you will determine the unknown concentration of the HCl. This is referred to as standardizing the HCl solution. HCl reacts with NaOH in a one-to-one mole ratio to produce water in the overall reaction:

HCl (aq) + NaOH (aq) ( H2O (l) + NaCl (aq)

The HCl ionizes 100% and the NaOH dissociates completely in water solution giving the ionic equation:

H+1(aq) + Cl-1(aq) + Na+1(aq) + OH-1(aq) ( H2O (l) + Na+1(aq) + Cl-1 (aq)

Removing the spectator ions gives the net ionic equation:

H+1(aq + OH-1(aq) ( H2O (l)

The equivalence point for a strong acid strong base titration will be pH 7 since the Na+1 ion and the Cl-1 (the conjugate base of a strong acid) do not undergo hydrolysis. The pH of the solution is therefore the pH of pure water. The volume of NaOH solution used to reach the equivalence point will be used to determine the molarity of the HCl.

A monoprotic weak acid reacts with NaOH in a one-to-one ratio to produce water and the conjugate base in the overall reaction:

HA (aq) + NaOH (laq ( H2O (l) + NaA (aq)

A weak acid does not ionize completely in aqueous solution giving the ionic equation:

HA (aq) + Na+1(aq) + OH-1(aq) ( H2O (l) + Na+1(aq) + A-1 (aq)

Removing the spectator ions gives the net ionic equation:

HA (aq) + OH-1(aq) ( H2O (l) + A-1(aq)

The equation for the ionization of the weak acid is shown below.

HA (aq) + H2O (l) ( H3O+1 (aq) + A-1 (aq)

The equilibrium constant for this reaction, Ka is:

[pic]

[A-1] is the molar concentration of the conjugate base and [HA] is the molar concentration of the weak acid.

Ka can be calculated using the initial concentration of the acid and the initial pH of the solution. The initial pH gives the [H3O+1] which equals the [A-1] in the initial weak acid solution. If the weak acid is only slightly ionized, the [HA] is assumed to be approximately equal to its initial concentration.

Ka can be calculated from the pH at the half-equivalence point. At this point in the titration, half of the moles of HA have been converted to A-1. The [A-1] = [HA], the ratio [A-1]/[HA] equals one, the [H3O+1] equals Ka, and the pH of the solution equals the pKa of the weak acid.

Ka can be calculated from the pH at the equivalence point. The moles of HA have been completely converted to A-1. The conjugate base, A-1, undergoes hydrolysis by the reaction:

A-1 (aq) + H2O (l) ( HA (aq) + OH-1 (aq)

This gives an equivalence point in the basic region (pH>7). The equilibrium for this reaction is described by the Kb for the conjugate base:

[pic]

The [A-1] is calculated from the intial moles of HA and the total volume of the reaction mixture at the equivalence point. The pH gives the [H3O+1] and the [OH-1] by using

Kw = [H3O+1] [OH-1] = 1.0 X 10-14

Since [OH-1] = [HA] at the equivalence point, the Kb for the conjugate base can be calculated. An important relationship between Ka for the acid and Kb for its conjugate base is: Kw = Ka x Kb.

[pic]

A monobasic weak base reacts with HCl in a one-to-one mole ratio to produce chloride ion and the conjugate acid in the overall reaction:

B(aq) + HCl (laq ( BH+1 (aq) ) + Cl-1 (aq)

A weak base does not ionize completely in aqueous solution giving the ionic equation:

B(aq) + H+1(aq) + Cl-1 (laq ( BH+1 (aq) ) + Cl-1 (aq)

Removing the spectator ions gives the net ionic equation:

B(aq) + H+1(aq) ( BH+1 (aq) )

The equation for the ionization of the weak base is shown below.

B(aq) + H2O (l) ( BH+1 (aq) ) + OH-1 (aq)

The equilibrium constant for this reaction, Kb is:

[pic]

[BH+1] is the molar concentration of the conjugate acid and [B] is the molar concentration of the weak base.

Kb can be calculated using the initial concentration of the base and the initial pH of the solution. The initial pH gives the [H3O+1] from which the [OH-1] can be calculated. The [OH-1] equals the [BH+1] in the initial weak base solution. If the weak base is only slightly ionized, the [B] is assumed to be approximately equal to its initial concentration.

Kb can be calculated from the pH at the half-equivalence point. At this point in the titration, half of the moles of B have been converted to BH+1. The [BH+1] = [B], the ratio [BH+1]/[B] equals one, the [OH-1] equals Kb, and the pOH of the solution equals the pKb of the weak base. Remember that pH + pOH = 14.

Kb can be calculated from the pH at the equivalence point. The moles of B have been completely converted to BH+1. The conjugate acid, BH+1, undergoes hydrolysis by the reaction:

BH+1 (aq) + H2O (l) ( H3O+1 (aq) + B (aq)

This gives an equivalence point in the acidic region (pH ................
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