Lab 3 (O)

Question 1

If an acid is strong

Answer 1

it completely dissociates in solution

HA(aq) → H+(aq) + A–(aq)

Question 2

If an acid is weak,

Answer 2

the acid reaches equilibrium in solution

HA(aq) ⇌ H+(aq) + A–(aq)

Question 3

During a titration,

Answer 3

the acid and base neutralize each other

H+(aq) + OH–(aq) → H2O(l)

Question 4

The pH of the solution during a titration is a function of

Answer 4

• the excess acid before the equivalence point,
• the conjugate base at the equivalence point, and
• the excess base after the equivalence point.

Question 5

In an acid-base titration, the change in the concentration of H+ is very large, often changing
concentration by a factor of 1012 (a trillion-fold change).

Answer 5

It is quite inconvenient to plot such
numbers; the pH scale is a more convenient way to represent these extreme changes.

Since pH is a logarithmic relation, when [H+] changes from 1.0 mol/L to 1.0·10–12 mol/L, the pH changes from 0.00 to 12.00.

The result of plotting pH versus volume of base added is called a titration curve.

Question 6

sample pH titration curve of a strong acid with a strong base, specifically a 0.50 mol/L HCl is titrated with 0.50 mol/L NaOH.

Answer 6

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

The initial concentration of hydronium ion is 0.50 mol/L (pH ~ 0.30) because HCl is fully
dissociated.

The pH changes slowly with the addition of base until near the equivalence point.

Further addition of base at that point causes the pH to change sharply, resulting in a sharp change
in the slope of the titration curve.

To obtain accurate data correlating pH to the volume of base added, it is important to add the titrant dropwise in the equivalence region.

If the acid is weak, the acid does not fully dissociate and the conjugate base, A– , is weak.

Question 7

Weak acids

Answer 7

The extent to which HA dissociates in solution can be described by two quantities:

• the equilibrium constant expression:

k_a = [H⁺][A^-] / [HA]

• the percent dissociation:

% dissociation = (amount dissociated / original amount) × 100%

• Relating the percent dissociation to the weak acid equilibrium gives

% dissociation = [A^-]_eq / [HA]_i × 100%

or

[H₃O⁺]_eq / [HA]_i × 100%

where [A- ]eq and [H3O+]eq denote the equilibrium concentration before any sodium hydroxide is added and [HA]i is the initial concentration of the weak acid before dissociation occurred.

Thus, the percent dissociation of a weak acid in pure water can be calculated from the initial concentration of acid and the initial pH of the solution.

Question 8

The equivalence point of the titration

Answer 8

is the point in a titration where the moles of titrant added are stoichiometrically equal to the moles of substance being titrated.

The original concentration of
the analyte can be calculated knowing the volume of titrant at the equivalence point and its known
concentration.

For a weak acid, the pH at the equivalence point is greater than 7 because the conjugate base is weak and will hydrolyze in water.

The equivalence point for the titration can be determined from the pH titration curve.

The volume of titrant added to reach this point is called the equivalence volume.

The concentration of the conjugate base can be determined using the following relationship:

[A-] at equivalence point = (moles A- / equivalence volume)

Question 9

The value of Ka for a weak acid can also be determined from the titration curve

Answer 9

On the titration curve, the point at which the initial acid has been half neutralized gives the pH of the
solution where the number of moles of A– formed through the neutralization reaction is exactly
equal to the number of moles of HA remaining in the solution.

In the equilibrium expression

k_a = [H⁺][A^-] / [HA]

[HA] will equal [A–] at the half equivalence point; therefore, these will cancel in the above relation
leaving Ka = [H+] and pH = pKa.

(This is not true for a strong acid titration.)

The procedure for weak acids depends on the existence of a point along the titration curve where [A–] = [HA].

When a strong acid is added to water, it is fully dissociated and the Ka, by definition, approaches infinity. There is never a point where [A–] = [HA].

Question 10

The application of le Châtelier’s principle.

Answer 10

A solution of a weak acid is indirectly neutralized by strong base: the strong base reacts with
hydronium ions, H+(aq).

As the concentration of H+(aq) decreases in solution, more of the weak acid will dissociate to compensate for this stress on the equilibrium.

Owing to the large equilibrium
constant for the reaction of OH–(aq) with H+(aq), at the equivalence point, the neutralization reaction is essentially complete.

This is an application of le Châtelier’s principle.

Often you might know the identity of the acid in solution, but not its concentration. To determine
the concentration, you need to titrate the sample of acid to its equivalence point using a
standardized base solution.

Question 11

Acid-Base Indicators

Answer 11

If you know how the pH curve is changing in the vicinity of the equivalence point, these titrations can be carried out without a pH meter. Instead, an acid-base indicator can be used.

Acid-base indicators are weak organic acids that have conjugate bases of a different color.

They change color in specific pH ranges. In fact, this color change occurs when the pH is close to the pKa value of the indicator, pKin. The rapid color change of the indicator signals the end point of the titration.

Question 12

There are a number of different indicators

Answer 12

and it is important to select the proper one for the
particular titration.

For the endpoint to be useful, it must occur at a volume of titrant very close to that of the equivalence point of the titration and the color change of the indicator must be dramatic enough to be detected.

Since indicators are weak acids, they are only partially dissociated in
aqueous solution.

HIn(aq) ⇌ H+(aq) + In–(aq)

k_in = [H⁺][in^-] / [Hin]

Rearranging for [H+] gives

[H⁺] = k_in x [Hin] / [in^-]

The indicator color depends on the relative proportions of the In– and the HIn forms, which is controlled by [H+].

Choose an appropriate indicator whose pKin is close to this pH.

EX:
- If the pH at the equivalence point is expected to be near 7, bromothymol blue or phenol red
would be suitable indicators.
- If the equivalence point is expected near pH = 9, thymol blue or phenolphthalein would be suitable.