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Question 1

Common Ion Effect

Answer 1

occurs when the ionisation of an acid (or base) is limited by the presence of a significant concentration of its conjugate base (or acid)

The addition of sodium acetate pushes the equilibrium to the left, creating
acetic acid and raising the pH (sodium does not effect pH).
– Before adding a common ion, the equilibrium condition Q = Ka is satisfied.
– After adding a common ion, Q > Ka
– Restoration of equilibrium would be achieved by net reaction in the direction that reduces Q.
The presence of conjugate base limits ionisation of the acid

Question 2

Controlling pH: Buffer Solutions

Answer 2

• The pH meter indicates the pH of water that contains a trace of acid, as well as some bromophenol blue acid-base indicator. (pH = 5.04)
• The solution at left is a buffer
  solution with a pH of about 7,
  also with the indicator.
• 5 mL of 0.1 mol L- 1 HCl is
  added to each solution
• The pH of the water drops by
  several units (pH = 2.33) also
  shown by the indicator colour
  change).
• The unchanged indicator
  colour indicates the pH of the
  buffer solution hardly changes.

Question 3

Composition and Mode of Operation of Buffers

Answer 3

• Minimize the change in pH when a strong acid or base is added.
• Contain large amounts of both a weak acid and its conjugate base, relative to the amounts of acid or base that are added.
• Function because the weak acid and base are in equilibrium:
  – Added H3O+ (aq) ions are “mopped up” by the conjugate base
  – Added OH- (aq) ions are “mopped up”’ by the weak acid.

Question 4

Commonly used buffering reagents

Answer 4

phthalic acid - hydrogenphthalate ion, 1.3x10-3 (2.89), 1.9-3.9

acetic acid, acetate ion, 1,8x10-5 (4.74), 3.7-5.8

hydrogenphosphate ion, phosphate ion, 3.6x10-13 (12.44), 11.3-13.3

Question 5

Quantitative Calculations of Buffer
Solution pH

Answer 5

[H3O+] = [HA]/[A-] x Ka

Question 6

Henderson-Hasselbalch equation (Log form):

Answer 6

pH = pKa + log10 [A-]/[HA] (conjugate base/acid)

determined largely by the pKa of the acid - adjusted by the ratio acid:conjugate base

actual concentrations of [acid] and [conjugate base] are not important, but the molar ratio is.

Diluting a buffer solution does not change its pH

Question 7

Design of a Buffer Solution of Specified pH

Answer 7

To design a buffer solution for a particular pH, choose a weak acid with pKa within 1 unit of the target pH.

• Calculate the [weak acid]/[conjugate base] ratio that will adjust the solution pH to the target value.
• Make a solution containing both the weak acid and its conjugate base with the appropriate ratio.
• Use amounts of the weak acid and conjugate base well in excess of the amounts of strong acid or base that are likely to be added.

(Henderson-Hasselbalch equation)

Question 8

pH Change of Buffer Solutions

Answer 8

To calculate how much pH changes on addition of a specified amount of H3O+
(aq) to a buffer solution, calculate

a) the amounts of weak acid and conjugate base in the buffer solution,
b) the amount of H3O+ (aq) added
c) the amount of acid that reacts and base that is formed,
d) the new ratio n(acid)/n(base), and
e) the new pH using the above ratio, and Ka of HA(aq).

Question 9

Buffer Capacity

Answer 9

the ability of a buffer solution to minimize pH change upon addition of strong acids or bases.

• The more concentrated the weak acid and base are in a solution, the higher the buffer capacity, and:

(a) for a given amount of added strong acid or base, the less is the pH change;
(b) the more strong acid or base can be added before the solution loses its effectiveness.

Question 10

Acid-Base Titrations

Answer 10

a method of quantitative analysis, based on a reaction of known stoichiometry. It depends on finding the volume of a solution containing the amount of acid (or base) that reacts exactly with a known amount of base (or acid).

Question 11

The Equivalence point

Answer 11

• Occurs when acid or base has been titrated with an equivalent amount of the corresponding species

– For strong acid and base this results in a neutral solution at a pH of 7.0
– For other combinations, the pH at equivalence is not 7
* Strong acid + weak base, pH < 7
* Strong base + weak acid, pH > 7

Question 12

Strong Acid- strong base (HCL and NaOH titration

Answer 12

Initially the solution is 0.1M HCl so pH=1.
* As NaOH is added the amount of HCl
decreases;
* that which remains is in an ever
increasing volume; the pH slowly increases.
* Just before the equivalence point the
solution is still quite acidic e.g. at 49
mls pH=3.
* The equivalence point in any titration is the midpoint of the vertical section.
* In this case the pH rises very rapidly
around the midpoint: 8 pH units
between 49 and 50 mls (8 orders of magnitude)

At the equivalence point in titration of a solution of a strong base into a solution of a strong acid, pH = 7.0.

• The pH at any point prior to the
  equivalence point can be calculated
  from the amount of unreacted acid.
• Immediately after the equivalence point the addition of a very small amount of base causes a dramatic increase in pH:
  essentially adding a strong base to
  water.
• After a bit more base has been added, the rate of increase decreases.

Question 13

Weak Acid- Strong base (Acetic acid and NaOH

Answer 13

• The initial pH is found from the [acid] and the acid pKa
• At the equivalence point the acetic acid and NaOH have been consumed.
  – pH is controlled by the acetate ion (conjugate base)
• At the ½ equivalence point ½ of the acid has been neutralised.
  – [CH3COOH]=[CH3COO-]
  (CH3COO-Na+ 100% dissociates)
• Because:

[H3O+] = [HA]/[A-] x Ka

at the half equivalence point pH=pKa
i.e. we have worked out the acid pKa
* The slow change in pH between 0 and 90 mls is because the solution is buffered, known as the buffer region.
* Above equiv. point pH rises sharply as before

pH > 7 at the equivalence point
for a weak acid with a strong
base. The pH can be calculated
using Kb of the conjugate base.

Question 14

Polyprotic weak acids and strong base (H2C2O4 and NaOH)

Answer 14

• The initial pH comes from the acid
  concentration and pKa1.
• The first rise occurs at about 100mls
  – the first H has been titrated.
  – HC2O4
• controls the pH
  (conjugate base)
• When the 2nd H is titrated the pH again rises, this time more sharply
  – C2O4 2- controls the pH
  (conjugate base)

C2O42- + H20 → HC2O4- + OH
Kb = 1.6 x 10-10
Therefore pH = 8.5

Kb =Kw/Ka

In diprotic weak acid with Ka1&raquo_space; Ka2 titration, the most acidic proton is removed from nearly all molecules before the second one is removed.

• Prior to the first equivalence point, the solution is a buffer solution, and when the volume of NaOH solution is
  exactly half that needed to
  reach the first equivalence point, pH = pKa1.

Between the first and second equivalence points, the solution is a buffer solution, and when the volume added is midway between the first and
second equivalence points, pH = pKa2.

Question 15

Weak Base - Strong Acid (NH3 and HCl)

Answer 15

• The initial pH is a function of the base
  concentration and its pKa (or pKb)
• At the half equivalence point half of the NH3 has been converted to NH4+Cl-

– Therefore [NH4+] = [NH4+Cl-]
* As (Henderson-Hasselbalch equation)

at this concentration pH = pKa
* As more HCl is added the pH falls slowlydue to the buffer effect.
* At the equivalence point, the pH of the solution is due to the weakly acidic NH4+(aq) ions, and the pH is about 5.
* As more HCl is added NH4+Cl- is formed and the pH falls do to the dissociation of NH4+

Question 16

Biochemical Acid-Base Speciation

Answer 16

• Maintenance of pH in cellular fluids of living organisms is vital, because enzyme activity is influenced by pH.
• The main protection against harmful
  pH changes in cells is provided by buffers.
• In most cells, the pH is maintained
  between 6.9 and 7.4 through the
  action of phosphate (H2PO4-/HPO4 2-)
  and carbonate (H2CO3/HCO3-) buffers.
• Just as increasing CO2(g) in the
  atmosphere causes acidification of sea water, increased CO2 (g) concentration
  in the lungs can affect blood pH.

Question 17

Colligative Properties

Answer 17

When water contains dissolved substances, the properties of the solution are affected:
– Equilibrium vapour pressure
– Boiling point
– Freezing point
– Osmotic pressure

depend only on the conc of solute particles and are independent of the identity of the particles in solution.

Question 18

Lowering of vapour pressure by non-volatile solutes

Answer 18

• is the pressure of the vapour above it
• when the liquid and the vapour are in equilibrium.

If we form a solution by dissolving a nonvolatile solute, the equilibrium vapour pressure of the solvent above the solution is lower than for the pure solvent.

For an ideal solution, the vapour pressure of the solvent is proportional to the mole fraction of solvent (x) at a specified temperature (Raoult’s law).

Question 19

Raoult’s law

Answer 19

p(solvent) = x(solvent) × p°
solvent

a good approximation at
low solute concentration, and in solutions where solute and solvent are similar, e.g. octane and octanol.

Question 20

An ideal solution…

Answer 20

a good approximation at
low solute concentration, and in solutions where solute and solvent are similar, e.g. octane and octanol.

Question 21

Lowering of vapour pressure by non-volatile solutes

Answer 21

Vapour pressure: determined only by volatile solvent

In 1:1 solvent:solute solution, surface is covered equally by the solvent and
non-volatile solute.

This means the number of solvent molecules able to escape the liquid and contribute to the vapour pressure is reduced.

This lowers the vapour pressure. The reduction is calculated using Raoults Law.

The decrease depends on the quantity of the non-volatile solvent, not its nature.

Question 22

Freezing Point Depression is

Answer 22

lower than that of the pure solvent.

• At low solute concentrations, the difference between the freezing point of solvent and that of solution is directly proportional to the molality (m-solute) of the solution:
  ∆ Tf = Kf × m(solute)

Question 23

Kf (in kg mol-1)

Answer 23

molal freezing point depression constant - the characteristic of the solvent.

Question 24

Fractional crystallization

Answer 24

As more ice forms, the solute conc in the residual solution increases, and the freezing point falls lower.

Cooling is stopped before freezing and residual solution is run off. Process is repeated to purify.

Question 25

Solutions of Electrolytes

Answer 25

Colligative properties depend only on the concentration of solute particles, not their identity

When the solute is an electrolyte, the freezing point depression depends on the total concentration of ions in solution.

Question 26

van’t Hoff factor

Answer 26

∆ Tf (measured)/∆ Tf (calculated) = ∆ Tf (measured) / Kf x m

∆ Tf = Kf x m x i

• approaches whole numbers (2, 3, and so on) only with very dilute solutions.

In more concentrated solutions, the experimental freezing point depressions tell us that the solutions behave as if there are fewer ions in solution than expected.

Attributed to: ion attractions in solutions. If some of the positive and negative ions are paired, decreasing the total concentration of particles.
Ion pairing reduces particle conc, so the freezing point depression is less than expected number of moles of ions .

Question 27

Determinants of van’t Hoff

Answer 27

– Ionization of the solute
– The effective concentration of the solute (or Activity, a, which accounts for ion pairing). The activity is different from the labelled concentration.

Question 28

Osmosis

Answer 28

the movement of solvent molecules through a semipermeable membrane from a solution of lower solute conc (or pure solvent) to a solution of higher solute concentration.

Question 29

Osmotic Pressure

Answer 29

the pressure that must be applied against the spontaneous direction of osmosis just sufficient to prevent osmotic flow.

For dilute solutions, the osmotic
pressure is related to the molar
concentration by the temperature
and the gas constant:

Π = cRT

Question 30

Determine Molecular Weight from Osmosis

Answer 30

Strategy
Calculate the solution concentration from its osmotic pressure. Then, use the volume and concentration of the solution to calculate the amount of solute.
Finally, find the molar mass of the solute from its mass and amount.

Question 31

Osmotic Pressures of Solutions of Electrolytes

Answer 31

Osmotic pressures depend on the total activities of ions in solution, rather than the labelled solute concentration.

Π = i × cRT
where c is the solution labelled concentration and i is the van’t Hoff factor.

The van’t Hoff factor
corrects for things like
ion pairing.

Question 32

Reverse Osmosis in Water Purification

Answer 32

Applying sufficient pressure can overcome the natural osmotic
flow across a semipermeable
membrane.

Question 33

Osmosis in biological systems

Answer 33

a) A cell placed in an isotonic solution would have zero net movement of water into and out of the cell because the solute concentration inside and outside the cell is equal.

b) In a hypertonic solution, the solute concentration outside the cell is greater than inside. There is a net flow of water out of the cell, causing the cell to shrivel (crenate).

c) In a hypotonic solution, the solute concentration outside the cell is less than inside. There is a net flow of water
into the cell, causing the cell to swell and perhaps burst (or lyse).

Question 34

Speciation and Plots

Answer 34

the distribution between
any weak acid-conjugate base pair as pH is changed.
* The generalized equation for weak acid–conjugate base equilibrium:
HA(aq) + H2O(ℓ) ⇌ A- (aq) + H3O- (aq)

Question 35

A zwitterion…

Answer 35

The isoelectric pH for an amino acid is the pH at which the species has a site with positive charge and a site with a negative charge, but zero overall charge