Chemistry of aqueous solutions

Question 1

Arrhenius Theory

Answer 1

Acid produces hydrogen ions in water.
A base reacts with an acid to give a salt and water only. A base is a metal oxide or hydroxide.

HA -> H+ +A-

• Acid + alkali
• Acid + ins metal oxide
• Acid + ins metal hydroxide

Question 2

Bronsted-Lowry Theory

Answer 2

Acid is a proton (H+) donor
Base is a proton (H+) acceptor

HA -> H+ +A-
:B + H+ -> BH+

• Hydrogen atom carries a delta positive charge
• Base contains lone pair of electrons that it can use to form a dative bond with the proton

Question 3

Lewis Theory

Answer 3

Acid receives a pair of electrons
Base donates a pair of electrons

Question 4

Conjugate acid-base pairs are related by..

Answer 4

transfer of one H+

Question 5

Strong/ Weak acids/bases

Answer 5

Strong acids/bases completely ionise in aq solution to produce H+ / OH- ions

Weak acids/bases partially ionise in aq solution and an equilibrium is achieved

Question 6

Defn. pH

Answer 6

It is the negative logarithm to the base ten of the hydrogen ion concentration

pH = -lg [H+]
pH = -lg [H3O+]

Question 7

Defn. Ka

Answer 7

For a weak monoprotic acid, HA:
HA (aq) + H2O (l) <-> H3O+ (aq) + A- (aq)

Acid dissociation constant, Ka = [H3O+][A-] / [HA]

The larger the Ka value, the stronger the acid

Question 8

Defn. pKa

Answer 8

pKa of an acid is the negative logarithm to the base ten of the acid dissociation constant, Ka.

The smaller the value of pKa, the stronger the acid

Question 9

Defn. Kb

Answer 9

For a weak monoprotic base, B:
B (aq) + H2O (l) <-> BH+ (aq) + OH- (aq)

Base dissociation constant, Kb = [BH+][OH-] / [B]

The larger the Kb value, the stronger the base

Question 10

Defn. pKb

Answer 10

The pKb of a base is the negative logarithm to the base ten of the base dissociation constant, Kb. The smaller the value of pKb, the stronger the base

Question 11

Defn. Kw

Answer 11

Kw = [H3O+][OH-] = 1.0x10^ -14 mol2dm-6 at 25 degree celsius

pKw = 14

Value of Kw is temperature dependent

Question 12

formulas

Answer 12

pH = -lg [H+]
pKa = -lg Ka
Kw = [H3O+][OH-]
pKw = -lg Kw = 14 at 25 degree celsius
Kw = (Ka)(Kb)
pKa + pKb = pKw
pH = pKw- pOH
pH = pKa + lg[salt]/[acid]
pOH = pKb + lg[salt]/[base]

[H+] = squareroot (Kac) where c is [HA]initial. assumption: [HA]initial≈[HA]eqm

Question 13

degree of dissociation

Answer 13

depends on the concentration of the acid

increases with decreasing conc.

Question 14

Acid-base properties of salt solutions (Salt hydrolysis)

Answer 14

The cations and anions formed may react with water molecules to undergo hydrolysis to produce neutral, basic or acidic solutions

Question 15

Defn. Buffer solution

Answer 15

is a solution that is able to maintain a fairly constant pH when a small amount of acid or base is added

Question 16

How does acidic buffer work

Answer 16

1. When a small amount of acid is added, the added H+ ions are removed by the large reservoir of A-, keeping the pH of the solution fairly constant
   H+ + A- -> HA
2. When a small amount of base is added, the added OH- ions are removed by the large reservoir of HA, keeping the pH of the solution fairly constant
   OH- + HA -> A- +H2O

Question 17

How does alkaline buffer work

Answer 17

1. When a small amount of acid is added, the added H+ ions are removed by the large reservoir of NH3, keeping the pH of the solution fairly constant
   H+ + NH3 -> NH4+
2. When a small amount of base is added, the added OH- ions are removed by the large reservoir of NH4+, keeping the pH of the solution fairly constant.
   OH- + NH4+ -> NH3 + H2O

Question 18

Importance of buffers in biological systems

Answer 18

1. Many enzymes in the body function efficiently inly at specific pH values
2. The pH of blood must be maintained at about 7.4. Any change in pH by 0.5 unit, especially for intravenous injection, can be fatal
3. Buffer systems in the body includes proteins, phosphates and hydrogencarbonates

Question 19

H2CO3 / HCO3 - Buffer system in the blood

Answer 19

1. Buffer system consists of carbonic acid H2CO3 and the hydrogencarbonate ion HCO3-, its conjugate base
   H2CO3 <-> H+ + HCO3-
2. An increase in [H+] is removed by combining with the large reservoir of HCO3- ion to form the molecular acid H2CO3
   H+ + HCO3- -> H2CO3
3. An increase in [OH-] is removed by the large reservoir of H2CO3 molecules to give HCO3- ions
   OH- + H2CO3 -> HCO3- + H2O

Question 20

Indicator colours

Answer 20

Phenolphthalein:
colourless - pale pink - red
(8 - 9.3 - 10)

Bromothymol blue:
yellow - green - blue
(6 - 7.0 - 8)

Litmus:
red -purple - blue
(5 - - - 8)

Methyl orange:
red - orange - yellow
(3 - 3.7 - 4)

Question 21

Indicator colours in different conditions

Answer 21

HIn <-> H+ +In-
(Kln)

In acidic conditions:
1. [H+] is high
2. By LCP, the position of equilibrium shifts to the left, producing more Hln-
3. The colour of the solution will be the colour of the undissociated indicator, Hln

In alkaline conditions,
1. [H+] is low
2. By LCP, the position of equilibrium shifts to the right, producing more H+ and ln- ions
3. The colour of the solution will be the colour of the ln- ion

At end point,
1. [HIn] = [In-]
2. The colour of the solution will be a mixture (midway) of colours of HIn and In-
3. pH is not necessarily =7 but pH=pKln

(Assuming the solutions used in the titration are colourless)

Question 22

2 conditions for accurate titration results (end point of titration coincide with equivalence point of reaction)

Answer 22

1. The range of the indicators lies within the pH change at equivalence point
2. The pH change at equivalence point must take place over 2 pH units (pKIn +- 1)

Question 23

Strong acid - strong base titration

Answer 23

Initial pH:
- Acid present
- Initial pH is low (pH < 2) due to high [H+] from the strong acid
- pH = - lg[H+] where [H+] = [HCl]

Midway of titration (1/2 Veq):
- Salt and unreacted acid present
- Initial addition of alkali has little effect on pH (flat portion) as water and a neutral salt are formed
- pH = - lg [H+] where [H+] = [HCl]unreacted

Equivalence point:
- Salt present
- Nearing the equivalence point, 1/2 drops of alkali causes a sharp increase in pH from pH 3 to pH 11
- At equivalence point, pH = 7 as a neutral salt is formed

Final pH:
- Salt and excess alkali present
- After the equivalence point, the graph flattens out at high pH (pH > 11 for strong bases) due to high [OH-] from excess strong base
- pOH = -lg [OH-] where [OH-] = [NaOH]unreacted

Choice of indicator:
- methyl orange, phenolphthalein, almost any indicator because the pH range of these indicators lie within the pH change at equivalence point (pH 3 to 11)

Question 24

Weak acid - strong base titration

Answer 24

Initial pH:
- Weak acid present
- Initial pH is higher than 2 as [H+] is low due to partial dissociation of a weak acid
- [H+] = squareroot(Kac)
- pH = - lg[H+]

Midway of titration:
- Salt and unreacted acid present
- Initial addition of strong base causes pH to rise rapidly, but soon remains fairly constant at pH = 5. due to presence of weak acid and salt, forming a buffer solution. This region with a slow rise in pH is called the buffer region
- When half of the weak acid is neutralised, half the original amount of acid is converted to salt, thus [weak acid] = [salt]. Thus a solution of maximum buffering capacity is formed
- pH = pKa + lg [salt]/[acid]

Equivalence point:
- Salt present
- Nearing the equivalence point, 1/2 drops of alkali causes a sharp increase in pH from pH 7 to pH 11
- At equivalence point, pH > 7 because salt hydrolysis of basic salt can take place to give OH- ions
- [OH-] = squareroot(Kbc)
- pH = - lg[OH-]

Final pH:
- Salt and excess alkali present
- After the equivalence point, the curve flattens out at high pH (pH > 11 for strong bases) due to high [OH-] from large excess of strong base
- pOH = -lg [OH-] where [OH-] = [strong base]unreacted

Choice of indicator:
- phenolphthalein because only the range of phenolphthalein (pH 8 to 10) lies within the pH change at equivalence point (pH 7 to 11)
Eg. colour of methyl orange would have changed way before the neutralisation point is reached

Question 25

Strong acid - weak base titration

Answer 25

Initial pH:
- Acid present
- Initial pH is low (pH < 2) due to high [H+] from the strong acid (complete dissociation of the strong acid)
- pH = - lg[H+] where [H+] = [HCl]

Midway of titration (1/2 Veq):
- Salt and unreacted acid present
- Initial addition of alkali has little effect on pH (flat portion) as water and a neutral salt are formed
- pH = - lg [H+] where [H+] = [HCl]unreacted

Equivalence point:
- Salt present
- Nearing the equivalence point, 1/2 drops of weak base causes a sharp increase in pH from pH 3 to pH 7
- At equivalence point, pH < 7 because salt hydrolysis of acidic salt can take place to give H+ ions

Final pH:
- Salt and excess alkali present
- After the equivalence point, the curve flattens out at high pH (pH <12 for weak bases) as the solution contains mainly the undissociated weak base , producing an alkaline buffer
- pOH = -lg [OH-] where [OH-] = [strong base]unreacted [OH-] = [NaOH]unreacted

Choice of indicator:
- methyl orange because only the pH range of methyl orange (pH 3-4) lies within the pH change at equivalence point (pH 3-7)
Eg. Phenolphthalein will change colour only when excess base is added beyond the equivalence point

Question 26

Weak acid - weak base titration

Answer 26

…

Choice of indicator:
- no suitable indicator as there is no vertical portion for the curve i.e. pH change at equivalence point is less than 2 pH units. no indicator will give a distinct colour change with 1 drop of titrant. A pH meter will be needed to find the equivalence point

Note: no calculation of pH is expected for this case

Question 27

DRAW TITRATION CURVES

Answer 27

!!!! >:)