Chemistry of aqueous solutions Flashcards
Arrhenius Theory
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
Bronsted-Lowry Theory
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
Lewis Theory
Acid receives a pair of electrons
Base donates a pair of electrons
Conjugate acid-base pairs are related by..
transfer of one H+
Strong/ Weak acids/bases
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
Defn. pH
It is the negative logarithm to the base ten of the hydrogen ion concentration
pH = -lg [H+]
pH = -lg [H3O+]
Defn. Ka
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
Defn. pKa
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
Defn. Kb
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
Defn. pKb
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
Defn. Kw
Kw = [H3O+][OH-] = 1.0x10^ -14 mol2dm-6 at 25 degree celsius
pKw = 14
Value of Kw is temperature dependent
formulas
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
degree of dissociation
depends on the concentration of the acid
increases with decreasing conc.
Acid-base properties of salt solutions (Salt hydrolysis)
The cations and anions formed may react with water molecules to undergo hydrolysis to produce neutral, basic or acidic solutions
Defn. Buffer solution
is a solution that is able to maintain a fairly constant pH when a small amount of acid or base is added
How does acidic buffer work
- 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 - 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
How does alkaline buffer work
- 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+ - 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
Importance of buffers in biological systems
- Many enzymes in the body function efficiently inly at specific pH values
- 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
- Buffer systems in the body includes proteins, phosphates and hydrogencarbonates
H2CO3 / HCO3 - Buffer system in the blood
- Buffer system consists of carbonic acid H2CO3 and the hydrogencarbonate ion HCO3-, its conjugate base
H2CO3 <-> H+ + HCO3- - An increase in [H+] is removed by combining with the large reservoir of HCO3- ion to form the molecular acid H2CO3
H+ + HCO3- -> H2CO3 - An increase in [OH-] is removed by the large reservoir of H2CO3 molecules to give HCO3- ions
OH- + H2CO3 -> HCO3- + H2O
Indicator colours
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)
Indicator colours in different conditions
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)
2 conditions for accurate titration results (end point of titration coincide with equivalence point of reaction)
- The range of the indicators lies within the pH change at equivalence point
- The pH change at equivalence point must take place over 2 pH units (pKIn +- 1)
Strong acid - strong base titration
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)
Weak acid - strong base titration
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
Strong acid - weak base titration
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
Weak acid - weak base titration
…
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
DRAW TITRATION CURVES
!!!! >:)
