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Summary Pharmaceutical Analysis A (WBFA035-05)
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Summary Pharmaceutical Analysis A of all the lectures with examples of important exam questions.
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Pharmaceutical Analysis A 2021-2022Exam: there is a formula sheet, mostly calculations and solving things, also some theory.
Final grade: exam: practical 75%: 25%
Lecture 1
CHAPTER 6
Involves analysis of medicinal drugs and compounds
1. Quality control
2. Measurement of bio accessibility ( release of active substance from the drug)
3. Determation of bioavailability (distribution of active substances and it metabolites in
the body)
4. Bioanalysis
Classic, wet-chemical quantitative methods
Gravimetry
Titrimetry: there is acid we know, but the concentration is unknown. In the buret there is a
known concentration.
Determination of the unknown amount of analyte, A, in a solution by measuring the
consumption of a reagent, T, that reacts with A.
A + T AT
- Choose a reaction in which all the T added each time is completely reacted with A to
produce AT.
Shortly after last amount of T is added, there is a equilibrium:
A + T AT
- When a reaction has reached equilibrium, the forward and reverse rated of reaction
are equal. Other words: the concentrations of A, T and AT remain the same.
The concentration of T is very very small.
Chemical reaction chemical equilibrium.
1) Reaction must go to completion
a. All titrant (T) added must react with available analyte (A)
2) Reaction must be fast
Laws
1. Law of mass action: at a given temperature, a chemical system reaches a state in
which a particular ratio of reactant and product concentration has a constant value.
2. Le chatelier’s principle or the equilibrium law: When a system at equilibrium is
subjected to a change that disturbs it, it will re-adjust itself to partly counteract the
effect of applied change so it can proceed back to an equilibrium.
BrO3- + 2Cr3+ + 4H2O Br- + Cr2O72- + 8H+
At constant temperature, T: K remains constant
Add dichromate: What happens?
,The reaction system re-adjusts to THE LEFT (reactant and product concentrations shift) until
the system’s K value is reestablished
An equilibrium don’t wont disturbing, but want to stay in an equilibrium.
Equilibrium constant, K:
aA + bB cC + dD K = [C]c [D]d
[A]a [B]b
K has no units
When K > 1, the numerator is greater than the denominator. The equilibrium lies to the right.
The reaction is favoured.
Stoichiometry: Ratios of substances participating in a chemical reaction. (Stoicheion =
element, Metron = size)
[A]stan = 1 M
- therefore: [A] is [A]/ (1 M), B is [B]/(1 M)
System in equilibrium:
O O
reactant product
Rate of forward reaction (reactant to product) is the same as the rate of the reverse
reaction (product to reactant)
Concentrations of reactant and product don’t change
Remove product, peturb equilibrim reaction proceeds to the right.. (until equilibrium is
re-established
OOOO
reactant product
If you add reactant, the reaction proceeds to the right.
Add product reaction proceeds to the left
O OOOO
reactant product
How can we tell if a reaction is at equilibrium or not? • We can confirm this by calculating the
reaction quotient, Q.
Q > K, we can conclude that the reaction will prodeed to the left. Product concentration will
go down (numerator) and the reactants will go up (dominator)
Q =K, the reaction is at equilibrium
Q > K, the reaction will proceed to the left until equilibrium is re-established.
Q < K, the reaction will proceed to the right until equilibrium is re-stablished
, What happens to the value of K when the reaction is performed at a different
temperature
Heat can also be regarded as a reagent or product
Standard temperature is 25 grades Celsius.
Endothermal reaction:
Heat + reagent product
Increase temp reaction proceeds to the right K gets bigger
Exotherminal reaction:
Reagent 1 + reagent 2 product + heat
Temp goes up reaction to the left K gets smaller
What kinds of equilibria are good for titrimetric analysis:
1) Solubility
a. Solubility product, Ksp
Sparingly soluble salt is immersed in water:
AB2 (s) A2+ (aq) 2 B- (aq)
Ksp = [A2+][B-]^2
Ksp is used when an excess of a sparingly soluble salt is immersed in water.
- both reactants and products are available
- equilibrium can be established
The salt dissolves until the concentrations of cations and anions have reached
their maximum values solution is saturated.
2) Acid-Base chemistry
a. Acid dissiociation constant, Ka
b. Base hydrolysis constant, Kb
3) Complex formation
a. Stepwise formation constants, K
b. Cumulative formation constants, B
4) Reduction-oxidation (redox) reactions
a. Ox1 + Red2 ⇋ Ox2 + Red1
b. Exchange of electrons: oxidation means reduction! Red2 gives up an electron
when it is oxidized to Ox2. Ox1 picks up that electron to be reduced to Red1.
Brønsted - Lowry Acids and bases:
Acid: proton donor – gives H+ away
Base: proton acceptor – picks H+ up
Amphiprotic : proton donor and proton acceptor molecule
An acid only loses a proton if there is a compound present capable of picking that proton up
(in other words, a base)
Base hydrolysis reaction Acid dissociation reaction
B(aq) + H2O ⇋ BH+ (aq) + OH- (aq) HA(aq) + H2O ⇋ H3O+ (aq) + A- (aq)
, The difference between a conjugate base and acid is just one H+.
Water can undergo self-ionization (autoprotolysis).
Kw = 1.01 × 10-14 @ 25.0°C
Pure water: Kw = 1.0 × 10-14 = [H+] [OH- ] [H+] = [OH- ] = 1.0 × 10-7
To satisfy Kw, a constant, the [H+] will increase as [OH- ] decreases.
Definition of pH = -log ([H+]) SO: pH of H2O = -log(1.0 × 10-7) = 7.00
Lecture 2
CHAPTER 8
Common ion effect
Starting condition: reaction in a vessel (e.g beaker) is at equilibrium.
Example:
Hg2Cl2 (s) Hg22+ (aq) + 2 Cl- (aq) (sparingly soluble salt in water)
-x x 2x
(- says that it might dissolves)
Ksp = [Hg] [Cl]2 = x (2x) 2 = 4x 3 = 1,2x 10 -18
In a saturated solution: [Hg2 2+] = 6,7 x 10 -7 and [Cl-] = 1,3 x 10 -6
Add chloride
An additional amount of a reaction component is added to a reaction vessel
Equilibrium is perturbed, system is disturbed.
Cl- increases [Hg2 2+] must decrease to satisfy the condition that Ksp is
constant at constant T. Hg2 2+ thus precipitates as Hg2Cl2
Reaction goes to the left
Reaction proceeds to correct for the change in concentration of chloride.
Effect of added electrolyte (dissolved ions)
iodate.
Hg2(IO3)2 Hg2 2+ + 2 IO3-
If you add KNO3 number of ions in solution increases, but they don’t play a role in the
reaction. It is a so called inert salt (K+ and NO3 don’t participate).
But still there is happening something, what and how? However, the solubility of Hg2(IO3)2
(s) increases! (i.e. more Hg2(IO3)2 (s) dissolves)
Positive ion (Hg) has a negative atmosphere and the negative ion (IO) has a positive
atmosphere. Normally the positive ion is attractive to the negative one. But in this case the
attraction is not that strong. The ions behave different dependent on other ions.
SO: attraction between ions goes down
- Ionic atmosphere is a spherical cloud of charge, δ- or δ+. This shields the charge of the central ion,
which appears less positively or negatively charged, respectively.
- The more ions in solution ⇒ the larger δ- or δ+ ⇒ the smaller the attraction between oppositely
charged ions
- Tendency for Hg2 2+ to directly interact with IO3 - to form precipitate decreases accordingly –
equilibrium shifts to the right
Hg2(IO3)2 (s) Hg2 2+ (aq) + 2 IO3 - (aq)
- This means that more Hg2(IO3)2 (s) can dissolve ⇒ the solution can accommodate higher
[Hg2 2+] and [IO3 - ]
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