(A Level Chem) A question about blood

Look at this question (taken from GCE A Level Nov 2004 P3 and Nov 2009 P3) that tests on equilibria calculation skills and on the knowledge of the structure of proteins. Don't faint when you see “blood”.

(a) The main buffering agent in blood plasma is the carbon dioxide/hydrogen carbonate system. H2O +CO2 = HCO3- + H+ Ka = 7.90 x 10-7 mol dm-3 What is meant by the term buffer? [2] Explain how the system above acts as a buffer when OH- ions are added to the blood plasma. [1] Use the data above to calculate the pH of normal blood plasma if the [HCO3-]/[CO2] ratio is 20:1. [2] The total CO2 + HCO3- content of blood plasma can be found by boiling a plasma sample with an excess of dilute sulfuric acid and absorbing the carbon dioxide evolved in a solution of barium hydroxide. The precipitated barium carbonate can be filtered off, dried and weighed. When 100 cm3 sample of plasma was treated in this way, 0.600 g of barium carbonate was produced. Write a balanced equation for the reaction between carbon dioxide and barium hydroxide. [2] Use the data given to calculate the total number of moles of CO2 + HCO3- in the blood plasma sample, and use the ratio given in (iii) to calculate the [CO2(aq)] in blood plasma. [3] (b) One molecule of haemoglobin can bind up to four molecules of oxygen: Hb(aq) + 4O2(aq) = Hb(O2)4(aq) Write an expression for Kc for this reaction, stating its units. [2] Experiments showed that when [O2] = 7.6 x 10-6 mol dm-3, the concentration of Hb and Hb(O2)4 are equal. Calculate the value of Kc. [1] Calculate [O2] necessary for 99% of the Hb to be converted to Hb(O2)4. [2] Myoglobin Mb is an oxygen carrier protein that occurs in muscle fibres. It has a higher affinity for O2 than does Hb but only binds one O2 molecule per Mb molecule. Mb(aq) + O2(aq) = MbO2(aq) Kc = 106 mol-1 dm3 Calculate the percentage of MbO2 in an Mb-MbO2 mixture when [O2] = 7.6 x 10-6 mol dm-3. [2] (c) With reference to the haemoglobin (Hb) molecule, describe and explain what is meant by the terms primary, secondary, tertiary and quaternary structures of proteins. In each case you should state the type of bonding or interactions involved. [8] In patients suffering from sickle cell anemia, the ß-chains of their haemoglobin (abbreviated as Hb-S) differ from those of normal haemoglobin through the replacement of a glutamic acid residue (-CH2-CH2-COOH) by a valine residue (-CH(CH3)2). This substitution has the effect if making the deoxygenated Hb-S molecules attract each other and aggregate, resulting in the formation of insoluble fibres. Draw the structural formula of a glutamic acid residue in its most stable state at pH 7. Glutamic acid has a pKa of 4.3. [1] Suggest in terms of the intermolecular forces present, give a reason for the aggregation of Hb-S molecules. [1] Suggest a reason why molecules of normal haemoglobin do not attract each other. [2] (d) Once the primary structures of the polypeptide chains of haemoglobin have been formed from their constituent amino acids, and in the presence of haem group, the chains spontaneously coalesce into a haemoglobin molecule. By considering the chemical interactions involved at each step, suggest what the signs of ∆G, ∆H and ∆S would be for the overall process. Explain your reasoning. [3]

Let's have a brief overview of a few concepts before we look at the answers.

• For acetic acid (ethanoic acid), a weak acid, dissolved in water, the equilibrium equation is

CH₃CO₂H(aq) = H⁺(aq) + CH₃CO₂^-(aq)
and the acid dissociation constant is K_a = [H⁺][CH₃CO₂^-]/[CH₃CO₂H].

Suppose 0.500 mol dm^-3 of acetic acid is present in water. At equilibrium, [CH₃CO₂H] = 0.500 – x ≈ 0.500 (since 0.500 >> x), [H⁺] = [CH₃CO₂^-] = x.
So, K_a = x²/0.500.

• For aqueous ammonia, the equilibrium equation is

NH₃(aq) + H₂O(l) = NH₄⁺(aq) + OH^-(aq)
and the base dissociation constant is K_b = [NH₄⁺][OH^-]/[NH₃]
(Ignore the species that're in solid and liquid states.)

Suppose 0.100 mol dm^-3 of ammonia is present in water. At equilibrium, [NH₃] = 0.100 – x ≈ 0.100 (since 0.100 >> x), [NH₄⁺] = [OH^-] = x.
So, K_b = x²/0.100.

• A buffer solution consists of a mixture of a weak acid (e.g. CH₃CO₂H) + conjugate base (e.g. CH₃CO₂^-) OR a weak base (e.g. NH₃) + conjugate acid (e.g. NH₄⁺). It resists changes in pH when a strong acid or a strong base is added. The equilibrium CH₃CO₂H(aq) = H⁺(aq) + CH₃CO₂^-(aq) shifts to the left to remove H⁺(aq) added, and to the right to removed excess OH^-(aq) added. The equilibrium NH₃(aq) + H₂O(l) = NH₄⁺(aq) + OH^-(aq) shifts to the rightto remove H⁺(aq) added, and to the left to removed excess OH^-(aq) added.
  
• In describing the structure of proteins, you need to state the types of bonds involved in each stage.
  (1) Primary structure refers to the polypeptide chain that's made up of a unique sequence of amino acids (NH₃-CHR-CO₂H) held together by covalent peptide (-CONH-) bonds.
  (2) Secondary structure is the coiling (alpha helixes) and bending (beta-pleated sheets) of the polypeptides due to hydrogen bonds between C=O and N-H along the backbone. We can now call this a protein. The side chains R not considered here.
  (3) Tertiary structure is the 3-dimensional shape of polypeptide due to ionic bonds (between -NH₃⁺ and -CO₂^-), disulfide bonds (between -CH₂SH and -CH₂SH in cysteine R to form -CH₂S-SCH₂-), hydrogen bonds (between -OH and -OH) and hydrophobic interactions (in H₂O molecules) between the side chains R. This significantly adds to the protein's stability.
  (4) Quaternary structure is the association of two or more sub-units of tertiary structures. The bonds involved in the quaternary structure are the same as those in the tertiary structure.
  
• Like all proteins, haemoglobin has a primary, secondary, tertiary and quaternary structures. The haemoglobin is made up of four polypeptides called globins coming together. Each globin is associated with a heme group that's a molecule with an iron atom. The iron atoms give the haemoglobin the oxygen-attracting property.
  
• In sickle-cell anemia, the red blood cells have an abnormal shape and cannot carry oxygen very well. This is because one of the amino acids has been changed causing a change in the interactions in the secondary and tertiary structures.
  

The diagram below shows

• some interactions between the different side chains in different parts on a polypeptide
  
• an amino acid (i.e. glutamic acid which has the negatively charged -CO₂^-) in a polypeptide of a normal red blood cell and that (i.e. valine which is hydrophobic) in a sickle cell

Now, here's the answer to the above question.

(a)(i) A buffer solution is a mixture of weak acid and conjugate base or weak base and conjugate acid [1] which resists changes in pH when small quantities of an acid or an alkali are added to it [1].

(ii) When OH^-(aq) is added to the system, H₂O +CO₂ = HCO₃^- + H⁺, the equilibrium shifts to the right to produce more H⁺(aq) which removes the the excess OH^-(aq). [1]

(iii) H₂O +CO₂ = HCO₃^- + H⁺ K_a = 7.90 x 10^-7 mol dm^-3
Given: [HCO₃^-]/[CO₂] = 20/1
K_a = [HCO₃^-][H⁺]/[CO₂] = 7.90 x 10^-7
20 x [H⁺] = 7.90 x 10^-7 [1]
[H⁺] = (7.90 x 10^-7)/20
pH = -lg [H⁺] = -lg [(7.90 x 10^-7)/20] = 7.40 (3 sig fig) [1]

(iv) CO₂(g) + Ba(OH)₂(aq) --> BaCO₃(s) +H₂O(l) [2]
Number of moles of BaCO₃ produced
= 0.600 g / {[137 + 12.0 + 3(16.0)] g mol-1} [1]
= 0.600/197

Number of moles of CO₂ + HCO₃^- present in 100 cm³ of plasma = 0.600/197
[CO₂] + [HCO₃^-] = (0.600/197) mol/100 cm³
[CO₂] + [HCO₃^-] = (0.600/197) mol/0.100 dm³ [1]
Let [CO₂] = x,
x + 20x = (0.600/197) / 0.100 (since [HCO₃^-]/[CO₂] = 20/1) [1]
21x = (0.600/197) / 0.100
x = 1.45 x 10^-3 mol dm^-3 (3 sig fig)

(b)(i) Hb(aq) + 4O₂(aq) = Hb(O₂)₄(aq)
K_c = [Hb(O₂)₄]/[Hb][O₂]⁴ mol^-4 dm¹² [1 for expression + 1 for unit]

(ii) Given: [O₂] = 7.6 x 10^-6 mol dm^-3, [Hb] = [Hb(O₂)₄]
K_c = [Hb(O₂)₄]/[Hb][O₂]⁴
= 1/(7.6 x 10^-6)⁴ [1]
= 2.997 x 10²⁰
= 3.00 x 10²⁰ mol^-4 dm¹² (3 sig fig)

(iii)                  Hb(aq) + 4O₂(aq) = Hb(O₂)₄(aq)
Initial:             100%                         0%
Equilibrium:  1%x = 0.01x              99%x = 0.99x

K_c = [Hb(O₂)₄]/{[Hb][O₂]⁴}
3.00 x 10²⁰ = 0.99x/{(0.01x)[O₂]⁴} [1]
[O₂]⁴ = 0.99/(0.01 x 3.00 x 10²⁰) = 2.39 x 10^-5 mol dm^-3 [1]

(iv) Mb(aq) + O₂(aq) = MbO₂(aq) K_c = 10⁶ mol^-1 dm³
[O₂] = 7.6 x 10^-6 mol dm^-3
K_c = [MbO₂]/([Mb][O₂])
10⁶ = [MbO₂]/([Mb] x 7.6 x 10^-6) [1]
[MbO₂]/[Mb] = 10⁶ x 7.6 x 10^-6 = 7.6
[MbO₂] = 7.6[Mb]

[MbO₂]/{[Mb] + [MbO₂]} x 100%
= 7.6[Mb]/{[Mb] + 7.6[Mb]} x 100% [1]
= 7.6/(1 + 7.6) x 100%
= 88.4%

(c)(i) The haemoglobin is made up of 4 polypeptides called globins (two alpha-globins and two beta-globins). [1]
In the primary structure, each long polypeptide is made up of amino acids [1] stringed together by peptide bonds [1].
In the secondary structure, each of the polypeptide is folded into alpha-helixes and beta-pleats [1] due to hydrogen bonds between C=O and N-H along the backbone [1].
In the tertiary structure, each polypeptide is arranged in a 3-dimensional globular shape [1] due to ionic bonds, disulfide bonds, hydrogen bonds and hydrophobic interactions associated with the side chains R [1].
In the quaternary structure, the four globins are then put together by the same types of interactions as in the tertiary structure [1]. Each globin is associated with a heme group (an organic molecule that holds an iron atom in its centre) in the centre.

(ii) -CH₂-CH₂-COOH(s) = -CH₂-CH₂-COO^-(s) + H⁺(aq)
pKa = 4.3 ==> -lg Ka = 4.3 ==> K_a = 10^-4.3
[H⁺] = 10^-4.3
pH = -lg [H⁺] = -lg (10^-4.3) = 4.3
When OH^-(aq) is added to achieve pH 7, the equilibrium shifts to the right to produce more H⁺(aq).
Therefore, at pH 7, -CH₂-CH₂-COO^- exists.
Structure: [1]

(iii) The side chain -CH₂-CH₂-COO^- is polar and is attracted to water molecules. If the side chain -CH₂-CH₂-COO^- is replaced by -CH(CH₃)₂, the side chain becomes non-polar and hydrophobic causing aggregation. [1]

(iv) The molecules of normal haemoglobin do not attract one another because of the repulsion between the negatively charged side chain -CH₂-CH₂-COO^-. [1]

(d) In each step, bonds (ionic bonds, hydrogen bonds, Vander Waals, disulfide bonds) are formed between the side chains and so heat is released i.e. ∆H = -ve. [1] ∆S is -ve because the polypeptide becomes more ordered. [1] Since the coalescing is spontaneous, ∆G is -ve. [1] 
In the primary structure of the protein, the bonds involved are the peptide bonds. In the secondary structure, the bonds involved are the hydrogen bonds between N-H and C=O along the backbone. In the tertiary structure, the bonds involved are the interactions between the side chains. The bonds involved in the quaternary structure are the same as those in the tertiary.
The haemoglobin consists of four polypeptides (two alpha and two beta). Each polypeptide possesses an iron-containing heme group.