Biochemistry 46a – Water, pH and Ionic Equilibria
Reading - Chapter 2
Practice problems - Chapter 2 - 6,7,8,9,12; Physical Chemistry extra problems; buffer problems
Properties of Water
Unusual Properties
· Life as we know it could not exist without water. All cells, whether from terrestrial or aquatic organisms, are more than 70% water.
· Water has unusual properties when compared to molecules with similar size or structure, e.g., NH3, HF or H2S.
o Higher boiling point
o Higher melting point
o Higher surface tension
o Higher heat of vaporization
· These properties suggest that the interaction between individual water molecules is unusually strong.
Structure of Water - the hydrogen bond
The bent nature of the water molecule (H-O-H bond angle is 104.5o) and the electronegativity difference between H and O confers a strong ionic character to the O-H bond, which results in a large dipole moment for water, making it a highly polar solvent.
The unique properties of water arise from directional interactions between the H on one water molecule and a lone pair of electrons on the O of another water molecule. This interaction is known as the hydrogen bond.
A hydrogen bond is an interaction between a covalently bonded hydrogen on a donor group (-O-H or –N-H) and a lone pair of non bonded electrons on an acceptor group (O=C, O-H). In biological systems, only O and N have the appropriate electronegativity to serve as a hydrogen bond donor. This table summarizes the properties of a number of hydrogen bonds. The energy of a hydrogen bond is small (@ 20 kJ/mol) compared to the O-H covalent bond strength (460 kJ/mol). However, because most biomolecules form many hydrogen bonds, they make a significant contribution to the stability of biomolecules. Because of its highly polar character, water also interacts favorably with ions.
Hydrophilic or Polar molecules are soluble in water because they interact favorably with water by formation of hydrogen bonds or via ionic interactions. Hydrophobic or Nonpolar molecules do not have ionic groups and cannot form hydrogen bonds. Also, the presence of hydrophobic molecules in water causes the water molecules to organize around the hydrophobic molecule in the form a Clathrate. Clathrate formation causes the order of the water molecules to increase. This corresponds to a decrease in Entropy, an unfavorable process. The combination of the lack of favorable hydrogen bonding interactions between hydrophobic molecules and water, and the unfavorable entropy combine to explain the very low solubility of hydrophobic molecules in water.
Amphipathic molecules have a hydrophilic polar head group and a hydrophobic nonpolar tail. Such molecules have an identity crisis - the polar head group interacts favorably with water but the hydrophobic tail does not. This crisis is solved by the formation of a Monolayer, a Micelle or a Bilayer. These self-assembled structures remove the nonpolar tail from interaction with water while allowing the polar head group to interact with water. Amphipathic molecules with a single tail, e.g., detergents or fatty acid salts, usually form micelles, while molecules with two tails, e.g., phospholipids, form bilayers, which, in turn, form closed structures like vesicles. Two-tailed amphipathic molecules are the basis for the formation of Cellular Membranes.
Colligative Properties are properties of the solution that depend on the number of solute molecules present. Examples are freezing point depression, boiling point elevation and osmotic pressure. Osmotic pressure is particularly important in biological systems. Osmotic pressure develops when a semipermeable membrane separates two solutions, one of which contains a solute that cannot pass through the membrane. A biological example would be the membrane of a cell. Water can pass through the membrane and will pass into the compartment containing the nonpermeable solute in an attempt to make the concentration (activity) of water the same on both sides of the membrane. As illustrated below, this will cause a column of water to rise in the tube containing the solute. The osmotic pressure (p) is the amount of pressure that must be exerted on the tune to prevent the column of solution from rising; p=RTm, where R is the gas constant, T is the absolute temperature and m is the molality = mol of solute/kg of solvent.
· For a cell, if water moves into the cell from the surrounding solution, the solution is said to be hypotonic - the concentration of dissolved material is less in the solution than in the cell. In such a solution the cell will expand and probably burst.
· For a cell, if water moves out of the cell into the surrounding solution, the solution is said to be hypertonic - the concentration of dissolved material is more in the solution than in the cell. In such a solution the cell will contract.
Noncovalent Interactions
· The structures and properties of the biomolecules on which life depends, proteins, nucleic acids, lipids and complex carbohydrates, are entirely dependent on the unique properties of water and no other solvent has properties similar to water.
· The noncovalent interactions between biomolecules, which are essential for the proper functioning of biomolecules, depend entirely on the properties of water.
o Noncovalent interactions are much weaker (10-100 times) than covalent bonds.
o It is this very weakness which holds the key to their importance, because it means that noncovalent interactions are continually being formed and broken in the dynamic processes that characterize life.
o Noncovalent interactions involve electrical charges.
o These electrical interactions can be direct, as in ionic or dipole interactions.
o They can be indirect, as in induced dipole or dispersion forces, in which distortion of the electrons in nonpolar molecules is the basis for the interaction.
o van der Waals interactions depend on outer electron orbital overlap.
o The Hydrogen bond is also a noncovalent interaction.
· The strength of interaction between any two ions is given by Coulomb’s Law (F=k[q1q2/e r2], in which it can be seen that the force is inversely proportional to the dielectric constant (e) of the solvent.
· Water has a high dielectric constant, which means that ionic interactions in water are much weaker than in other solvents.
· This is due to the fact water solvates ions through charge-dipole interactions.
· Small ions in water can influence the interaction between biomolecules, by shielding the effective charge of the biomolecules.
· The strength of the effect of the small ions is given by the ionic strength (I) of the solution - I=½S MiZi2, where Mi is the molarity and Zi is the charge of the ith ion in the solution.
Acids, Bases, Buffers
· For biological systems, one of the most important parameters of an aqueous solution is the concentration of protons ([H+]). Although the [H+] is quite low, typically 10-6 to 10-8 M, it must be maintained within this range for life to exist.
· Weak acids dissociate in water according to the following equation:. The actual reaction is , but because the concentration water is constant, the shorthand equation shown is used.
· The acid dissociation constant for this reaction is given by Ka = [A-][H+]/[HA].
· The Ka values for weak acids of biological importance are much less than 1 and for convenience are reported as pKa values, where pKa = -log Ka.
· In biological systems, the concentration of protons, [H+], is very low. For convenience the [H+] is reported as pH, where pH = -log [H+].
Titration Curve
GLYCINE contains two weak acid groups: the a-carboxyl group, pKa 2.4, and the a-amino group, pKa=9.8. This figure also shows the net charge of glycine in its three ionization states.
This is the titration curve for glycine. The green areas are the regions of buffering. In these areas there is a relatively small change in pH even though the percent titration changes significantly.
· The +, +/-, and - shows pH regions where these ionic forms (see above) predominate.
· Any point along the titration curve can be calculated using the Henderson-Hasselbalch Equation - pH = pKa + log{[A-]/[HA]}.
Derivation of the Henderson-Hasselbalch equation.
Practice Problems
As is true with any quantitative concept, the only was to truly make the concept of pH and buffers your own is to work problems. So, here are four common type of pH-buffer problems with solutions.
1. Describe the preparation of 1000 ml of a 0.1 M phosphate buffer pH 7.2 starting with 0.5 M H3PO4 and 0.5 M NaOH. Assume the following pKas: H3PO4 ® H2PO4- = 2.2; H2PO4- ® HPO4-2 = 6.8; HPO4-2 ® PO4-3 = 12.4.
Answer:
· The total phosphate is 100 mM = 200 ml of 0.5 M H3PO4.
· At pH 7.2, the two predominant forms of phosphate will be H2PO4- and HPO4-2.
· The ratio of HPO4-2 to H2PO4- is calculated from the Henderson-Hasselbalch equation. 7.2 = 6.8 + log {[HPO4-2]/[H2PO4-]}, which means that [HPO4-2]/[ H2PO4-] = 2.5.
· The sum of [HPO4-2] + [ H2PO4-] = 100 mM and [HPO4-2]/[ H2PO4-] = 2.5.
· Therefore, [HPO4-2] = 71.4 mM and [H2PO4-] = 28.6 mM
· To prepare the buffer
o Take 200 ml of 0.5 M H3PO4.
o Add 200 ml of 0.5 M NaOH to convert all the H3PO4 to H2PO4-.
o Add 142.8 ml 0f 0.5 M NaOH to make 71.4 mM HPO4-2.
o Add water to 1000 ml.
2. What is the final pH of a solution made by mixing 100 ml of 0.05 M acetic acid and 100 ml of 0.1 M sodium acetate? Assume the pKa for acetic acid is 4.76
Answer:
· 100 ml of 0.05 M acetic acid = 5 mmol.
· 100 ml of 0.1 M sodium acetate = 10 mmol.
· Using the Henderson-Hasselbalch equation
· pH = pKa + log{[acetate]/[acetic acid]}.
· pH = 4.76 + log{[10]/[5]}.
· pH = 5.06
3. Describe the preparation of 1000 ml of a 0.2 M acetate buffer pH 5. Assume the pKa of acetic acid is 4.76.
Answer:
· Acetic acid + acetate = 0.2 M
· Using the Henderson-Hasselbalch equation, pH = pKa + log{[acetate]/[acetic acid]}.
· 5.0 = 4.76 + log{[acetate]/[acetic acid]}.
· [acetate]/[acetic acid] = 1.74.
· Therefore, [acetate] = 0.126 M; and [acetic acid] = 0.074 M.
· To prepare the buffer
· Take 800 ml of water and add 0.2 mole of acetic acid.
· Add 126 ml of 1.0 M NaOH to produce 0.126M sodium acetate.
· Add water to 1000 ml.
4. For a solution that is 0.08 M K2HPO4 and 0.12 M KH2PO4, calculate the concentration of H3PO4, H2PO4-, HPO4-2, PO-3; H+, OH-, and K+. Use the pKas given above.
Answer:
· [H2PO4-] = 0.12 M (the initial value).
· [HPO4-2] = 0.08 M (the initial value).
· pH = 6.8 + log {[0.08]/[0.12]} = 6.62.
· [H+] = 2.4 x 10-7 M.
· [OH-] = 1 x 10-14/2.4 x 10-7 = 4.17 x 10-8.
· For H3PO4. pH = pKa + log{[H2PO4-]/[H3PO4]}; 6.62 = 2.2 + log{[H2PO4-]/[H3PO4]}; [H2PO4-]/[H3PO4] = 2.63 x 104; Therefore, [H3PO4] = 4.56 x 10-6 M.
· For PO4-3. pH = pKa + log{[PO4-3]/[HPO4-2]}; 6.62 = 12.4 + log{[PO4-3]/[HPO4-2]}; [PO4-3]/[HPO4-2] = 1.66 x 10-6; Therefore, [H3PO4] = 1.32 x 10-7 M.
· K+ = 0.12 M + 2x 0.08 M = 0.28 M.
Maintaining blood pH with the bicarbonate buffer system
Conventional wisdom states that a buffer maintains pH best when the pH = pKa ± 1.0. Yet the major buffer of the blood is the carbonic acid/bicarbonate buffer system which has a pKa=6.1, more than one pH unit below the pH of blood at pH = 7.4. Does this mean that Mother Nature didn't understand buffers? Indeed not!
The carbonic acid/bicarbonate buffer system has the unique property that the acid, H2CO3, is in equilibrium with a gas, CO2.
· Because of this, the concentration of H2CO3 in blood is fixed by the concentration of CO2 in the gas phase.
· The concentration of CO2 in the gas phase is determined by the concentration of CO2 in the lungs.
· The CO2 content in the lungs depends on the rate ofCO2 production in metabolism and the rate of breathing.
· Slow breathing (hypoventilation) raises the CO2 content and rapid breathing (hyperventilation) lowers the CO2 content.
· The concentration of CO2 in the lungs is reported as the partial pressure of CO2 = pCO2.
· Partial pressure is the way the composition of gases is reported. At sea level, atmospheric pressure = 760 mm Hg. IfCO2 comprises 5.25% of the gases, pCO2 = 40 mmHg - the normal value in the lungs.
· For this system, the Henderson-Hasselbalch equation is, where 0.03 is a constant that defines the solubility of CO2 in water.
· So why is this system so great?
o If we add acid to the system it reacts with HCO3- to produce H2CO3 (HCO3- + H+ ® H2CO3). But, the excess H2CO3 is converted to CO2, which leaves the solution. Thus, [H2CO3] remains constant.
o If we add base to the system it reacts with H2CO3 to produce HCO3- (H2CO3 + OH- ® HCO3- + H2O). But more CO2 dissolves in the solution. Thus, [H2CO3] remains constant.
o The net effect is that the denominator of the Henderson-Hasselbalch equation remains constant, which means that pH changes only slightly as H+ or OH- are added.
o Also the rate of breathing can also be adjusted, which changes pCO2: slower breathing (hypoventilation) causes an increase in pCO2; rapid breathing (hyperventilation) causes a decrease in pCO2. This means that pH=7.4 can be maintained even in the face of large amounts of acid or base.
These effects are illustrated quantitatively in the table below, which shows the change in pH in response to addition of H+ or OH- to different buffer systems. All buffers initially at pH=7.4.
· A - Buffer with pKa = 7.4 containing 24 meq each of HA and A-
· B - Bicarbonate buffer, no CO2 in the gas phase; [H2CO3] = 1.2 mM and [HCO3-]=24 mM