Download Eukaryotic Cell Structure - Human Biochemistry I - Medical | BMS 9265 and more Study notes Geriatrics in PDF only on Docsity!
CHAPTER 1 EUKARYOTIC CELL STRUCTURE
I. OVERVIEW
The topics discussed will be acids, bases and buffers. The student is to review the cellular structure as presented in the textbook.
II. PROPERTIES OF WATER
A. Structure of water
1. Constituent atoms.
2 H atoms and 1 O atom.
2. Polarity
Partial positive charge on H; partial negative charge on O. H-O-H bond angle of 104.5 º Electrically asymmetric, creating a dipole.
3. Hydrogen bonds
A hydrogen bond is a non-covalent bond formed by specific attraction between an electronegative atom and a hydrogen atom.
A Requirements Two electronegative atoms, e.g., O, N and S. One hydrogen atom. Two bonds formed: 1. covalent bond; 2. H bond.
B Solvent properties of water: Polar nature and ability to form H bonds enable water to dissolve many solutes. Water acts as: I H donor II H acceptor
III. Electrolytes
Dissolved substances in the cell may be classified according to their electrical charge as:
A. Nonelectrolytes
These substances bear no charge, e.g., sugars and alcohols.
B. Electrolytes
These substances bear charge and may be divided into:
1. Strong electrolytes
100% dissociated. Examples: Salts, e.g., NaCl, strong acids, e.g., HCl and strong base, e.g., NaOH.
2. Weak electrolytes
< 100% dissociated. Examples: Weak acids, e.g., CH3-COOH and weak bases, e.g., NH 4 OH
IV. REVIEW OF LOGARITHM THEOREMS:
log (M x N) = log M + log N
log M
n
= n log M
log (M/N) = log (M x1/N) = log M + log N-1^ = log M – log N
[H+] [OH-] = 1 x 10-
[H+] = [OH-] = 1 x 10-
pH = - log [H+] (by definition);
pOH = - log [OH-^ ] (by definition)
In pure water: pH 7. pOH 7. pH + pOH = 14
VII. Relationship between [H
] and pH:
[H+] pH
10 -1^1
10 -2^2
10 -3^3 acidic: 0 < pH < ~ 7. 10 -4^4 10 -5^5 10 -6^6 10 -7^7 neutral: pH of ~ 7. 10 -8^8 10 -9^9 10 -10^10 10 -11^11 alkaline: ~ 7.0 < pH < 14 10 -12^12 10 -13^13 10 -14^14
Conclusion: High acidity = high [H
] = low pH Low acidity = low [H+] = high pH
VIII. Dissociation of phosphoric acid and relationship
between acid strength, K eq, and p K
Dissociation of H 3 PO 4 is described by the following equation
K 1 K 2 K 3
H 3 PO 4 H 2 PO- 4 HPO 4 2-^ PO 4 3-
Phosphoric acid undergoes three ionizations characterized by K 1 , K 2 , and K 3.
Acid K , M p K H 3 PO 4 7.25 x 10-3^ 2. H 2 PO 4 -^ 1.38 x 10-7^ 6. HPO 4 2-^ 3.98 x 10-13^ 12.
Comparing the values in the table: Acid strength: H 3 PO 4 > H 2 PO 4 -^ > HPO 4 2- Dissociation constants: K 1 > K 2 > K 3 p K : p K1 < p K2 < p K
IX. Dissociation of carbonic acid
Carbonic acid (H 2 CO 3 ) is in equilibrium with dissolved carbon dioxide (CO 2 ) as well as bicarbonate (HCO 3 -^ ) according to the following equation:
K 2 K 1
CO 2 + H 20 H 2 CO 3 H+^ + HCO 3 -
Bicarbonate controls blood pH at 7.4. A drop in blood pH is referred to as metabolic acidosis and could lead to serious complications in patient management. Therefore, it is important to monitor the pH of blood and to be able to adjust it properly if required.
A. Dissociation of H 2 CO 3 :
[H+] [HCO 3 - ]
K 1 = = 1.7 x 10-4^ ; pK 1 = 3.
[H 2 CO 3 ]
B. Hydration of CO 2 :
log K = log ([H+] x [A-]/[HA])
log K = log [H+] + log ([A - ]/[HA]) or
- log [H+] =-log K + log ([A - ]/[HA])
Since pH = -log [H+] and pK = - log K , the equation becomes the Henderson- Hasselbalch equation:
pH = p K + log ([A - ]/[HA])
XI. Buffers
Organisms maintain specific and constant pH for optimal physiologic functions. This is done through the action of buffers, e.g., phosphate and bicarbonate buffers.
A. Definition
A buffer is a mixture of a weak acid and its conjugate base. It has the ability to resist large changes in pH when small amounts of another acid or base is added.
B. Illustration using acetate buffer (CH 3 COOH + CH 3 COO
1. Qualitative explanation
A Acid buffering When an acid, H+, is added, the CH 3 COO-^ component of the buffer neutralizes the H+, pH remains essentially unchanged. The slight decrease in pH is due to the dissociation of the newly formed CH 3 COOH. B Base buffering When a base, OH-, is added, the CH 3 COOH component of the buffer neutralizes the OH-, pH remains essentially unchanged. The slight increase in pH is due to the hydration of the newly formed CH3COO-.
2. Quantitative explanation
A Given: 0.05M CH 3 COOH 0.05M CH 3 COO- pK of acetic acid = 4. Question: What is the pH of this solution? Answer: Since pH = p K + log ([A-]/[HA]); therefore pH = 4.74 + log (0.05/0.05) = 4.74 + log 1 = 4.74 + 0 = 4.
B Given: 0.05M CH 3 COOH 0.05M CH 3 COO- Addition of 0.01M HCl Question: What is the pH of this solution? Answer: Due to neutralization of 0.01M CH 3 COO-^ by 0.01M HCl, the new conjugate base and conjugate acid concentrations are: [CH 3 COO-] = 0.04M [CH 3 COOH] = 0.06M Since pH = p K + log ([A-]/[HA]); therefore pH = 4.74 + log (0.04/0.06) = 4.74 + log 0.6667 = 4.74 – 0.18 = 4.
C Given: 0.05M CH 3 COOH 0.05M CH 3 COO- Addition of 0.01M NaOH Question: What is the pH of this solution? Answer: Due to neutralization of 0.01M CH 3 COOH by 0.01M NaOH, the new conjugate base and conjugate acid concentrations are: [CH 3 COO-] = 0.06M [CH 3 COOH] = 0.04M Since pH = p K + log ([A-]/[HA]); therefore
[H+] [CH 3 COO-] X^2 X^2
K = 1.82 X 10-5^ = = =
[CH 3 COOH] 0.1M – X 0.1M
X = 1.35 X 10-
pH = 2.
Note: Percent ionization of a 0.1M acetic acid: % Ionization = [(1.35 X 10-3^ M)/(0.1M)] X 100% = 1.35% G Given: 0.05M CH 3 COOH 0.05M CH 3 COO- Addition of 0.05M NaOH Question: What is the pH of this solution? Answer: Due to neutralization of 0.05M CH 3 COOH by 0.05M NaOH, the new conjugate base and conjugate acid concentrations are approximately: [CH 3 COO-] = 0.1M [CH 3 COOH] = 0M
In actuality, a finite, albeit low, concentration of CH 3 COOH exits due to hydration of CH 3 COO-. This situation is illustrated below:
CH 3 COO-^ + H 2 O + CH 3 COOH + OH- 0.1M – X X X
[CH 3 COOH][OH-] K = 9.89 X 10-12^ = [CH 3 COO-][ H 2 O]
K [ H 2 O] = (9.89 X 10-12)(55.6)
[CH 3 COOH][OH-] X^2 X^2 = 5.49 X 10-10^ = = = [CH 3 COO-] 0.1M – X 0.1M
X^2 = 5.49 X 10-
X = 7.40 X 10-
pOH = 5.
pH = 8.
3. Summary:
A p K is the pH at which the acid is 50% dissociated, i.e., ([A-]/[HA]) = 1. B Weak acids/bases will resist changes in pH in the buffering range near its p K (within one pH unit range, e.g. p K ±±±± 1.0). C When pH = p K + 1, then 91% of weak acid is deprotonated; when pH = p K - 1, then 91% of weak acid is protonated. D In a titration curve, showing pH vs. OH-^ equivalents, the midpoint of the plateau represents: I Along the pH axis, the p K of the acid being titrated. II Along the OH-^ equivalents axis, ½ equivalent of OH-^ added.
