University:
California State University, Northridge
Course:
CHEM 101 | General Chemistry I
Academic year:

2023
Views:

35

Pages:

18
Author:

Everly J.

Water • Most biochemical reactions occur in an aqueous environment. • Water is highly polar because of its bent geometry. • Water is highly cohesive because of intermolecular hydrogen bonding. • Water participates in H-bonding with biomolecules. Ionization of water: H2O + H2O ó H3O+ + OH- pH, Acids and Bases • • • • • • • • • pH = -log [H+] pOH = -log [OH-] ([H +] and [OH-] in M) [H+] x [OH -] = 1 x 10-14 M2 / pH + pOH = 14 An acid is defined as a proton donor AH = A - + H+ AH is the acid and A- is its conjugate base. A base is defined as a proton acceptor B + H2O = BH + + OHB is the base and BH+ is its conjugate acid The pH scale An acidic solution is one in which [H+] > [OH-] •In an acidic solution, [H+] > 10-7, pH < 7. •A basic solution is when [OH-] > [H+]. •In a basic solution, [OH-] > 10-7, pOH < 7, and pH >7. • When the pH = 7, the solution is neutral. •Physiological pH range is 6.5 to 8.0 Weak Acids and pKa • The strength of an acid can be determined by its dissociation constant, Ka. • Acids that do not dissociate significantly in water are weak acids. • The dissociation of an acid is expressed by the following reaction: HA = H + + A- and the dissociation constant Ka = [H+][A-] / [HA] • When Ka < 1, [HA] > [H +][A-] and HA is not significantly dissociated. Thus, HA is a weak acid when ka < 1. • The lesser the value of Ka, the weaker the acid. • Similar to pH, the value of Ka can also be represented as pKa. • pKa = -log Ka. • The larger the pKa, the weaker the acid. • pKa is a constant for each conjugate acid and its conjugate base pair. • Most biological compounds are weak acids or weak bases. Polyprotic Acids • Some acids are polyprotic acids; they can lose more than one proton. • In this case, the conjugate base is also a weak acid. • For example: Carbonic acid (H2CO3) can lose two protons sequentially. • Each dissociation has a unique Ka and pKa value. Ka1 = [H+][HCO3-] / [H2CO3] Ka2 = [H+][CO3-2] / [HCO3-] Note: (The difference between a weak acid and its conjugate base differ is one hydrogen) Some weak acids and their conjugate bases The Henderson-Hasselbalch equation Dissociation of a weak acid is mathematically described by the Henderson-Hasselbalch equation Ka = [H+][A-] / [HA] or Ka = [H+] x [A-] / [HA] logKa = log[H+] + log {[A-] / [HA]} -log[H+] = -logKa + log {[A-] / [HA]} pH = pKa + log {[A-] / [HA]} So, if CB = conjugate base and WA = weak acid, then: pH = pKa + log {[CB] / [WA]} This is the Henderson-Hasselbalch equation Note: pH = pKa when [CB] = [WA] Applications of the Henderson-Hasselbalch equation • Calculate the ratio of CB to WA, if pH is given • Calculate the pH, if ratio of CB to WA is known • Calculate the pH of a weak acid solution of known concentration • Determine the pKa of a WA-CB pair • Calculate change in pH when strong base is added to a solution of weak acid. This is represented in a titration curve • Calculate the pI Titration curve for weak acids •Initially, [WA] >>> [CB] •When [WA]=[CB], pH=pKa • The central region of the curve (pH+1) is quite flat because: When [CB]/[WA] = 10, pH = pKa +1; When [CB]/[WA] = 0.1, pH = pKa - 1 •Titration curve is reversible, if we start adding acid, [WA] increases Titration of a weak acid with a strong base • A weak acid is mostly in its conjugate acid form • When strong base is added, it removes protons from the solution, more and more acid is in the conjugate base form, and the pH increases • When the moles of base added equals half the total moles of acid, the weak acid and its conjugate base are in equal amounts. The ratio of CB / WA = 1 and according to the HH equation, pH = pKa + log(1) or pH = pKa. • If more base is added, the conjugate base form becomes greater till the equivalance point when all of the acid is in the conjugate base form. Buffers • Biological systems use buffers to maintain pH. • Definition: A buffer is a solution that resists a significant change in pH upon addition of an acid or a base. • Chemically: A buffer is a mixture of a weak acid and its conjugate base • Example: Bicarbonate buffer is a mixture of carbonic acid (the weak acid) and the bicarbonate ion (the conjugate base): H2CO3 + HCO3• All OH- or H+ ions added to a buffer are consumed and the overall [H+] or pH is not altered H2CO3 + HCO3- + H+ ßà 2H2CO3 H2CO3 + HCO3- + OH- ßà 2HCO3- + H2O • For any weak acid / conjugate base pair, the buffering range is its pKa +1. Mechanism by which Buffers Operate Example: CH3COOH + CH3COO- + OH- = 2CH3COO- + H 2O (you get more conjugate base) CH3COOH + CH3COO- + H+ = 2CH3COOH (you get more weak acid) Ampholytes • A molecule containing ionizing groups with both acidic and basic pKa values is called an ampholyte. • The ionic form of each group in the compound is dependent on the pH of the solution. • If the pH of solution is greater than the pKa, the group is in the conjugate base form (deprotonated). • If the pH of solution is less than the pKa, the group is in the conjugate acid form (protonated). Ionic forms of Glycine •Glycine is H2N-CH2-COOH. •pKa of carboxylate group is 2.3 ; pKa of amino group is 9.6 (Note: glycine can serve as a buffer in 2 different buffer ranges). •The ionic form with a net charge of zero is called a zwitterion •The isoelectric point (pI) is the pH at which the net charge on the ampholyte is zero (or equal number of + and – charged ions). Titration of ampholyte glycine Carboxylate and amino groups lose their protons successively. The first mole equivalent of added base converts the carboxylate to its conjugate base; next, the amino group gets deprotonated. Note the steep jump in pH around the pI. Calculation of pI for Glycine • Use the Henderson-Hasselbalch equation to calculate the pI. • At isoelectric point, pH = pI • pI = pKCOOH + log [H3N+CH2COO-] [H3N+CH2COOH] • pI = pKNH3+ + log [H2NCH2COO-] [H3N+CH2COO-] • Adding up: 2pI = pKCOOH + pKNH3+ + log [H2NCH2COO-] [H3N+CH2COOH] • When pH=pI, [H2NCH2COO-]=[H3N+CH2COOH] • 2pI = pKCOOH + pKNH3+ or pI = {pKCOOH + pKNH3+}/2