Lecture 2 Acid Base 2024 PDF

LECTURE II  The Law of Mass Action: The velocity of a chemical reaction is proportional to the active masses of the reacting substances. A+B C + D Vf ɑ [A].[B] or Vf = Kf [A].[B] Where Kf is the velocity constant of the forward reaction. Vb ɑ [C].[D] or Vb = Kb [C].[D]...

LECTURE II  The Law of Mass Action: The velocity of a chemical reaction is proportional to the active masses of the reacting substances. A+B C + D Vf ɑ [A].[B] or Vf = Kf [A].[B] Where Kf is the velocity constant of the forward reaction. Vb ɑ [C].[D] or Vb = Kb [C].[D] Where Kb is the velocity constant of the backward reaction. At Equilibrium: Vf = Vb Kf [A].[B] = Kb [C].[D] [C].[D] / [A].[B] = Kf / Kb = Keq Keq = “Equilibrium Constant" Acid- Base Equilibria in Water  Acid- Base Equilibria in Water: The equilibrium of dilute solutions of weak acids at constant temperature can be represented by the law of mass action: Mono-protic acids: Example: HAc HAc ⇋ H+ + Ac- Applying the law of mass action: Ka = [H+].[Ac-] / [HAc] Where Ka is the ionization of dissociation constant of the acid.  Acid- Base Equilibria in Water: Di-protic acids: Example: H2S H2S ⇋ H+ + HS- HS- ⇋ H+ + S2- Applying the law of mass action: Ka1 = [H+].[HS-] / [H2S] Ka2 = [H+].[S2-] / [HS-] Where Ka1, Ka2 are the primary and secondary dissociation constants, respectively. The Dissociation of Water: The dissociation of water is reversible and occurs to a very limited extent as shown by its very weak conductivity: H2O ⇋ H+ + OH- Applying the law of mass action: K = [H+][OH-]/[H2O] Since water is very slightly ionized, the value of H2O can be regarded as unity, hence: Kw = [H+][OH-]  Ionic Product of Water: It was found that the product of [H+][OH-] is constant at a given temperature and termed ionic product of water, or Kw and it is equal to 1 X 10-14 at 25oC. In pure water, the hydrogen ion and the hydroxyl ion concentrations are equal. [H+] = [OH-] = 1X 10-7 (At 25oC) Therefore: In Neutral Solutions: [H+] = [OH-] In Acidic Solutions: [H+] > [OH-] In Alkaline Solutions: [H+] < [OH-]  Hydrogen Ion Exponent "pH" Instead of actual concentration of H+ or OH-, it is more convenient to use the negative logarithms, pH and pOH or in other words: pH = - log [H+] pOH = - log [OH-] As explained before, in pure water: pH = pOH pH + pOH = 14 Therefore: In Neutral Solutions: pH = pOH = 7 In Acidic Solutions: pH < 7 In Alkaline Solutions: pH > 7  pH of strong Acids and Bases: pH of strong Acids and Bases Strong acids and bases are assumed to be completely dissociated; therefore the concentration of the acid or the base represents the concentration of [H +] or [OH-]  pH of strong Acids and Bases: 1- pH of Strong Acids: 0.1N HCl gives [H+] = 0.1 = 10-1 pH = - log [H+] = - log 10-1 = 1  pH of strong Acids and Bases: 2- pH of Strong Bases: 0.1N NaOH gives [OH-] = 0.1 = 10-1 pOH = - log [OH-] = - log 10-1 = 1 pH = pKw – pOH = 14 -1 = 13  pH of weak Acids and Bases: 1- pH of Weak Acids: Consider the dissociation of the weak acid HA: HA ⇋ H+ + A- Ka = [H+] [A-] / [HA] But since [H+] = [A-], and since the degree of dissociation is very small, we can assume that [HA] = Ca, where Ca is the total acid concentration. pH of weak Acids and Bases: 1- pH of Weak Acids: Therefore: Ka = [H+]2 / Ca [H+]2 = Ka. Ca [H+] = √Ka. Ca pH = 1/2 pKa + 1/2 PCa  pH of weak Acids and Bases: 2- pH of Weak Bases: Similarly, weak bases dissociate to a small extent giving OH -. Consider the dissociation of the weak base BOH: BOH ⇋ B+ + OH- Kb = [B+] [OH-] / [BOH] But [B+] = [OH-], and [BOH] = Cb, where Cb is the total base concentration. pH of weak Acids and Bases: 2- pH of Weak Bases: Therefore: Kb = [OH-]2 / Cb [OH-]2 = Kb. Cb [OH-] = √Kb. Cb pOH = 1/2 pKb + 1/2 pCb pH = 14 - 1/2 pKb - 1/2 PCb pH of Strong Acids pH = -log [H+] pH of Strong Bases pH = 14-(-log [OH-]) pH of Weak Acids pH = ½ pKa+ ½ pCa pH of Weak Bases pH = 14-(½ pKb+ ½ pCb)  pH of Salts: 1- Salts of Strong Acids and Strong Bases: e.g. NaCl pH = 7 2- Salts of Strong Acids and Weak Bases: e.g. NH4Cl pH = 1/2 PKw - 1/2 pKb + 1/2 PCs 3- Salts of Strong Bases and Weak Acids: e.g. NaAc pH = 1/2 PKw + 1/2 pKa - 1/2 PCs 4- Salts of Weak Acids and Weak Bases: e.g. NH4Ac pH = 1/2 PKw + 1/2 pKa - 1/2 PKb  Buffer Solutions: Definition: These are solutions which resist the change in pH upon addition of small amount of acids or bases.  Types of Buffers: 1- Acidic Buffers: They consist of weak acid and its salt. Example: Acetic acid – Sodium acetate. 2- Basic Buffers: They consist of weak base and its salt. Example: Ammonium hydroxide – Ammonium chloride.  Mechanism of Action of Buffers: 1- Acidic Buffers (HAc – NaAc): Upon addition of strong acid: H+ + Ac- ⇋ HAc "weak acid" The strong acid, therefore, is converted to a weakly dissociated acid. So the pH remains almost constant. Upon addition of strong base: OH- + HAc ⇋ Ac- + H2O "neutral" The strong base, therefore, is converted to water. So the pH remains almost constant. Mechanism of Action of Acidic Buffers: Upon adding strong acid (H+): pH is almost constant H+ HAC HACAC- H+ HAc Mechanism of Action of Acidic Buffers: Upon adding strong base (OH -): pH is almost constant OH- HAC H2O +AC- AC- OH- H2O+Ac-  Mechanism of Action of Buffers: 2- Basic Buffers (NH4OH - NH4Cl): Upon addition of strong acid: H+ + NH4OH ⇋ NH4+ + H2O "neutral" The strong acid, therefore, is converted to water. So the pH remains almost constant. Upon addition of strong base: OH- + NH4+ ⇋ NH4OH "weak base" The strong base, therefore, is converted to a weakly dissociated base. So the pH remains almost constant. Mechanism of Action of Basic Buffers: Upon adding strong acid (H+): pH is almost constant H+ NH H42OH O + NH4+ NH4+ H+ H2O + NH4+ Mechanism of Action of Basic Buffers: Upon adding strong base (OH -): pH is almost constant OH- NH4OH NH4+ NH4OH OH- NH4OH Henderson Equation for Calculating the pH of Buffers: 1- pH of Acidic Buffers: Example: HAc - NaAc HA ⇋ H+ + A- Ka = [H+] [A-] / [HA] -log Ka = -log [H+] - log [A-] / [HA] pKa = pH - log [salt] / [acid] pH = pKa + log [salt] / [acid] Henderson Equation for Calculating the pH of Buffers: 2- pH of Basic Buffers: Example: NH4OH – NH4Cl BOH ⇋ B+ + OH- Kb = [OH-] [B+] / [BOH] -log Kb = -log [OH-] - log [B+] / [BOH] pKb = pOH - log [salt] / [base] pOH = pKb + log [salt] / [base] pH = pKw – pKb - log [salt] / [base]  Examples: 1- Calculate the pH of a buffer solution containing 0.1M acetic acid and 0.1M sodium acetate,(pKa = 4.76). Solution: pH = pKa + log [salt] / [acid] pH = 4.76 + log 1/1 pH = 4.76 Examples: 2- Calculate the pH of a buffer solution containing 0.01M ammonium hydroxide and 0.1M ammonium chloride,(pKb = 4.76). Solution: pH = pKw – pKb - log [salt] / [base] pH = 14 – 4.76 - log 0.1/ 0.01 pH = 8.24  Buffer Capacity: It is the magnitude of the resistance of a buffer to the change in pH. β = Δ B/Δ pH Where: β is the buffer capacity. Δ B is the change in mg equivalent of strong acid or base added. Δ pH is the change in pH. The higher the buffer capacity, the more efficient the buffer. The higher the concentration of the two components of the buffer, the higher the buffer capacity. Buffer Ratio, Maximum and Adequate Buffer Capacity: Buffer ratio: It is the [salt] / [acid] ratio. Maximum Buffer Capacity: It occurs when the [salt] = [acid]. Thus, pH = pka Adequate Buffer Capacity: The ratio of [salt] / [acid] should not exceed 1/10 or 10/1 to obtain adequate buffer capacity. Thus, pH = pKa ± 1