University:
Kentucky State University
Course:
CHE 315 | Biochemistry
Academic year:

2020
Views:

201

Pages:

28
Author:

salcignofi17

Acids and Bases 23 Definitions of Acids and Bases 24 Common Acids • • • • • • HCl- hydrochloric- stomach acid H2SO4- sulfuric acid - car batteries HNO3 – nitric acid - explosives HC2H3O2- acetic acid - vinegar H2CO3-carbonic acid – sodas H3PO4- phosphoric acid -flavorings 25 Common Bases • NaOH- sodium hydroxide (LYE) soaps, drain cleaner • Mg (OH)2 - magnesium hydroxide-antacids • Al(OH)3-aluminum hydroxide-antacids, deodorants • NH4OH-ammonium hydroxide- “ammonia” 26 Water as both an acid and a base Amphoterism - an ion or molecule can act as an acid or base depending upon the reaction conditions Water in NH3 serves as an acid H2O + NH3 acid base NH4+ + OH- acid base Water in acetic acid serves as a base H2O + base CH3CO2H acid H3O+ + acid Conjugate acid of H2O CH3CO2base Conjugate base of acetic acid 27 Acid and Base Strength • Some acids can cause burns if come in contact with skin, other acids are safe. why? • How easy can the acid produce proton – Strong acid: gives up H+ easily (100% dissociated in water) HCl H+ + Cl– Weak acid: gives up H+ with difficulty ( less than 100% dissociated) CH3COOH H+ + CH3COO- Acid dissociation constant, Ka – The general ionization of an acid is as follows: H3O+ + A- HA + H2O So the acid dissociation constant is as follows: [H 3O  ][A  ] Ka  [HA][H2O] [H2O] = 55.5 M and is constant in all equations   [H3O ][A ] Ka  [HA] 29 Acid dissociation constant Ka = [H+][A-] [HA] Larger Ka means: More dissociation Smaller pKa Stronger acid For simplicity: pKa = log 1/Ka = - log10 Ka The larger the value of the pKa, the smaller the extent of dissociation. 30 31 Decrease acid strength Increase acid strength Base dissociation constant B + H2O BH+ + OH [BH][OH ] Kb  [B] 32 Strong Acids - Dissociate readily - Ka is very large - Examples: Hydrochloric, Nitric; Sulfuric Ex. HCl → H+ + ClKa = [H+] [Cl-] [HCl] [H+] = [acid] Ex.1 M solution of HCl has a [H+] of 1 M 1 mM HCl solution has a [H+] of 1 mM 0.1 M H2SO4 solution has a [H+] of 0.2 M 33 Weak Acids - Dissociate slightly - Ka is smaller than strong acids - Examples: Acetic, Boric, Nitrous, Phosphoric, Sulfurous Ex. CH3COOH + H2O CH3COO- + H3O+ Ka = [H+] [CH3COO-] [CH3COOH] What is the H+ of a 0.1 M solution of acetic acid? Ka= 1.74 X 10-5 1.74 X 10-5 = x2/[0.1 ] x=1.3 X 10-3 M 34 Equilibrium constant and the pH of water H2O dissociates to a slight extent to form hydrogen (H+) and hydroxyl (OH-) ions. H2O H+ + OH- [H+] of pure water is only 10-7M 35 Dissociation of water • Kw = [H+] [OH-] = 10-14 Example: A solution has an [OH-] = 10-9 M [H3O+] = 10-5 M Problem solving Example: What is the [H+] of a 0.01 M NaOH solution? Kw = [H+] x [OH-] = [H+] x 10-2 = 10-14 [H+] = 10-12 M Example: What is the [OH-] of a 0.01 M HCl solution? Kw = [H+] x [OH-] = 10-2 x [OH-] = 10-14 [OH-] = 10-12 M Example : The [H+] of a 0.03 M weak base solution is 1 x 10-10 M. Calculate pKb? B + H 2O BH+ + OH- [OH-] = 10-4 Kb = (10-4 x 10-4 ) / 0.03 = 3.33 x 10-7 M pKb = -log Kb = 6.48 38 Measuring the acidity of solutions, pH pH = -log10 [H+] [H3O+] = 10-pH Example 1: [H3O+] in household bleach is 10-12 M pH = - log [10-12] = 12 Example 2: Orange juice has a pH of 4 [H3O+] = 10-4 M 39 Acidic more H+ pH in=solutions Basic = more OHAcids Neutral Bases 40 pH The pH scale is a logarithmic scale. One pH unit difference implies a 10-fold difference in [H+]. Example: lemon juice at pH 2.0 contains more than 100 times as much H+ as orange juice at pH 4.0 pH problem solving Example 1: What is the pH of a solution whose hydrogen ion concentration is 3.2 x 10–4 mol/L? pH= -log [H+] = -log (3.2 x 10-4) = -log (3.2) – log (10-4) = -0.5 + 4.0 = 3.5 42 Enzyme activity and pH 43 Example: Find the Ka of a 0.04 M weak acid HA whose pH is 4.0? HA H+ + A- Ka = [A-] [H+] / [HA] = [H+]2 / [HA] = 10-4 x10-4 / 0.04 = 2.5 x 10-7 Example 2: What is the pH of a 0.05 M Ba(OH)2? Ba(OH)2 Ba + 2OH- [OH-] = 2x 0.05 = 0.10 M = 1 x 10-1 pOH = - log 1 x 10-1 = 1 Remember pOH = -log [OH-] pH= 14-1 = 13 [H+] = 1x 10-13 44 Example: The Kb for ammonia is 1.8 x 10-5 M. What is the pH of 1 x 10-2 M of ammonia? NH3 + H2O NH4+ + OH- Kb = [NH4+] [OH-] / [NH3] 1.8 x 10-5 = [OH-]2 / 0.01 [OH-] = 4.24 x 10-4 M pOH = -log 4.24 x 10-4 = 3.37 pH = 14 - 3.37 = 10.63 45 Henderson-Hasselbalch Equation – The dissociation of a weak acid is as follows: HA H+ + A- The acid dissociation constant is as follows: Ka = [H+][A-] [HA] Rearranging this expression in terms of the parameter of interest [H+] gives the following: [H+] = Ka [HA] [A-] 46 Henderson-Hasselbalch Equation Take the log of both sides: log[H+] = [HA] log Ka + log [A-] Change the signs , remember pKa = -log Ka : pH = pKa - log [HA] [A-] or pH = pKa + log [A-] [HA] 47 What happens when ….. pH = pKa Substance protonation and deprotanation are in equillibruim. pH < pKa H+ on, substance protonated pH = pKa + log [A-] [HA] pH > pKa H+ off, substance deprotonated pH= pKa -1 Base/acid = 0.1 pH=pKa + 1 Base/acid = 10 48 Monoprotic- and PolyproticAcids 1. Monoprotic acids have only one ionizable proton. 2. Polyprotic acids have more than one ionizable proton. 3. The protons are removed in steps, not all at once. CH3COOH + NaOH pKa1 = 2.14 H3PO4 CH3COO-Na + + H2O pKa2 = 7.20 H2PO4+ H+ pKa3 = 12.4 HPO4-2 + H+ PO4-3 + H+ - It is always easier to remove the first proton in a polyprotic acid than the second. - Ka1 > Ka2 > Ka3 49 Polyprotic Acids 50

Acids and Bases in Biochemistry

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