2 Hydrogen Ions and Acidity Water molecules go through self ionization to form hydrogen ion and hydroxide ions.H2O(l)H+(aq) + OH-(aq)In aqueous solution, hydrogen ions H+ are always joined to a water molecule as hydronium ionsIn pure (neutral) water, the self-ionization of water results in 1 x 10-7 M of H+ ions and 1 x 10-7 M of OH- ionsH2O (l) + H2O (l)H3O+(aq) + OH-(aq)
3 Keq = [H3O+]*[OH-] Called the Kw or the ion product constant H2O (l) + H2O (l)H3O+(aq) + OH-(aq)You can write an equation for an equilibrium constant for this!!Called the Kw or the ion product constantKw ALWAYS equals 1x10-14 at 25ºC, regardless of what else is in the solution, like an acid or baseIn a basic/alkaline solution, the hydroxide ion (OH-) is greater than 1x10-7 M and the hydrogen ion (H+) is less 1x10-7 MIn a acidic solution, the hydrogen ion (H+) is greater than 1x10-7 M and the hydroxide ion (OH-) is less 1x10-7 MKeq = [H3O+]*[OH-]You can use [H+] as an alternative here as well.
4 Kw = [H+]*[OH-] 1x10-14 = (2.6*10-3 M)* [OH-] 3.85* 10-12 = [OH-] A can of Coke has [H+]= 2.6*10-3 M. What is the concentration of OH-?Kw = [H+]*[OH-]1x10-14 = (2.6*10-3 M)* [OH-]3.85* = [OH-]
5 KW = [H+]*[OH-] 1x10-14 = (5.99*10-8 M)* [H+] 1.67* 10-7 = [H+] Calculate the [H+] concentration in each of the following:[OH-] = 5.99*10-8M[OH-] = 1.43 * MKW = [H+]*[OH-]1x10-14 = (5.99*10-8 M)* [H+]1.67* 10-7 = [H+]1x10-14 = (1.43 * M)* [H+]6.99* 10-3 = [H+]
6 The pH Concept pH + pOH = 14 pH = -log[H+] pOH = -log[OH-] pH is the negative logarithm of the hydrogen ion concentrationA neutral solution H+ = 1x10-7 has a pH = -log[1x10-7]= 7A solution in which the [H+] is greater than 1x10-7 M and has a pH less than 7.0 is acidicA solution with a pH greater than 7 is basic and has a [H+] concentration of less than 1x10-7 MYou can also calculate pOH which is the negative logarithm of hydroxide ion concentrationpOH = -log[OH-]pH + pOH = 14
7 pH and significant figures Hydrogen ion concentrations should always be reported to two significant figurespH and pOH calculations should always be reported to two decimal placesRules are due to the sensitivity of pH metersHow to solve pH/pOH problems:pH[H+][OH-]pOH
8 pH practice Calculate the pH of a solution with [H+] = 7.42*10-5M Calculate the pOH of a solution with [OH-] = 4.21*10-9MpH = -log[H+]pH = -log[7.42*10-5]pH = 4.13pOH = -log[OH-]pH + pOH = 14pOH = -log[4.21*10-9]pOH = 8.38pH = ?pH = 5.62
9 pH practice The pH of a solution is 3.85. What is the [H+]? What is the [OH-] concentration of a solution with pOH = 10.52?pH = -log[H+]3.85 = -log[H+]-3.85 = log[H+]Find the inverse log (inv log) or 10x of this number on calculator1.41*10-4 = [H+]pOH = -log[OH-]10.52 = -log[OH-]= log[OH-]3.02*10-11 = [OH-]
10 Calculate the pOH of a solution with [H+] = 6.32*10-5M. Kw = [H+]*[OH-]pH = -log[H+]pH = -log[6.32*10-5]1x10-14 = (6.32*10-5M)* [OH-]1.58* = [OH-]pH = 4.20pOH = -log[OH-]pH + pOH = 14pOH = -log[1.58* ]pOH = 14pOH = 9.80pOH = 9.80
11 Strong and Weak Acids and Bases A strong acid completely ionizes in water – simple KeqWeak acids ionize only slightly in aqueous solution, which means there is a still an aqueous component to them – need to rewrite equilibrium constantRemember – concentrated ≠ strongHCl(aq) + H2O(l)H3O+(aq) + Cl-(aq) % ionizedCH3COOH(aq) + H2O(l)H3O+(aq) + CH3COO-(aq) <1% ionized
13 Acid Disassociation Constant The equilibrium constant for weak acids (HA) can be written as:AcidConjugate baseH3O+(aq) + A-(aq)HA(aq) + H2O(l)Keq[H3O+] X [A-][HA] X [H2O]=For dilute solutions, the concentration of water is a constant, and can be combined with Keq to give the acid dissociation constant (Ka)Keq X H2O = Ka[H3O+] X [A-][HA]=Ka is sometimes referred to as the ionization constantWeak acids have small Ka values, while stronger acids have larger Ka values
14 Base Disassociation Constant The equilibrium constant for weak Bases (B) can be written as:baseConjugate acidBH+(aq) + HO-(aq)B(aq) + H2O(l)Keq[BH+] X [HO-][B] X [H2O]=Again, for dilute solutions, the base dissociation constant (Kb)Keq X H2O = Kb[BH+] X [OH-][B]=The magnitude of Kb indicates the ability of a weak base to compete with the very strong base OH- for hydrogen ionsThe smaller the Kb the weaker the base
15 Calculating Dissociation Constants Problems involving Ka of weak acid or the Kb of a weak base:Write the expression for Ka or Kb using the actual compounds in the chemical equation. Rearrange for the unknown variable.Substitute the known numbers of all the variables at equilibrium into the expression for Ka or Kb.Solve for the unknown.
16 A M solution of acetic acid is only partially ionized and has a pH of What is the acid dissociation constant (Ka) or ethanoic acid?H3O+(aq) + C2H3O2-(aq)HC2H3O2(aq) + H2O(l)pH = -log[H+]Ka =2.87 = -log[H+]-2.87 = log[H+]1.35*10-3 = [H+]= [C2H3O2-]Ka =Ka = 1.82*10-5
17 The Ka of nitrous acid is 4. 10-4 The Ka of nitrous acid is 4.0 *10-4. What is the concentration of [H+] in a 0.85M solution of nitrous acid? What is the pH?H3O+(aq) + NO2-(aq)HNO2(aq) + H2O(l)Ka =pH = -log[H+]pH = -log[1.84*10-2]pH = 1.73Ka * [HNO2] = [NO2-]*[H+]Ka * [HNO2] = x*x = x2√(Ka∗[HNO2] =x√(4.0∗ 10 −4 ∗0.85)=x1.84*10-2= x = [H+]
Your consent to our cookies if you continue to use this website.