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Hydrogen Ions and Acidity

pH. Hydrogen Ions and Acidity. Water molecules go through self ionization to form hydrogen ion and hydroxide ions. In aqueous solution, hydrogen ions H + are always joined to a water molecule as hydronium ions

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Hydrogen Ions and Acidity

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  1. pH

  2. Hydrogen Ions and Acidity • Water molecules go through self ionization to form hydrogen ion and hydroxide ions. • In aqueous solution, hydrogen ions H+ are always joined to a water molecule as hydronium ions • In 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- ions H2O(l) H+(aq) + OH-(aq) H2O (l) + H2O (l) H3O+(aq) + OH-(aq)

  3. You can write an equation for an equilibrium constant for this!! • Called the Kw or the ion product constant • Kw ALWAYS equals 1x10-14 at 25ºC, regardless of what else is in the solution, like an acid or base • In a basic/alkaline solution, the hydroxide ion (OH-) is greater than 1x10-7M and the hydrogen ion (H+) is less 1x10-7M • In a acidic solution, the hydrogen ion (H+) is greater than 1x10-7M and the hydroxide ion (OH-) is less 1x10-7M Keq = [H3O+]*[OH-] You can use[H+] as an alternative here as well. H2O (l) + H2O (l) H3O+(aq) + OH-(aq)

  4. 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* 10-12 = [OH-]

  5. Calculate the [H+] concentration in each of the following: • [OH-] = 5.99*10-8M • [OH-] = 1.43 * 10 -12M KW = [H+]*[OH-] 1x10-14 = (5.99*10-8 M)* [H+] 1.67* 10-7 = [H+] 1x10-14 = (1.43 * 10 -12M)* [H+] 6.99* 10-3 = [H+]

  6. The pH Concept pH = -log[H+] • pH is the negative logarithm of the hydrogen ion concentration • A neutral solution H+ = 1x10-7 has a pH = -log[1x10-7]= 7 • A solution in which the [H+] is greater than 1x10-7 M and has a pH less than 7.0 is acidic • A solution with a pH greater than 7 is basic and has a [H+] concentration of less than 1x10-7M • You can also calculate pOH which is the negative logarithm of hydroxide ion concentration pOH = -log[OH-] pH + pOH = 14

  7. pH and significant figures • Hydrogen ion concentrations should always be reported to two significant figures • pH and pOH calculations should always be reported to two decimal places • Rules are due to the sensitivity of pH meters • How to solve pH/pOH problems: [H+] pH [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*10-9M pH = -log[H+] pH = -log[7.42*10-5] pH = 4.13 pOH= -log[OH-] • pH + pOH = 14 pOH= -log[4.21*10-9] pH = ? pOH = 8.38 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 calculator 1.41*10-4 = [H+] pOH= -log[OH-] 10.52 = -log[OH-] -10.52 = 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-] pH = 4.20 1.58* 10-10 = [OH-] • pH + pOH = 14 pOH= -log[OH-] pOH= -log[1.58* 10-10 ] • 4.20 + pOH = 14 pOH = 9.80 pOH = 9.80

  11. H3O+(aq) + CH3COO-(aq) <1% ionized CH3COOH(aq) + H2O(l) Strong and Weak Acids and Bases • A strong acid completely ionizes in water – simple Keq • Weak acids ionize only slightly in aqueous solution, which means there is a still an aqueous component to them – need to rewrite equilibrium constant • Remember – concentrated ≠ strong HCl(aq) + H2O(l) H3O+(aq) + Cl-(aq) 100% ionized

  12. pH of 0.10 M Solutions of Common Acids and Bases • CompoundpH • HCl (hydrochloric acid)  1.1 • H2SO4 (sulfuric acid)  1.2 • H2SO3 (sulfurous acid)  1.5 • H3PO4 (phosphoric acid)  1.5 • HF (hydrofluoric acid)  2.1 • CH3CO2H (acetic acid) 2.9 • H2CO3 (carbonic acid) 3.8 (saturated solution) • H2S (hydrogen sulfide)  4.1 • NH4Cl (ammonium chloride)  4.6 • HCN (hydrocyanic acid)  5.1 • Na2SO4 (sodium sulfate)  6.1 • NaCl (sodium chloride)  6.4 • NaHCO3 (sodium bicarbonate)  8.4 • Na2SO3 (sodium sulfite)  9.8 • NaCN (sodium cyanide)  11.0 • NH3 (aqueous ammonia)  11.1 • Na2CO3 (sodium carbonate)  11.6 • Na3PO4 (sodium phosphate)  12.0 • NaOH (sodium hydroxide, lye)  13.0 http://www.cartage.org.lb/en/themes/sciences/chemistry/Inorganicchemistry/AcidsBases/Common/Common.htm

  13. HA(aq) + H2O(l) H3O+(aq) + A-(aq) [H3O+] X [A-] Keq = [HA] X [H2O] [H3O+] X [A-] Keq X H2O =Ka = [HA] Acid Disassociation Constant • The equilibrium constant for weak acids (HA) can be written as: • For dilute solutions, the concentration of water is a constant, and can be combined with Keqto give the acid dissociation constant (Ka) Acid Conjugate base • Kais sometimes referred to as the ionization constant • Weak acids have small Ka values, while stronger acids have larger Kavalues

  14. B(aq) + H2O(l) BH+(aq) + HO-(aq) [BH+] X [HO-] Keq = [B] X [H2O] [BH+] X [OH-] Keq X H2O = Kb = [B] Base Disassociation Constant base Conjugate acid • The equilibrium constant for weak Bases (B) can be written as: • Again, for dilute solutions, the base dissociation constant (Kb) • The magnitude of Kb indicates the ability of a weak base to compete with the very strong base OH- for hydrogen ions • The 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. HC2H3O2(aq) + H2O(l) H3O+(aq) + C2H3O2-(aq) • A 0.1000M solution of acetic acid is only partially ionized and has a pH of 2.87. What is the acid dissociation constant (Ka) or ethanoic acid? pH = -log[H+] 2.87 = -log[H+] Ka = Ka = -2.87 = log[H+] 1.35*10-3= [H+] = [C2H3O2-] Ka = 1.82*10-5

  17. HNO2(aq) + H2O(l) H3O+(aq) + NO2-(aq) • 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? pH = -log[H+] pH = -log[1.84*10-2] Ka = pH = 1.73 Ka* [HNO2] = [NO2-]*[H+] Ka* [HNO2] = x*x = x2 1.84*10-2= x = [H+]

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