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PH. Hydrogen Ions and Acidity Water molecules go through self ionization to form hydrogen ion and hydroxide ions. In aqueous solution, hydrogen ions H.

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Presentation on theme: "PH. Hydrogen Ions and Acidity Water molecules go through self ionization to form hydrogen ion and hydroxide ions. In aqueous solution, hydrogen ions H."— Presentation transcript:

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 M of H + ions and 1 x M of OH - ions H 2 O(l)H + (aq) + OH - (aq) H 2 O (l) + H 2 O (l)H 3 O + (aq) + OH - (aq)

3 H 2 O (l) + H 2 O (l)H 3 O + (aq) + OH - (aq) You can write an equation for an equilibrium constant for this!! Called the K w or the ion product constant K w ALWAYS equals 1x 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 -7 M and the hydrogen ion (H + ) is less 1x10 -7 M In a acidic solution, the hydrogen ion (H + ) is greater than 1x10 -7 M and the hydroxide ion (OH - ) is less 1x10 -7 M K eq = [H 3 O + ]*[OH - ] You can use [H+] as an alternative here as well.

4 A can of Coke has [H+]= 2.6*10 -3 M. What is the concentration of OH - ? K w = [H + ]*[OH - ] 1x = (2.6*10 -3 M)* [OH - ] 3.85* = [OH - ]

5 Calculate the [H+] concentration in each of the following: [OH-] = 5.99*10 -8 M [OH-] = 1.43 * M K W = [H + ]*[OH - ] 1x = (5.99*10 -8 M)* [H + ] 1.67* = [H + ] 1x = (1.43 * M)* [H + ] 6.99* = [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 -7 M 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: pH[H+][OH-]pOH

8 pH practice Calculate the pH of a solution with [H+] = 7.42*10 -5 M Calculate the pOH of a solution with [OH-] = 4.21*10 -9 M pH = -log[H + ] pH = -log[7.42*10 -5 ] pH = 4.13 pOH = -log[OH-] pOH = -log[4.21*10 -9 ] pOH = 8.38 pH = ? pH + pOH = 14 pH = 5.62

9 pH practice The pH of a solution is What is the [H+]? What is the [OH-] concentration of a solution with pOH = 10.52? pH = -log[H + ] 3.85 = -log[H+] 1.41*10 -4 = [H+] = log[H+] Find the inverse log (inv log) or 10 x of this number on calculator pOH = -log[OH-] = -log[OH-] 3.02* = [OH-] = log[OH-]

10 Calculate the pOH of a solution with [H+] = 6.32*10 -5 M. 1x = (6.32*10 -5 M)* [OH - ] 1.58* = [OH - ] pH = -log[H + ] pH = -log[6.32*10 -5 ] pH = 4.20 pOH = -log[OH-] pOH = -log[1.58* ] pOH = 9.80 pH + pOH = 14 K w = [H + ]*[OH - ] pOH = 14 pOH = 9.80

11 Strong and Weak Acids and Bases A strong acid completely ionizes in water – simple K eq 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) + H 2 O(l)H 3 O + (aq) + Cl - (aq) 100% ionized CH 3 COOH(aq) + H 2 O(l) H 3 O + (aq) + CH 3 COO - (aq) <1% ionized

12 pH of 0.10 M Solutions of Common Acids and Bases Compound pH HCl (hydrochloric acid) 1.1 H 2 SO 4 (sulfuric acid) 1.2 H 2 SO 3 (sulfurous acid) 1.5 H 3 PO 4 (phosphoric acid) 1.5 HF (hydrofluoric acid) 2.1 CH 3 CO 2 H (acetic acid)2.9 H 2 CO 3 (carbonic acid)3.8 (saturated solution) H 2 S (hydrogen sulfide) 4.1 NH 4 Cl (ammonium chloride) 4.6 HCN (hydrocyanic acid) 5.1 Na 2 SO 4 (sodium sulfate) 6.1 NaCl (sodium chloride) 6.4 NaHCO 3 (sodium bicarbonate) 8.4 Na 2 SO 3 (sodium sulfite) 9.8 NaCN (sodium cyanide) 11.0 NH 3 (aqueous ammonia) 11.1 Na 2 CO 3 (sodium carbonate) 11.6 Na 3 PO 4 (sodium phosphate) 12.0 NaOH (sodium hydroxide, lye) 13.0

13 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 K eq to give the acid dissociation constant (K a ) K eq [H 3 O + ] X [A - ] [HA] X [H 2 O] = K eq X H 2 O = K a [H 3 O + ] X [A - ] [HA] = K a is sometimes referred to as the ionization constant Weak acids have small Ka values, while stronger acids have larger K a values H 3 O + (aq) + A - (aq)HA(aq) + H 2 O(l) AcidConjugate base

14 Base Disassociation Constant The equilibrium constant for weak Bases (B) can be written as: Again, for dilute solutions, the base dissociation constant (K b ) K eq [BH + ] X [HO - ] [B] X [H 2 O] = K eq X H 2 O = K b [BH + ] X [OH - ] [B] = The magnitude of K b indicates the ability of a weak base to compete with the very strong base OH- for hydrogen ions The smaller the K b the weaker the base BH + (aq) + HO - (aq)B(aq) + H 2 O(l) baseConjugate acid

15 Calculating Dissociation Constants Problems involving K a of weak acid or the K b of a weak base: 1. Write the expression for K a or K b using the actual compounds in the chemical equation. Rearrange for the unknown variable. 2. Substitute the known numbers of all the variables at equilibrium into the expression for K a or K b. 3. 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 (K a ) or ethanoic acid? H 3 O + (aq) + C 2 H 3 O 2 - (aq)HC 2 H 3 O 2 (aq) + H 2 O(l) K a = pH = -log[H + ] 2.87 = -log[H+] = log[H+] 1.35*10 -3 = [H+]= [C 2 H 3 O 2 - ] K a = 1.82*10 -5

17 The K a of nitrous acid is 4.0 * What is the concentration of [H+] in a 0.85M solution of nitrous acid? What is the pH? H 3 O + (aq) + NO 2 - (aq)HNO 2 (aq) + H 2 O(l) K a = 1.84*10 -2 = x = [H + ] K a * [HNO 2 ] = [NO 2 - ]*[H + ] K a * [HNO 2 ] = x*x = x 2 pH = -log[H + ] pH = -log[1.84*10 -2 ] pH = 1.73


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