Reactions of Acids & Bases Reactions of Acids & Bases

Self-Ionization of Water H 2 O + H 2 O H 3 O + + OH - K w = [H 3 O + ][OH - ] = 1.0   In pure water at 25ºC, both H 3 O + and OH - ions are found at concentrations of 1.0 x M.

 In all solutions at 25ºC, the product of the concentrations of H 3 O + and OH - ions is equal to 1.0 x M.

if [H 3 O + ] > 1.0  It’s an acid …

if [OH - ] > 1.0  It’s a base …

if [H 3 O + ] = [OH - ] = 1.0  It’s neutral …

pH = -log[H 3 O + ] pH Scale 0 7 INCREASING ACIDITY NEUTRAL INCREASING BASICITY 14 pouvoir hydrogène (Fr.) “hydrogen power”

pH Scale pH of Common Substances

pH Scale The pH scale runs from 0 to 14, but each unit represents a tenfold change in the concentration. The H 3 O + concentration of 1 x 10 0 is not 14 times a concentration of 1 x , but it is a factor of 10 14, which is 100 trillion times! The pH scale runs from 0 to 14, but each unit represents a tenfold change in the concentration. The H 3 O + concentration of 1 x 10 0 is not 14 times a concentration of 1 x , but it is a factor of 10 14, which is 100 trillion times!

pH Scale

 What is the concentration of OH - ions in saturated lime if [H 3 O + ] = 3.98 x M? Is lime acidic, basic, or neutral? BASIC Practice Problem #1 [OH - ] > 1.0  [H 3 O + ][OH - ] = 1.0  [3.98  ][OH - ] = 1.0  [OH - ] = 2.5  M or M

 Analysis of a sample of maple syrup reveals that the concentration of OH - ions is 5.0 x M. What is the pH? Is it acidic, basic, or neutral? [H 3 O + ][OH - ] = 1.0  Practice Problem #11 Acidic pH = -log[H 3 O + ] pH = 6.7 [H 3 O + ][5.0 x ] = 1.0  [H 3 O + ] = 2.0  M

More Sample Problems  What is the pH of M HNO 3 ?  What is the [H 3 O + ] and the [OH - ]? pH = -log[H 3 O + ] pH = -log[0.080] pH = 1.1 Acidic [H 3 O + ] = M [H 3 O + ][OH - ] = 1.0 x [ OH - ] = 1.0 x [H 3 O + ] [ OH - ] = 1.3 x

More Sample Problems  What is the pH of M H 2 SO 4 ?  What is the [H 3 O + ] and the [OH - ]? pH = -log[H 3 O + ] pH = -log[0.0246] pH = 1.61 Acidic [H 3 O + ] = 2(0.0123) M = M [H 3 O + ][OH - ] = 1.0 x [ OH - ] = 1.0 x [H 3 O + ] [ OH - ] = 4.07 x

More Sample Problems  The pH of a solution is 4.29  What is the [H 3 O + ] and the [OH - ]? pH = -log[H 3 O + ] Acidic [H 3 O + ][OH - ] = 1.0 x [ OH - ] = 1.0 x [H 3 O + ] [ OH - ] = 1.95 x = [H 3 O + ] [H 3 O + ] = 5.13 x pH

Buffers

 A buffer is a mixture that is able to release or absorb H + ions, keeping a solution’s pH constant.

 Most common buffers are mixtures of weak acids and their conjugate bases.

Example H 3 O + + C 2 H 3 O 2 - H 2 O + HC 2 H 3 O 2 Acetic acid and acetate ion When H 3 O + ions are added to this solution, they react with the acetate ion. pH changes only slightly

Example OH - + HC 2 H 3 O 2 H 2 O + C 2 H 3 O 2 - Acetic acid and acetate ion When OH - ions are added to this solution, they react with the acetic acid. pH changes only slightly

 Buffer Capacity is the amount of acid or base that a buffer can neutralize.  All buffers have a limited capacity to neutralize added H 3 O + or OH - ions.

 If you add H 3 O + or OH - ions beyond the buffer capacity, the ions will remain in solution, and the pH will change.

 The greater the concentration of buffer in the solution, the greater the buffer capacity.

 The human body must maintain the pH of blood between 7.35 and  A pH outside this range can cause extreme illness or death.

Acid-Base Titration Section 19-3

Titration An acid base titration is a carefully controlled neutralization reaction. Find concentration of an unknown solution by using a known “standard” solution standard solution unknown solution

 Equivalence point When enough standard solution is added to neutralize all the acid or base in the unknown solution. dramatic change in pH Determined by the Endpoint indicator color change Titration

 Strong Acid with Strong Base  Equivalence Point: pH = 7  phenolphthalein Titration

 Weak Acid with Strong Base  Equivalence Point: pH > 7  Phenolphthalein Titration

 Weak with Strong Acid  Weak Base with Strong Acid  Equivalence Point: pH < 7  Methyl red Titration

moles H 3 O + = moles OH - M  V  # a = M  V  # b M:Molarity V:volume #:# of H + ions in the acid or OH - ions in the base

Titration  42.5 mL of 1.3M KOH are required to neutralize 50.0 mL of H 2 SO 4. Find the molarity of H 2 SO 4. H3O+H3O+ M = ? V = 50.0 mL n = 2 OH - M = 1.3M V = 42.5 mL n = 1 MV# a = MV# b M(50.0mL)(2) =(1.3M)(42.5mL)(1) M = 0.55M H 2 SO 4