Chapter 15.5-15.6: Acids and Bases Libby Mensch

Properties of Water Water is amphoteric - can act as either an acid or a base HCl (aq) + H2O (l) ➝ H3O+ (aq) + Cl- (aq) *Here water is acting as a base * NH3 (aq) + H2O (l) ⇌ NH4+ (aq) + OH- (aq) *Here water is acting as an acid*

Autoionization of Water Autoionization - Process when something acts as an acid and base with itself One is acting as an acid and the other is acting as a base What is happening in an autoionization reaction: H2O (l) + H2O (l) ⇌ H3O+ (aq) + OH- (aq) How it is written: H2O (l) ⇌ H+ (aq) + OH- (aq)

Equilibrium Constant for Autoionization Reactions KW - Ion product constant for water (equilibrium constant for water) Aka: Dissociation constant for water Can calculate KW in different levels of pH Can determine if it is an acidic, neutral, or basic solution * [H3O+] [OH-] = KW = 1.0 X 10-14 * Always 1.0 X 10-14 at 25 oC This can be found in your reference table:

How to Use KW In these problems you will be given the value of either [H3O+] or [OH-] and will have to find the other to determine whether the solution is acidic, basic, or neutral. Neutral Relationships: Concentrations are equal so KW is easily calculated, if you are given one value at 1.0 X 10-7 then the solution is neutral and the other concentration is also 1.0 X 10-7 [H3O+] = [OH-] = 1.0 X 10-7 pH Level Neutral Acidic Basic Relationship [H3O+] = [OH-] [H3O+] > [OH-] [H3O+] < [OH-]

Practice Problem [H3O+] = 1.5 X 10-9 Calculate [OH-] at 25 oC for the solution and determine if the solution is acidic, basic, or neutral. [H3O+] [OH-] = KW = 1.0 X 10-14 (1.5 X 10-9) [OH-] = 1.0 X 10-14 1.5 X 10-9 1.5 X 10-14 [OH-] = 6.7 X 10-6 M [H3O+] < [OH-] ⟶ Basic Solution

Practice Problem [H3O+] = 7.5 X 10-5 Calculate [OH-] at 25 oC for the solution and determine if the solution is acidic, basic, or neutral. [H3O+] [OH-] = KW = 1.0 X 10-14 (7.5 X 10-5) [OH-] = 1.0 X 10-14 7.5 X 10-5 7.5 X 10-5 [OH-] = 1.3 X 10-10 M [H3O+] > [OH-] ⟶ Acidic Solution

The pH Scale: A Way to Quantify Acidity/Basicity Another way to determine if a solution is basic or acidic is using the pH scale ⭑ pH = -log(H3O+) ⭑ In general at 25 °C: *pH is temperature dependent* pH < 7 The solution is acidic pH > 7 The solution is basic pH = 7 The solution is neutral *on reference table* Note: A one unit change in the pH scale is a 10-fold change in H+ concentration.

Practice Problem [H3O+] = 1.8 X 10-4 Calculate the pH of the solution at 25 °C and state if the solution is acidic or basic. pH = -log[H3O+] = -log (1.8 X 10-4) = -(-3.74) = 3.74 ➝ solution is acidic

Alternate pH scales: pOH Another way to write a pH scale is in terms of [OH-] rather than [H3O+]: pOH = -log[OH-] Relationship can be derived between pH and pOH at 25 °C: pH + pOH = 14.00 Ex. pH = 4 4 + pOH = 14.00 ⇢ pOH = 10 *14 = pH + pOH is on reference table*

Finding the [H3O+] and pH of Strong and Weak Acid Solutions Strong or weak acids added to a solution will shift equilibrium left due (think Le Châtelier’s Principle) due to the addition of H3O+ ions or OH- ions In most solutions, the autoionization of water contributes barely any H3O+ ions compared to the ionization of a strong or weak acid ➝ focus on the effect of the strong/weak acid

Adding a Strong Acid The concentration of H3O+ in a strong acid solution is equal to the concentration of the strong acid Ex. 0.1 M HCl (strong acid) solution: [H3O+] = 0.10 M pH = -log[H3O+] pH = -log(.01) pH = 1.00 (very acidic) Side note: from here we can determine the pOH: pH + pOH = 14.00 pOH = 13.00 (very acidic on pOH scale)

Adding a Weak Acid More complicated than a strong acid because the concentration of H3O+ is not equal to the concentration of the weak acid (since weak acids do not completely ionize) Set up an equilibrium expression (similar to the ones from last chapter) using Ka Ka is the equilibrium constant for a weak acid and the formula is: (on reference table) Ka = [H+] [A-] [HA] Set up an ICE table, check if x is small and then solve! ⭑This is according to the generic equation: HA (aq) + H2O (l) ⇌ H3O+ (aq) + A- (aq) ⭑

Practice Problem Find the pH of a 0.200 M HNO2 solution. HNO2 (aq) + H20 (l) ⇌ H3O+ (aq) + NO2- (aq) Ka = 4.6 X 10-4 [HNO2 ] [H3O+] [NO2-] I 0.200 ≈ 0.00 0.00 C -x +x E 0.200 - x x

Practice Problem cont. √ √ [H30+] = 9.6 X 10-3 M pH = -log(9.6 X 10-3) = 2.02 (acidic) Ka = [H+] [A-] [HA] = x2 0.200 - x √ 4.6 X 10-4 = x2 0.200 √ (0.200) (0.200) [HNO2] = 0.2 - (9.6 X 10-3) = 0.19 M [H3O+] = 0.2 M [NO2+] = 0.2 M = 9.6 X 10-3 .200 X 100% = 4.8 % (x is small)

When to Use Each Equation Want to find the value of H3O+ or OH- and given one of these values: use KW Want to find pH and given H3O+ : use pH = -log(H3O+) Want to find pOH and given OH-: use pOH = -log(OH-) Given either pOH or pH and want to find the other: use pOH + pH = 14 Want to find the effect of a strong acid in a solution: pH = -log(H3O+) Want to find the effect of a weak acid in solution: Ka = [H+] [A-] [HA]

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