Ionic Equilibria (Acids and Bases) Chapter 18

Phase I STRONG ELECTROLYTES

Strong Electrolytes 1. Strong Acids – 7 strong acids 2. Strong Bases – Group 1 and 2 (Ca, Sr, Ba) hydroxides 3. Soluble Salts 4. Some molecular compounds – requires extensive ionization

Calculating Ion Concentration Calculate the concentration of each ion and the pH in 0.050M HNO3 Calculate the concentration of each ion and the pH in 0.050M HNO3 HNO 3 (aq) + H 2 O (l)  H (aq) + NO 3 - (aq) Assume 100% dissociation (all strong electrolytes) Assume 100% dissociation (all strong electrolytes) H (aq) = 0.050M H (aq) = 0.050M NO 3 - (aq) = 0.050M NO 3 - (aq) = 0.050M pH = -log(H ) = -log(0.050) = 1.30 pH = -log(H ) = -log(0.050) = 1.30

Calculating Ion Concentration Calculate the concentraion of ions and the pH of 0.020M Ba(OH)2 solution. Calculate the concentraion of ions and the pH of 0.020M Ba(OH)2 solution. Ba(OH) 2  Ba OH - Assume strong electrolyte = 100% dissociation Assume strong electrolyte = 100% dissociation Ba 2+ = 0.020M Ba 2+ = 0.020M OH - = 2(0.020M) = 0.040M OH - = 2(0.020M) = 0.040M [OH - ][H ] = 1 x [OH - ][H ] = 1 x [H ] = 2.5 x [H ] = 2.5 x pH = -log(2.5 x ) = pH = -log(2.5 x ) = 12.60

Auto-ionization of Water 2H 2 O (l)  H 3 O + (aq) + OH - (aq) Kc = [H 3 O + ][OH - ] *temp dependent Kc = Kw = [H 3 O + ][OH - ] = 1.0 x o C Table 18-2 (p755) has values at different temperatures

Auto-ionization of Water Calculate the concentration of H 3 O + and OH - in 0.050M HCl Calculate the concentration of H 3 O + and OH - in 0.050M HCl HCl + H 2 O  H 3 O + + Cl - H 3 O + = 0.050M H 3 O + = 0.050M [H 3 O + ][OH - ] = 1.0 x [H 3 O + ][OH - ] = 1.0 x [0.050M][OH - ] = 1.0 x [0.050M][OH - ] = 1.0 x [OH - ] = 2.0 x [OH - ] = 2.0 x

Question Why don’t we add in the concentration of H 3 O + from the auto ionization of water? Why don’t we add in the concentration of H 3 O + from the auto ionization of water? Compare to 0.50…no contest Compare to 0.50…no contest So much H 3 O + that equilibrium is pushed to H 2 O So much H 3 O + that equilibrium is pushed to H 2 O

pH and pOH pH and pOH express the acidity and basicity of dilute solutions pH and pOH express the acidity and basicity of dilute solutions pH = -log [H 3 O + ] [H 3 O + ] = 10 -pH pOH = -log [OH - ] [OH - ] = 10 -pOH pKc = -log (kc) kc = 10 -pkc

Cool relationship between pH and pOH [H 3 O + ][OH - ] = 1.0 x log[H 3 O + ] + log[OH - ] = log(1.0 x ) and -log[H 3 O + ] - log[OH - ] = -log(1.0 x )  pH + pOH = 14.00

Complete the Table Below What are the ranges of pH and pOH? [H3O+][H3O+][H3O+][H3O+] [OH - ] pHpOH x x x x

Example Problem Calculate [H 3 O + ], pH, pOH, and [OH - ] for 0.020M HNO 3. Calculate [H 3 O + ], pH, pOH, and [OH - ] for 0.020M HNO 3. HNO 3 + H 2 O  H 3 O + + NO 3 - [H 3 O + ] = [0.020] [H 3 O + ] = [0.020] pH = -log(0.020) = 1.70 pH = -log(0.020) = 1.70 pOH = 14 – pH = 14 – 1.70 = pOH = 14 – pH = 14 – 1.70 = [OH - ] = 10 -pOH = = 5.0 x [OH - ] = 10 -pOH = = 5.0 x

Phase II WEAK ELECTROLYTES

Ionization of Weak Acids The activity of water is assumed to be 1 as it is nearly a pure liquid with this weak ionization. The activity of water is assumed to be 1 as it is nearly a pure liquid with this weak ionization.

Ionization of Weak Acids Small ka = weak acid Small ka = weak acid Large ka = strong acid Large ka = strong acid Ionization constants are measured experimentally Ionization constants are measured experimentally –Freezing point depression –Electrical conduction measurement –pH measurements

Example In a 0.12M solution of weak monoprotic acid (HA) is 5.0% dissociated, calculate ka. In a 0.12M solution of weak monoprotic acid (HA) is 5.0% dissociated, calculate ka. HA  H + + A -

Example The pH of weak acid H  is measured at 2.97 in a 0.10M solution. Calculate the Ka. The pH of weak acid H  is measured at 2.97 in a 0.10M solution. Calculate the Ka.

Example Calculate the concentration of all species in a 0.15M acetic acid solution if Calculate the concentration of all species in a 0.15M acetic acid solution if Ka = 1.8 x When x is <5% of the number being added or subtracted from the initial concentration, it may be NEGLECTED. When x is <5% of the number being added or subtracted from the initial concentration, it may be NEGLECTED. % Ionization = [CH3COOH ionized] % Ionization = [CH3COOH ionized] [CH3COOH initial] [CH3COOH initial]

Ionization of Weak Bases Review Appendix G, p. A-15

Example Calculate the concentration of various species in 0.15M NH3 and % Ionization. Calculate the concentration of various species in 0.15M NH3 and % Ionization.

Example An Ammonia solution has a pH of Calculate its molarity. An Ammonia solution has a pH of Calculate its molarity.

Polyprotic Acids Acids with more than one proton Acids with more than one proton Most are considered weak acids, however the first proton ionizes the strongest Most are considered weak acids, however the first proton ionizes the strongest Typically 10 4 to 10 6 difference between Ka Typically 10 4 to 10 6 difference between Ka

Example H 3 AsO 4, arsenic acid H 3 AsO 4, arsenic acid Ka1 = 2.5 x Ka1 = 2.5 x Ka2 = 5.6 x Ka2 = 5.6 x Ka3 = 3.0 x Ka3 = 3.0 x Calculate the concentration of all species in 0.100M solution of H 3 AsO 4. Calculate the concentration of all species in 0.100M solution of H 3 AsO 4.

Solvolysis Solvolysis – the reaction of a dissolved substance with a solvent Solvolysis – the reaction of a dissolved substance with a solvent –Hydrolysis – the reaction of a dissolved substance with WATER. –Generally, anions of STRONG acids do not hydrolyze –Generally, anions of WEAK acids do hydrolyze

Solvolysis KCl + H2O  NR (no pH change) Neither K nor Cl will Hydrolyze (from strong sources) NaClO + H2O  HClO + Na+ + OH- Anions of weak acids are relatively strong bases. Kb = [HClO][OH-] [ClO-]

Yet another scary math proof

Solvolysis Works for all conjugate acid/base pairs in water. Works for all conjugate acid/base pairs in water.

Salts Where things get murkey Yay!

Salts Objective: Objective: What happens to pH when I put a salt of a Strong Acid, Weak Acid, Strong Base, or Weak Base into water? What happens to pH when I put a salt of a Strong Acid, Weak Acid, Strong Base, or Weak Base into water?

Strong Acid and Strong Base Cation From: Strong Base (Na+) Strong Base (Na+) Na + + H 2 O  NR Na + + H 2 O  NR No Effect No Effect Anion From: Strong Acid (Cl-) Strong Acid (Cl-) Cl - + H 2 O  NR Cl - + H 2 O  NR No Effect No Effect

Strong Base and Weak Acid Cation From: Strong Base (K+) Strong Base (K+) K + + H 2 O  NR K + + H 2 O  NR No Effect No Effect Anion From: Weak Acid (ClO-) Weak Acid (ClO-) ClO - + H 2 O  HClO + OH- ClO - + H 2 O  HClO + OH- Increases pH Increases pH

Weak Base and Strong Acid Cation From: Weak Base (NH 4 + ) Weak Base (NH 4 + ) NH H 2 O  H 3 O + + NH 3 NH H 2 O  H 3 O + + NH 3 Decreases pH Decreases pH Anion From: Strong Acid (NO 3 -) Strong Acid (NO 3 -) NO H 2 O  NR NO H 2 O  NR No Effect No Effect

Weak Base and Weak Acid Cation From: Weak Base (NH 4 + ) Weak Base (NH 4 + ) NH H 2 O  H 3 O + + NH 3 NH H 2 O  H 3 O + + NH 3 Decreases pH Decreases pH Anion From: Weak Acid (ClO-) Weak Acid (ClO-) ClO - + H 2 O  HClO + OH- ClO - + H 2 O  HClO + OH- Increases pH Increases pH

Weak Base and Weak Acid How do you know if the pH will increase or decrease? How do you know if the pH will increase or decrease? If Kb = Ka, then  pH = 0 If Kb = Ka, then  pH = 0 If Kb > Ka, then  pH = + If Kb > Ka, then  pH = + If Kb < Ka, then  pH = - If Kb < Ka, then  pH = -