Ionic Equilibria Ionic Equilibria (Acids and Bases)(Acids and Bases)

Chapter 18Chapter 18

Phase IPhase I

STRONG ELECTROLYTESSTRONG ELECTROLYTES

Strong ElectrolytesStrong Electrolytes

1.1. Strong AcidsStrong Acids – 7 strong acids – 7 strong acids

2.2. Strong BasesStrong Bases – Group 1 and 2 (Ca, – Group 1 and 2 (Ca, Sr, Ba) hydroxidesSr, Ba) hydroxides

3.3. Soluble SaltsSoluble Salts

4.4. Some molecular compoundsSome molecular compounds – – requires extensive ionizationrequires extensive ionization

Calculating Ion Calculating Ion ConcentrationConcentration

Calculate the concentration of each Calculate the concentration of each ion and the pH in 0.050M HNO3ion and the pH in 0.050M HNO3

HNOHNO33 (aq)(aq) + H + H22O O (l)(l) H H3300+ + (aq)(aq) + NO + NO33

-- (aq)(aq)

Assume 100% dissociation (all strong Assume 100% dissociation (all strong electrolytes)electrolytes)

HH3300+ + (aq) (aq) = 0.050M= 0.050M

NONO33-- (aq) (aq) = 0.050M= 0.050M

pH = -log(HpH = -log(H3300++) = -log(0.050) = 1.30) = -log(0.050) = 1.30

Calculating Ion Calculating Ion ConcentrationConcentration

Calculate the concentraion of ions and the Calculate the concentraion of ions and the pH of 0.020M Ba(OH)2 solution.pH of 0.020M Ba(OH)2 solution.

Ba(OH)Ba(OH)22 Ba Ba2+2+ + 2OH + 2OH--

Assume strong electrolyte = 100% Assume strong electrolyte = 100% dissociationdissociation

BaBa2+ 2+ = 0.020M= 0.020M OHOH- - = 2(0.020M) = 0.040M= 2(0.020M) = 0.040M [OH[OH--][H][H3300++] = 1 x 10] = 1 x 10-14-14

[H[H3300++] = 2.5 x 10] = 2.5 x 10-13-13

pH = -log(2.5 x 10pH = -log(2.5 x 10-13-13) = 12.60) = 12.60

Auto-ionization of WaterAuto-ionization of Water

2H2H22O (l) O (l) H H33OO++ (aq) + OH (aq) + OH-- (aq) (aq)

Kc = [HKc = [H33OO++][OH][OH--] *temp dependent] *temp dependent

Kc = Kw = [HKc = Kw = [H33OO++][OH][OH--] = 1.0 x 10] = 1.0 x 10-14-14 @ @ 2525ooCC

Table 18-2 (p755) has values at different Table 18-2 (p755) has values at different temperaturestemperatures

Auto-ionization of WaterAuto-ionization of Water

Calculate the concentration of HCalculate the concentration of H33OO++ and and OHOH-- in 0.050M HCl in 0.050M HCl

HCl + HHCl + H22O O H H33OO++ + Cl + Cl--

HH33OO+ + = 0.050M= 0.050M [H[H33OO++][OH][OH--] = 1.0 x 10] = 1.0 x 10-14-14

[0.050M][OH[0.050M][OH--] = 1.0 x 10] = 1.0 x 10-14-14

[OH[OH--] = 2.0 x 10] = 2.0 x 10-13-13

QuestionQuestion

Why don’t we add in the Why don’t we add in the concentration of Hconcentration of H33OO++ from the auto from the auto ionization of water?ionization of water?

Compare 10Compare 10-7-7 to 0.50…no contest to 0.50…no contest So much HSo much H33OO++ that equilibrium is that equilibrium is

pushed to Hpushed to H22OO

pH and pOHpH and pOH

pH and pOH express the acidity and pH and pOH express the acidity and basicity of dilute solutionsbasicity of dilute solutions

pH = -log pH = -log [H[H33OO++]]

[H[H33OO++] = 10] = 10-pH-pH

pOH = -log pOH = -log [OH[OH--] ]

[OH[OH--] = 10] = 10-pOH-pOH

pKc = -log (kc)pKc = -log (kc) kc = 10kc = 10-pkc-pkc

Cool relationship between pH Cool relationship between pH and pOHand pOH

[H[H33OO++][OH][OH--] = 1.0 x 10] = 1.0 x 10-14-14

log[Hlog[H33OO++] + log[OH] + log[OH--] = log(1.0 x 10] = log(1.0 x 10-14-14))

andand

-log[H-log[H33OO++] - log[OH] - log[OH--] = -log(1.0 x 10] = -log(1.0 x 10-14-14))

pH + pOH = 14.00pH + pOH = 14.00

Complete the Table BelowComplete the Table BelowWhat are the ranges of pH and What are the ranges of pH and

pOH?pOH?[H[H33OO++]] [OH[OH--]] pHpH pOHpOH

1.01.0 1.0 x 101.0 x 10-14-14 0.000.00 14.0014.00

1.0 x 101.0 x 10-3-3

1.0 x 101.0 x 10-7-7

2.0 x 102.0 x 10-12-12

0.000.00

Example ProblemExample Problem

Calculate [HCalculate [H33OO++] , pH, pOH, and [OH] , pH, pOH, and [OH--] ] for 0.020M HNOfor 0.020M HNO33..

HNOHNO33 + H + H22O O H H33OO++ + NO + NO33--

[H[H33OO++] = [0.020]] = [0.020] pH = -log(0.020) = 1.70pH = -log(0.020) = 1.70 pOH = 14 – pH = 14 – 1.70 = 12.30pOH = 14 – pH = 14 – 1.70 = 12.30 [OH[OH--] = 10] = 10-pOH = -pOH = 1010-12.30-12.30 = 5.0 x 10 = 5.0 x 10-13-13

Phase IIPhase II

WEAK ELECTROLYTESWEAK ELECTROLYTES

Ionization of Weak AcidsIonization of Weak Acids

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

]][[

]][[

23

33

OHCOOHCH

COOCHOHkc

Ionization of Weak AcidsIonization of Weak Acids

Small ka = weak acidSmall ka = weak acid Large ka = strong acidLarge ka = strong acid Ionization constants are measured Ionization constants are measured

experimentallyexperimentally– Freezing point depressionFreezing point depression– Electrical conduction measurementElectrical conduction measurement– pH measurementspH measurements

5

3

33 108.1][

]][[

xCOOHCH

COOCHOHka

ExampleExample

In a 0.12M solution of weak In a 0.12M solution of weak monoprotic acid (HA) is 5.0% monoprotic acid (HA) is 5.0% dissociated, calculate ka.dissociated, calculate ka.

HA HA H H++ + A + A--

ExampleExample

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

ExampleExample

Calculate the concentration of all Calculate the concentration of all species in a 0.15M acetic acid solution species in a 0.15M acetic acid solution if if Ka = 1.8 x 10Ka = 1.8 x 10-5-5

When x is <5% of the number being When x is <5% of the number being added or subtracted from the initial added or subtracted from the initial concentration, it may be NEGLECTED.concentration, it may be NEGLECTED.

% Ionization = % Ionization = [CH3COOH ionized][CH3COOH ionized] [CH3COOH initial][CH3COOH initial]

Ionization of Weak BasesIonization of Weak Bases

Review Appendix G, p. A-15Review Appendix G, p. A-15

ExampleExample

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

ExampleExample

An Ammonia solution has a pH of An Ammonia solution has a pH of 11.37. Calculate its molarity.11.37. Calculate its molarity.

Polyprotic AcidsPolyprotic Acids

Acids with more than one protonAcids with more than one proton

Most are considered weak acids, Most are considered weak acids, however the first proton ionizes the however the first proton ionizes the strongeststrongest

Typically 10Typically 1044 to 10 to 1066 difference between difference between KaKa

ExampleExample

HH33AsOAsO44, arsenic acid, arsenic acid Ka1 = 2.5 x 10Ka1 = 2.5 x 10-4-4

Ka2 = 5.6 x 10Ka2 = 5.6 x 10-8-8

Ka3 = 3.0 x 10Ka3 = 3.0 x 10-13-13

Calculate the concentration of all Calculate the concentration of all species in 0.100M solution of Hspecies in 0.100M solution of H33AsOAsO44..

SolvolysisSolvolysis

SolvolysisSolvolysis – the reaction of a – the reaction of a dissolved substance with a solventdissolved substance with a solvent– HydrolysisHydrolysis – the reaction of a dissolved – the reaction of a dissolved

substance with WATER.substance with WATER.– Generally, anions of STRONG acids do Generally, anions of STRONG acids do

not hydrolyzenot hydrolyze– Generally, anions of WEAK acids do Generally, anions of WEAK acids do

hydrolyzehydrolyze

SolvolysisSolvolysis

KCl + HKCl + H22O O NR (no pH change) NR (no pH change)

Neither K nor Cl will Hydrolyze (from strong Neither K nor Cl will Hydrolyze (from strong sources)sources)

NaClO + HNaClO + H22O O HClO + Na HClO + Na++ + OH + OH--

Anions of weak acids are relatively strong Anions of weak acids are relatively strong bases.bases.

Kb = Kb = [HClO][OH[HClO][OH--]]

[ClO-][ClO-]

Yet another scary math Yet another scary math proofproof

][

][

][

]][[

H

Hx

ClO

OHHClOkb

]1[

]][[

]][[

][

OHHx

HClO

HClOkb kw

Ka

1

Ka

KwKb

SolvolysisSolvolysis

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

SaltsSalts

Where things get murkeyWhere things get murkey

Yay!Yay!

SaltsSalts

Objective:Objective: What happens to pH when I put a salt What happens to pH when I put a salt

of a Strong Acid, Weak Acid, Strong of a Strong Acid, Weak Acid, Strong Base, or Weak Base into water?Base, or Weak Base into water?

Strong Acid and Strong BaseStrong Acid and Strong Base

Cation From:Cation From: Strong Base (Na+)Strong Base (Na+) NaNa++ + H + H22O O NR NR

No EffectNo Effect

Anion From:Anion From: Strong Acid (Cl-)Strong Acid (Cl-) ClCl-- + H + H22O O NR NR

No EffectNo Effect

Strong Base and Weak AcidStrong Base and Weak Acid

Cation From:Cation From: Strong Base (K+)Strong Base (K+) KK++ + H + H22O O NR NR

No EffectNo Effect

Anion From:Anion From: Weak Acid (ClO-)Weak Acid (ClO-) ClOClO-- + H + H22O O HClO + HClO +

OH-OH-

Increases pHIncreases pH

Weak Base and Strong AcidWeak Base and Strong Acid

Cation From:Cation From: Weak Base (NHWeak Base (NH44

++))

NHNH44++ + H + H22O O HH33OO++

+ NH+ NH33

Decreases pHDecreases pH

Anion From:Anion From: Strong Acid (NOStrong Acid (NO33-)-)

NONO33- + H- + H22O O NR NR

No EffectNo Effect

Weak Base and Weak AcidWeak Base and Weak Acid

Cation From:Cation From: Weak Base (NHWeak Base (NH44

++))

NHNH44++ + H + H22O O HH33OO++

+ NH+ NH33

Decreases pHDecreases pH

Anion From:Anion From: Weak Acid (ClO-)Weak Acid (ClO-) ClOClO-- + H + H22O O HClO HClO

+ + OH-OH-

Increases pHIncreases pH

Weak Base and Weak AcidWeak Base and Weak Acid

How do you know if the pH will How do you know if the pH will increase or decrease?increase or decrease?

If Kb = Ka, then If Kb = Ka, then pH = 0pH = 0 If Kb > Ka, then If Kb > Ka, then pH = +pH = + If Kb < Ka, then If Kb < Ka, then pH = -pH = -