Neutralization: Strong Acid-Strong Base

Chapter 15 - Applications of Aqueous Equilibria

Molecular: HCl(aq) + NaOH(aq)  NaCl(aq) + H2O(l)  SA-SB rxn goes to completion (one-way )

• Write ionic and net ionic rxns H+(aq) + Cl-(aq) + Na+(aq) + OH-(aq) H2O(l) + Na+(aq) + Cl-(aq)

GCC CHM152

• Net: H+(aq) + OH-(aq)  H2O(l)  After neutralization, what’s in solution?  Do either of the salt ions react with water?  What is pH of the salt solution?

Neutralization: Weak Acid-Strong Base

Neutralization: Strong Acid- Weak Base

HF (aq) + KOH (aq)  KF (aq) + H2O (l)  Rxn goes to completion due to SB Write ionic and net rxns:

HCl (aq) + NH3 (aq)  NH4+ (aq) + Cl- (aq)

HF(aq) + K+(aq) + OH-(aq)  K+(aq) + F-(aq) + H2O(l)

H+(aq) + Br-(aq) + NH3(aq)  NH4+(aq) + Br-(aq)

Net: HF(aq) + OH-(aq)  H2O(l) + F-(aq)

Net Ionic: H+(aq) + NH3(aq)  NH4+(aq)

 Rxn goes to completion because of strong acid Write ionic and net rxns:

 After neutralization, what’s in solution?

 After neutralization, what’s in solution?

 Do either of the salt ions react with water?

 Do either of the salt ions react with water?

 What will the pH of the solution be, roughly?

 What will the pH of the solution be, roughly?

Neutralization: Weak Acid-Weak Base

Neutralization Reactions

 Reaction does not go to completion since no reactant is completely ionized.

• Predict whether the pH after neutralization will be greater than, less than, or equal to 7 for the following combinations:

Ionic: CH3COOH(aq) + NH3(aq)  NH4+(aq) + CH3COO-(aq)

• HNO2 and KOH

Net: same

• HCl and LiOH

The salt remaining after neutralization contains an acidic and a basic ion, so pH depends on their relative Ka and Kb values.

• HBr and NH3

CH3COOH(aq) + NH3(aq)  NH4CH3COO(aq)

1

Common Ion effect The shift in equilibrium caused by the addition of a substance having an ion in common with the equilibrium mixture.  Adding a common ion suppresses the ionization of a weak acid or a weak base.

Common Ion Concept Problem Given this reaction: CH3CO2H + H2O  H3O+ + CH3CO2-

What happens to the pH of the acetic acid solution if we add NaCH3CO2? [CH3CO2-] 

The source of the common ion is typically provided by adding a strong acid, a strong base or a soluble salt to the equilibrium reaction mixture.

Eq shifts  [H3O+] , thus pH 

Common Ion Effect CH3CO2H + H2O  CH3CO2- + H3O+ 0.100 M CH3CO2H , Ka = 1.810-5

Common Ion Problem • What is the pH of 1.00 M HF solution? Ka = 7.0 x 10- 4

• WA soln: pH = 2.87 Add common ion CH3CO2- , pH 

• What is the pH of 1.00 M HF solution after adding 0.500 M NaF? Ka = 7.0  10-4

0.100 M CH3CO2H, add 0.050 M NaCH3CO2 • Mix of WA and conj base: pH = 4.44

16.3 Buffer Solution A solution that resists changes in pH when a small amount of acid or base is added. •Buffers are used to control pH e.g. biological buffers maintain the pH of all body fluids Best buffer systems consist of either a) a weak acid and its conjugate base e.g. HC2H3O2 and NaC2H3O2 b) a weak base and its conjugate acid e.g. NH3 and NH4Cl

Which are buffer solutions? • Identify the solutions below that would make good buffer solutions (what criteria need to be met?):       

HF and NaF NH3 and NH4Cl KOH and KF CH3COOH and LiCH3COO NaNO3 and HNO3 NaOH and NaCl HCl and NaCH3COO

2

Buffer Capacity

Buffer Range

Buffers only work within a pH range set by the value of Ka: pH = pKa ± 1 Thus 0.1 < [HA]/[A-] < 10 • Outside this range, we see little buffering effect. • If you know the desired pH of a buffer solution, you can choose an acid with a pKa near the pH. • Also the [acid]:[base] ratio must be less than 10:1 (or 1:10)  Optimum buffer has same amount of HA & A-.

• Select an appropriate acid-base pair to get within the range of the desired pH. • Look at table of Ka values (some are shown on the next screen), calculate pKa, and select an appropriate combination. • How would you make a buffer solution with pH = 4.10? What could you use to make this?

Table of Ka values

Buffer Solutions

•

Acid

Ka

HF HNO2 C9H8O4 (aspirin) HCO2H (formic) C6H8O6 (ascorbic) C6H5CO2H (benzoic) CH3CO2H (acetic) HCN C6H5OH (phenol)

3.5 x 10 –4 4.5 x 10 –4 3.0 x 10 –4 1.7 x 10 –4 8.0 x 10 –5 6.5 x 10 –5 1.8 x 10 –5 4.9 x 10 –10 1.3 x 10 –10

Buffer Solutions • Figure 15.3

• HF(aq) + H2O(l)  H3O+(aq) + F-(aq) • Add HCl, which component of the buffer soln will HCl reaction with (acid or base)?  H+ will react with F- (conj. base)

• Add NaOH, which component of the buffer soln will NaOH react with (acid or base)?  OH- will react with HF (weak acid)

• There will be a pH change in each case, but not as much as if the HCl (or NaOH) were added to water.

Henderson-Hasselbalch Equation • When solving for [H3O+] at equilibrium and assuming x > 1.30 x 10-5, so [Cl-]= 0.100 M

• What is the new [Ag+]? Note this is x in ICE!

Common Ion Effect What will happen if I add solid NaF to a solution of saturated SrF2? SrF2(s)  Sr2+(aq) + 2F-(aq) a) Calculate the molar solubility of SrF2 in pure water (Ksp = 4.3 x 10-9).

b)Calculate the molar solubility of SrF2 in 0.010 M NaF.

Common Ion Effect • • • •

SrF2(s)  Sr2+(aq) + 2F-(aq) (in water) Equil: x 2x Ksp = [Sr2+][F-]2 = 4.3 x 10-9 x = 1.0 x 10-3 M

• • • •

SrF2(s)  Sr2+(aq) + 2F-(aq) (in NaF) Initial: 0 0.010 Equil: x 0.010 + 2x Ksp = [Sr2+][F-]2 = 4.3 x 10-9

• x = 4.3 x 10-5 M

• Worked ex 15.11, Problem 15.25

Group Quiz #15 • Calculate the molar solubility of Cu(OH)2. Ksp = 1.6 x 10-19 • Calculate the molar solubility of Cu(OH)2 when 0.10 M Cu(NO3)2 is added.

Effect of pH on Solubility Addition of an acid can increase the solubility of an insoluble basic salt. E.g. CaF2(s)  Ca2+(aq) + 2F-(aq) Add strong acid (e.g. HCl) provides H3O+ ions: H3O+ + F-  HF + H2O  Adding H+ causes [F-] , equilibrium shifts right to form more F- & solubility of CaF2.

9

pH and solubility • E.g. For the following salts, predict whether the salt will dissolve in an acidic solution. • A. AgBr _____________ • B. CdCO3 _____________ • C. PbCl2 _____________ • D. BaS _____________

Precipitation of Ionic Cmpds • Worked Example 15.14: Will a precipitate form when 0.150 L of 0.10 M lead (II) nitrate and 0.100 L of 0.20 M sodium chloride are mixed? • What is the precipitate that will form? What are Ksp and the equilibrium expression? • Find total concentration of each ion once mixed. • Problem 15.29

Precipitation of Ionic Cmpds • • • •

Q > Ksp Supersaturated; ppt forms Q = Ksp Saturated Q < Ksp Unsaturated; no ppt forms 2.45 mg of magnesium carbonate is placed in 1.00 L of water. Will it all dissolve? Ksp =

Precipitation of Ionic Cmpds • Note: book refers to IP, same idea as Q (value of concentration ratios NOT necessarily at equilibrium) • Compare Q and K. System will shift until Qsp = Ksp (a saturated solution) • Example: We want to know if precipitate will form when 0.150 L of 0.10 M lead (II) nitrate and 0.100 L of 0.20 M sodium chloride are mixed? • Find Qsp and relate it to Ksp.

Precipitation • PbCl2(s)  Pb2+(aq) + 2Cl-(aq) • Calculate moles of each ion. • Convert moles to concentration using total volume. • Plug concentrations into Qsp equation. • If Qsp > Ksp, rxn shifts left (toward solid) and precipitate will form. • If Qsp < Ksp, rxn shifts right (toward ions) and precipitate will not form.

Group quiz 16 Will a solution containing 4.0 x 10-5 M Cland 2.0 x 10-4 M Ag+ form a precipitate of AgCl? Ksp = 1.70 x 10-10 (Hint: think Q)

4.0 x 10-5

• Find molarity of solid. Write equilibrium expression for solubility. Calculate Qsp.

10

Complex Ion Equilibria Formation of complex ions can increase solubility of an insoluble salt Complex ion: ion containing a metal cation bonded to one or more molecules or ions which are called ligands. The ligands act as Lewis bases (e.g. NH3, H2O, OH-) • E.g. Al(H2O)63+, CuBr4-

Complex Ion Equilibria • Lewis acid-base reactions also reach a state of equilibrium • Metal ion + ligand  complex ion • Kf = [complex ion]/[metal ion][ligand] • Kf = formation constant • Hg2+ + 4I-  HgI42• Kf = [HgI42-]/[Hg2+][I-]4

• Kf values are usually pretty large

Adding NH3 to AgCl solution increases the solubility of AgCl: 1) AgCl(s)  Ag+(aq) + Cl-(aq) K1 = 1.7 x10-10 + + 2) Ag (aq) + 2NH3(aq)  Ag(NH3)2 (aq) K2 = 1.6 x 107 3) AgCl(s) + 2NH3(aq)  Ag(NH3)2+(aq) + Cl-(aq)

Knet = K1 x K2 = 2.7 x 10-3 •In qual lab we add ammonia to dissolve our AgCl precipitate so it can be separated from Hg.

Qualitative Analysis • Skipping section 15.14: Selective Precipitation • 15.15 Qualitative analysis is used to identify unknown ions in a solution.  152LL should read this section before qual lab!

• Each ion can be precipitated out by addition of selective reagents. • Purely qualitative research, like solving a puzzle.

11