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Buffer Solutions and Equilibria: Weak Acids and Bases, Study notes of Qualitative research

University of South Carolina (USC) - Union Qualitative research

The concepts of buffer solutions, their role in resisting pH changes, and the equilibria of weak acids and bases. Topics include the reaction of salts with water, the role of common ions, and the calculation of pH changes when acids or bases are added to buffer solutions.

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Chapter 15 - Applications of

Aqueous Equilibria

GCC CHM152

Neutralization: Strong Acid-Strong Base

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)

•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

HF (aq) + KOH (aq)  KF (aq) + H2O (l)

Rxn goes to completion due to SB

Write ionic and net rxns:

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

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

After neutralization, what’s in solution?

Do either of the salt ions react with water?

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

Neutralization: Strong Acid- Weak Base

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

Rxn goes to completion because of strong acid

Write ionic and net rxns:

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

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

After neutralization, what’s in solution?

Do either of the salt ions react with water?

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

Neutralization: Weak Acid-Weak Base

CH3COOH(aq) + NH3(aq)  NH4CH3COO(aq)

Reaction does not go to completion since no

reactant is completely ionized.

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

Net: same

The salt remaining after neutralization contains an

acidic and a basic ion, so pH depends on their relative

Ka and Kb values.

Neutralization Reactions

•Predict whether the pH after neutralization

will be greater than, less than, or equal to

7 for the following combinations:

•HNO2 and KOH

•HCl and LiOH

•HBr and NH3

Partial preview of the text

Download Buffer Solutions and Equilibria: Weak Acids and Bases and more Study notes Qualitative research in PDF only on Docsity!

Chapter 15 - Applications of

Aqueous Equilibria

GCC CHM

Neutralization: Strong Acid-Strong Base Molecular: HCl(aq) + NaOH(aq)  NaCl(aq) + H 2 O(l)  SA-SB rxn goes to completion (one-way )

Write ionic and net ionic rxns H+(aq) + Cl-(aq) + Na+(aq) + OH-(aq) H 2 O(l) + Na+(aq) + Cl-(aq)
Net: H+(aq) + OH-(aq)  H 2 O(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 HF (aq) + KOH (aq)  KF (aq) + H 2 O (l)  Rxn goes to completion due to SB Write ionic and net rxns: HF(aq) + K+(aq) + OH-(aq)  K+(aq) + F-(aq) + H 2 O(l) Net: HF(aq) + OH-(aq)  H 2 O(l) + F-(aq)  After neutralization, what’s in solution?  Do either of the salt ions react with water?  What will the pH of the solution be, roughly? Neutralization: Strong Acid- Weak Base HCl (^) (aq) + NH 3 (aq)  NH 4 +^ (aq) + Cl-^ (aq)  Rxn goes to completion because of strong acid Write ionic and net rxns: H+(aq) + Br-(aq) + NH 3 (aq)  NH 4 +(aq) + Br-(aq) Net Ionic: H+(aq) + NH 3 (aq)  NH 4 +(aq)  After neutralization, what’s in solution?  Do either of the salt ions react with water?  What will the pH of the solution be, roughly? Neutralization: Weak Acid-Weak Base CH 3 COOH(aq) + NH 3 (aq)  NH 4 CH 3 COO(aq)  Reaction does not go to completion since no reactant is completely ionized. Ionic: CH 3 COOH(aq) + NH 3 (aq)  NH 4 +(aq) + CH 3 COO-(aq) Net: same The salt remaining after neutralization contains an acidic and a basic ion, so pH depends on their relative Ka and Kb values.

Neutralization Reactions

Predict whether the pH after neutralization

will be greater than, less than, or equal to

7 for the following combinations:

HNO 2 and KOH
HCl and LiOH
HBr and NH 3

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.

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. Common Ion Concept Problem Given this reaction: CH 3 CO 2 H + H 2 O  H 3 O+^ + CH 3 CO 2 - What happens to the pH of the acetic acid solution if we add NaCH 3 CO 2? [CH 3 CO 2 - ]  Eq shifts  [H 3 O+]  , thus pH 

Common Ion Effect

CH 3 CO 2 H + H 2 O  CH 3 CO 2 -^ + H 3 O+

0.100 M CH 3 CO 2 H , Ka = 1.8 10 -^5

WA soln: pH = 2. Add common ion CH 3 CO 2 -^ , pH  0.100 M CH 3 CO 2 H, add 0.050 M NaCH 3 CO 2
Mix of WA and conj base: pH = 4.

Common Ion Problem

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

16.3 Buffer Solution

Best buffer systems consist of either a) a weak acid and its conjugate base e.g. HC 2 H 3 O 2 and NaC 2 H 3 O 2 b) a weak base and its conjugate acid e.g. NH 3 and NH 4 Cl 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

Which are buffer solutions?

Identify the solutions below that would make good buffer solutions (what criteria need to be met?):  HF and NaF  NH 3 and NH 4 Cl  KOH and KF  CH 3 COOH and LiCH 3 COO  NaNO 3 and HNO 3  NaOH and NaCl  HCl and NaCH 3 COO

Buffer Solutions + HCl

What is the pH of 500.0 mL of 0.10 M

formic acid combined with 500.0 mL of 0.

M sodium formate? Ka = 1.8 x 10-^4

pH = pKa + log ([A-]/[HA]) = 4.
Now add 10.0 mL of 0.50 M HCl  calculate moles of substances, then calculate changes  Use a CHANGE Table (initial, change, final)

Buffer Solutions + HCl

To find the new pH, first assume that all the added strong acid reacts with A-^ to form HA.  H 3 O+^ + A-^  H 2 O + HA  How many moles of HA, A-, and H 3 O+^ are in solution?
Initial mol HA = (0.10 M) (0.5000 L) = 0.050 mol
Initial mol A-^ = (0.20 M) (0.5000 L) = 0.10 mol
mol HCl = (0.0100 L) (0.50 M) = 0.0050 mol
Final mol A-^ = 0.10 mol – 0.0050 mol = 0.095 mol A-
Final mol HA = 0.050 mol + 0.0050 mol = 0.055 mol HA  Can use final moles in H-H equation. Why?
pH = pKa + log ([A-]/[HA]) = 3.98 (initially 4.04 w/o acid)
pH  slightly when SA added to buffer

Buffer +NaOH

What will happen when we add a strong

base to a buffer system?

 Added base reacts with HA to form A- :  HA and A-

 HA + OH-^  H 2 O + A-

Add 10.0 mL of 0.500 M NaOH to the

formic acid/sodium formate buffer. What

is the new pH?

Buffer Solutions + NaOH

First assume that all OH-^ reacts with HCO 2 H:
HCO 2 H + OH-^  H 2 O + HCO 2 -
How many moles of base were added?
mol OH-^ = (0.500 M) (0.0100 L) = 0.00500 mol OH-
final mol HA = 0.050 mol - 0.00500 mol = 0.045 mol
final mol A-^ = 0.10 mol + 0.00500 mol = 0.105 mol
pH = pKa + log ([A-]/[HA]) = 4.11 (initially 4.04 w/o SB)
pH slightly when SB added
In water, pH would change from 7 to 13 if same amount of SB was added.

Practice Ex. Try after class

1) Calculate the pH of a 1.25 L buffer

solution made up of 0.50 M acetic acid and

0.50 M sodium acetate.Ka = 1.8 x 10-^5

2) What is the pH of the solution when 25.

mL of 0.40 M NaOH is added to the buffer?

pH Titration Curves

Strong Acid-Strong Base Titrations:
The equivalence point is the point at which equimolar amounts of acid and base have reacted (review 1st semester notes on calcs).
Graph generated with pH probe 
Figure 15.

pH Titration Curves

4 key points on a titration curve  At the beginning (before titrant is added)  Before the equivalence point  At the equivalence point  After the equivalence point
Consider titrating 20.00 mL of 0.200 M HCl

with 0.100 M NaOH. Find equiv. pt. first!

M =

mol

L

1000 mmol

1000 mL

mmol

mL

Strong Acid-Strong Base

Draw 4 beakers.
1. Draw 2 moles HCl.
1. Draw what happens when 1 mole

NaOH is added.

1. Draw what happens when 2 moles

NaOH are added.

1. Draw what happens when 3 moles

NaOH are added.

HCl + NaOH Titration

1. 0 mL NaOH, pH is calculated from strong acid. [H 3 O+] = 0.200 M  pH = - log (0.200) = 0.700 initial pH of SA
1. 5.00 mL NaOH, acid is still in excess; but by how much?  mmol acid – mmol base  (0.200 M)(20.00 mL) – (0.100 M)(5.00 mL)  [H 3 O+] = mmol / total solution volume  [H 3 O+] = 3.50 mmol / 25.00 mL = 0.140 M  pH = - log (0.140) = 0.854 excess SA so pH still quite low!

HCl + NaOH Titration

1. 40.00 mL NaOH added; at the equivalence point, neutral salt and water in solution.  mmol acid = mmol base = 4.00 mmol  pH = 7 (always 7 for strong acid-strong base!)
1. 50.00 mL NaOH added, base in excess. Calc moles (or mmol) of excess base remaining  mmol base – mmol acid  (0.100 M)(50.00 mL) – (0.200 M)(20.00 mL)  1.00 mmol base excess / 70.00 mL = 0.01429 M NaOH  pOH = - log (0.01429) = 1.845, pH = 12.155 excess SB
Problems 15.13, 15.

Weak Acid-Strong Base

Acetic acid in flask, KOH is titrant in buret
Find pH at 4 pts.
Figure 15.8 shows strong and weak acids.
Titrate 50.00 mL of 0.100 M acetic w/ 0.150 M KOH.  Find eq. pt., Ka, pKa, and mmols of acid first

Weak Acid-Strong Base

Draw 4 beakers.
1. Draw 2 moles acetic acid.
1. Draw what happens when 1 mole KOH

is added.

1. Draw what happens when 2 moles KOH

are added.

1. Draw what happens when 3 moles KOH

are added.

Weak Acid-Strong Base

After the equivalence point:  50.00 mL of 0.100 M acid (5.00 mmol) and 50.00 mL of 0.150 M base (7.50 mmol)  7.50 mmol OH-^ added – 5.00 mmol acid present = 2.50 mmol OH-^ remaining  total mL = 50.00 + 50.00 = 100.00 mL  [OH-]= 2.5 mmol/100.00 mL = 0.0250 M OH-  pOH = - log (0.0250 M) = 1.  pH = 14 – 1.602 = 12.

Group Quiz

Exactly 50.00 mL of 0.20 M hydrazoic acid (Ka = 1.9 x 10-^5 ) are titrated with a 0.15 M KOH solution. Calculate the pH when 66.67 mL of base have been added:

Animations:

http://www.chembio.uoguelph.ca/educmat/chm19104/chemtoons/chemtoons.htm

Half Equivalence Point

33.34 mL base added (half the volume needed to get to eq pt for quiz #14 problem)
At this point half the acid has been neutralized mmol substances at half eq pt: mmol HA = (10.00 mmol – 5.00 mmol) = 5.00 mmol
mmol A-^ = 5.00 mmol
Thus, [HA] = [A-]
pH = pKa + log ([A-]/[HA]) = pKa

Weak Base-Strong Acid

Figure 15.
Calculations are similar to weak acid/strong base; opposite values; extra steps to find pH values. See pages 613 - 615 for outline.

Acid-Base Indicators Polyprotic Acid Titrations

One eq. pt.

for each

proton

2 eq. pts. and

2 half eq. pts.

Figure 15.11 -

Alanine with

NaOH

Solubility equilibria

Dissolution of slightly soluble salts in water

important biological examples:

 tooth decay - tooth enamel dissolves in acidic soln  formation of kidney stones - salts precipitate in kidney

15.10 Solubility Equilibria

Many ions combine to form solid precipitates

in aqueous solutions. (Solubility rules)

These “insoluble” salts dissolve to a small

extent and form a saturated solution.

The undissolved solid and the dissociated

ions in solution establish an equilibrium

system that is characterized by the solubility

product constant, Ksp.

AgBr(s)  Ag+(aq) + Br - (aq)

Ksp = [Ag+][Br - ]

Ksp is the solubility product constant - the equilibrium constant for insoluble salts

It is a measure of how soluble a salt is in H 2 O For salts with the same # of ions, the smaller the Ksp, the less soluble the salt. Table 15.2 shows some Ksp values Example. The dissolution of AgBr Write the solubility equilibrium^ reactions^ and Ksp expressions for a) MgF 2 b) Ca 3 (PO 4 ) 2.

A) MgF 2 (s)  Mg2+(aq) + 2F-(aq)

B) Ca 3 (PO 4 ) 2 ( s )  3Ca^2 +^ (aq) + 2PO 43 -^ (aq)

Ksp = [Ca2+]^3 [PO 43 - ]^2

Ksp = [Mg2+][F-]^2

Solubility

Ksp can be used to calculate the solubility , which can be compared for different salts.
Solubility (molar solubility) = molar concentration of dissolved salt; ion concentrations are related to this by their coefficients.
Solubility can also be expressed in g/L
Examples:
AgCl: [Ag+] = x [Cl-] = x Ksp = (x)(x)
Ag 2 S: [Ag+] = 2x [S^2 - ] = x Ksp = (2x)^2 (x)
Fe(OH) 3 : [Fe^3 +] = x [OH-] = 3x Ksp = (x)(3x)^3

Solubility/Ksp Practice

Ksp for AgBr is 7.7 x 10-^13. Calculate the molar and gram solubility (solve for x).
If a saturated solution prepared by dissolving CaF 2 in water has [Ca2+] = 3.3 10 -^4 M, what is the value of Ksp?
Ksp for Cu 2 O is 2.0 x 10-^15. Calculate the molar solubility.
The solubility of Ca(OH) 2 is 0.233 g/L. Calculate Ksp.
Worked example 15.8 – 15.10, Problems 15.21 –

pH and solubility

E.g. For the following salts, predict whether

the salt will dissolve in an acidic solution.

A. AgBr _____________
B. CdCO 3 _____________
C. PbCl 2 _____________
D. BaS _____________

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 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.

Precipitation

PbCl 2 (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.

Precipitation of Ionic Cmpds

Q > K sp Supersaturated; ppt forms
Q = K sp Saturated
Q < K sp Unsaturated; no ppt forms
2.45 mg of magnesium carbonate is placed

in 1.00 L of water. Will it all dissolve? Ksp =

4.0 x 10-^5

Find molarity of solid. Write equilibrium

expression for solubility. Calculate Qsp.

Group quiz 16

Will a solution containing 4.0 x 10 -^5 M Cl-

and 2.0 x 10-^4 M Ag+^ form a precipitate

of AgCl? Ksp = 1.70 x 10-^10

(Hint: think Q)

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. NH 3 , H 2 O, OH-)

E.g. Al(H 2 O) 6 3+, CuBr 4 -

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
Hg^2 +^ + 4I-^  HgI 42 -
Kf = [HgI 42 - ]/[Hg^2 +][I-]^4
Kf values are usually pretty large **Adding NH 3 to AgCl solution increases the solubility of AgCl:

AgCl(s)**  Ag+(aq) + Cl-(aq) K 1 = 1.7 x10-^10 2) Ag+(aq) + 2NH 3 (aq)  Ag(NH 3 ) 2 +(aq) K 2 = 1.6 x 10^7 3) AgCl(s) + 2NH 3 (aq)  Ag(NH 3 ) 2 + (aq) + Cl-(aq )

Knet = K 1 x K 2 = 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

Buffer Solutions and Equilibria: Weak Acids and Bases, Study notes of Qualitative research

Related documents

Partial preview of the text

Download Buffer Solutions and Equilibria: Weak Acids and Bases and more Study notes Qualitative research in PDF only on Docsity!

Chapter 15 - Applications of

Aqueous Equilibria

GCC CHM

Neutralization Reactions

will be greater than, less than, or equal to

7 for the following combinations:

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 Effect

CH 3 CO 2 H + H 2 O  CH 3 CO 2 -^ + H 3 O+

Common Ion Problem

16.3 Buffer Solution

Which are buffer solutions?

Buffer Solutions + HCl

formic acid combined with 500.0 mL of 0.

M sodium formate? Ka = 1.8 x 10-^4

Buffer Solutions + HCl

Buffer +NaOH

base to a buffer system?

 HA + OH-^  H 2 O + A-

formic acid/sodium formate buffer. What

is the new pH?

Buffer Solutions + NaOH

Practice Ex. Try after class

1) Calculate the pH of a 1.25 L buffer

solution made up of 0.50 M acetic acid and

0.50 M sodium acetate.Ka = 1.8 x 10-^5

2) What is the pH of the solution when 25.

mL of 0.40 M NaOH is added to the buffer?

pH Titration Curves

pH Titration Curves

with 0.100 M NaOH. Find equiv. pt. first!

M =

mol

L

1000 mmol

1000 mL

mmol

mL

Strong Acid-Strong Base

NaOH is added.

NaOH are added.

NaOH are added.

HCl + NaOH Titration

HCl + NaOH Titration

Weak Acid-Strong Base

Weak Acid-Strong Base

is added.

are added.

are added.

Weak Acid-Strong Base

Group Quiz

Animations:

Half Equivalence Point

Weak Base-Strong Acid

Acid-Base Indicators Polyprotic Acid Titrations

for each

proton

2 half eq. pts.

Alanine with

NaOH

Solubility equilibria

important biological examples:

15.10 Solubility Equilibria

Many ions combine to form solid precipitates

in aqueous solutions. (Solubility rules)

These “insoluble” salts dissolve to a small

extent and form a saturated solution.

The undissolved solid and the dissociated

ions in solution establish an equilibrium

system that is characterized by the solubility

product constant, Ksp.

AgBr(s)  Ag+(aq) + Br - (aq)