StuffHour Exam 2 Date: Monday Feb 18 Room: Schmitt Hall C114Time: 6-7:30PM

Strategy: Do Homework, Sample Problems and Practice Multiple Choice Quizzes in the Links of my blog.

Chapter 19: 632-635 (omit calculating effect of H+ on Buffer pH 635-636) 637-644 read and know.

Make-Up Exam ComprehensiveDate: Senior’s Final Week

2

Ionic Equilibria in Aqueous Systems

1. The Common-Ion Effect in Acid-Base Equilibria

2. Buffers

3. Acid-Base Indicators

4. Neutralization Reactions and Titration Curves

5. Ksp or “equilbrium for precipitation reactions”

EXAM 2 2012-13

EXAM 3 2012-13

The common ion effect is a LeChatlier equilibrium shift that occurs upon addition of a an ion common to an acid or its conjugate base.

Suppose a weak acid equilibrium in water:

Now add sodium acetate salt--which “COMMON” to the acetate produced by the acid dissociation.

What will happen to the acid equilibrium?

add H+ or add CH3COO-Na(s)

Shifts the equilibrium to reactants

CH3COOH (aq) H+ (aq) + CH3COO- (aq)

The presence of a common ion suppresses the ionization of a weak acid or a weak base.

Common Ion Effect (Strong acid + weak acid)• Consider a solution with 0.100M CH3CO2H and 0.100 M

HCl. CH3CO2H has Ka = 1.8 X 10-5

Initial AcAc(M)

Change (M)Equilibrium (M)

0.100 0.00

-x +x

0.100 - x

0.100+x

x 0.100 + x

CH3CO2H + H2O <===> CH3CO2- + H3O+

Initial HCl(M)0.00

0.00 -

Ka = 1.8 X 10-5 = x (0.100 + x)0.100 - x

CH3CO2H(aq) + H2O <=> CH3CO2-(aq) + H3O+(aq)

HCl(g) + H2O <=> Cl-(aq) + H3O+(aq) Strong acid

Weak acid

Simplfy? 100 x Ka << 0.1

x (0.100)0.100 = 1.8 X 10-5Ka =

Analysis of the Common Ion Effect

X = 1.8 X 10-5

0

x

Initial (M)Change (M)Equilibrium (M)

0.100 0.00-x +x

0.100 - x x

CH3CO2H + H2O <==> CH3CO2- + H3O+

x = [H+] = [CH3COO-] = 1.34 X 10-3

Can we simplify?

100 x Ka << 0.1, so we can drop x

x2 = 1.8 X 10-5 X 0.100

Ka =x2

0.100 - x= 1.8 X 10-5.

Ka =x2

0.100= 1.8 X 10-5.

+x

Same Problem---But No HCl Added

Comparison with (common ion) and without HCl

x = 1.34 X 10-3 x = 1.8 X 10-5

Weak Acid + Strong Acid

CH3CO2H + H2O <==> CH3CO2- + H3O+

--the addition of the strong acid or conjugate base to a weak acid suppresses the ionization of the

weak acid

x xSingle Solute ResultWeak Acid Only

IT DOES NOT CHANGE Ka!

A buffer is a an aqueous solution made from from a weak acid-conjugate base pair that will resist pH change upon the addition of H+ or OH-.

equal concentrations of weak acid and its conjugate base

Add OH-

Add H3O+

after addition of H3O+

upon addition of OH-

CH3CO2- + H3O+ <==> CH3CO2H + H2O

CH3CO2H + OH- <==> CH3CO2- + H2O

after addition of OH-

after addition of H3O+

The common ion effect makes buffers possible

• Buffers are aqueous solutions composed of a weak conjugate acid/base pair. – the combination of a weak acid + conjugate-base pair is

able to resist a change in pH upon addition of another acid (H+) or base (OH-) or both.

– The correct “buffer combination” is an ample concentration of :

• A weak acid and it’s conjugate base• OR a weak base and it’s conjugate acid

If you can recognize a weak acid + conjugate base pairs then you can understand buffers!

pH = pKa + log [base][acid]pH = pKa + log

[A-][HA]

HA(aq) + H2O(l) H3O+(aq) + A- (aq) Ka =[H3O+][A-]

[HA]

The A- and HA are concentrations of the weak acid-conjugate base. It’s a dilution/stoich problem.

acid baseconjugatebase

conjugateacid

Acid Formula Ka Conjugate Kb

Iodic acid HI 1.61 X 10-1 I- 6.1 X 10-14

Chlorous Acid HClO2 1.1 X 10-2 ClO2- 9.1 X 10-13

Phosphoric Acid H3PO4 7.6 X 10-3 H2PO4- 1.3 X 10-12

Nitrous Acid HNO2 7.2 X 10-4 NO2- 1.4 X 10-11

Hydrofluoric Acid HF 6.6 X 10-4 F- 1.5 X 10-11

Formic Acid HCOOH 1.8 X 10-4 HCO2- 5.6 X 10-11

Acetic Acid CH3CO2H 1.8 X 10-5 CH3CO2 - 5.6 X 10-10

Hydrosulfuric Acid H2S 1.32 X 10-7 HS- 7.6 X 10-8

Hypochlorous Acid HClO 3.5 X 10-8 ClO- 2.9 X 10-7

Hydrazine N2H4 1.1 X 10-8 OH- 8.9 X 10-7

Water H2O 1.0 X 10-7 OH- 1.0 X 10-7

Ammonia NH3 5.6 X 10-10 NH4+ 1.8 X 10-5

Hydrocyanic Acid HCN 4.0 X 10-10 CN- 2.5 X 10-5

Dimethyl Amine (CH3)2NH 1.9 X 10-11 (CH3)2NH2+ 5.4 X 10-4

Ka and Kb of Some Common Weak Acids and BaseAcid Formula Ka pKa Conjugate

Iodic acid HI 1.61 X 10-1 0.79 I-

Chlorous Acid HClO2 1.1 X 10-2 1.96 ClO2-

Phosphoric Acid H3PO4 7.6 X 10-3 2.1 H2PO4-

Nitrous Acid HNO2 7.2 X 10-4 3.1 NO2-

Hydrofluoric Acid HF 6.6 X 10-4 3.2 F-

Formic Acid HCOOH 1.8 X 10-4 3.7 HCO2-

Acetic Acid CH3CO2H 1.8 X 10-5 3.7 CH3CO2 -

Hydrosulfuric Acid H2S 1.32 X 10-7 6.87 HS-

Hypochlorous Acid HClO 3.5 X 10-8 7.5 ClO-

Hydrazine N2H4 1.1 X 10-8 7.9 OH-

Water H2O 1.0 X 10-7 7.0 OH-

Ammonia NH3 5.6 X 10-10 9.3 NH4+

Hydrocyanic Acid HCN 4.0 X 10-10 9.4 CN-

Dimethyl Amine (CH3)2NH 1.9 X 10-11 10.7 (CH3)2NH2+

Ka and Kb of Some Common Weak Acids and Base

(a) KF is a weak acid and F- is its conjugate base. It’s a buffer solution. (b) HBr is a strong acid, KBr is the salt of a strong base and strong acid. It can not be a buffer solution.(c) CO3

2- is the conjugate base of HCO3-. HCO3- is it

conjugate acid. It is a buffer solution.(d) HPO4

2- is a conjugate base of H3PO4. It is a conjugate acid--base pair and a buffer.

Which of the following are buffer systems? (a) KF/HF (b) KBr/HBr, (c) Na2CO3/NaHCO3 (d) Na2HPO4/H3PO4

To be a buffer we need a weak acid-conjugate base pair or the same combination generated in-situ.

(e) CH3COO- is conjugate base of CH3COOH. The HCl will generate the acid. It is a buffer.

The “Buffer Equation” (Henderson-Hasslebach) is key to making and understanding buffers.

pH = pKa + log [base][acid]

Know this equation and how to use it!

pH of the buffer

pKa of the acid

Molarity of conjugate base and the acid

pKa = −log(Ka)

pKa = −log

�Kw

Kb

�

The “Buffer Equation” (Henderson-Hasselbach) is derived from the Ka expression for a weak acid.

pH = pKa + log [base][acid]

Ka =[H3O+][A-]

[HA]

[H+] =Ka [HA]

[A-]

-log [H+] = -log Ka - log [HA][A-]

-log [H+] = -log Ka + log [A-][HA]

pH = pKa + log[A-][HA]

HA(aq) + H2O(l) H3O+(aq) + A- (aq)

Solve for [H+]

Take -log of both sides and distribute

Move the minus sign into the log

standard equilibriumfor HA

Ka for acid HA

OR

We can also prepare alkaline buffer solutions by mixing a weak base with its conjugate acid just like we do with an acid-conjugate base buffer system.

B + H2O <==> BH+ + OH-

Kb =[BH+][OH-]

[B]

[OH-] =Kb [B][BH+]

-log [OH-] = -log Kb - log [B][BH+]

-log [H+] = -log Kb + log [BH+][B]

pOH = pKb + log [conjugate acid][base]

pOH = pKb + log[BH+][B]

Solve for [OH-]

Take -log of both sides

Move the minus sign into the log

standard equilibriumfor base B

Kb for base, B

What is the pH of a solution containing 0.30 M HCOOH (pKa = 3.77) and 0.52 M HCOOK (potassium formate)?

pH = pKa + log [conjugate base][acid]

pH = 3.77 + log [0.52][0.30]

pH = 3.77 + log [0.52][0.30]

pH = 3.77 +0.2388 = 4.00

Buffer Capacity and Range and How To• Buffer range is the pH range over which a buffer works.– Buffer range is +/- 1 units from the pKa of the acid.

• Buffer capacity is the amount of “contaminate” acid and base a buffer can neutralize before the buffer solution pH changes.

– [HA] scavenges the addition of any added base (see equation below).– [A-] reacts with any added acid that spills into the buffer. – [HA] and [A-] are normally set up to handle equal amounts of acid or

base if they contaminate a buffer. – The higher the concentration of [HA] and [A] the larger the volumes

of acid/base they can cope with. upon addition

of OH-

CH3CO2- + H3O+ <==> CH3CO2H + H2O

CH3CO2H + OH- <==> CH3CO2- + H2O

after addition of H3O+

Steps In Preparing A Buffer:

pH = pKa + log [base][acid]

1. Answer: Determine the buffer pH you want to prepare?

2. Once the pH is chosen, then select a weak acid/conjugate base pair that has a pKa of +/- 1 of the pH you wish to buffer the pH. Look up the pKa of candidate weak acids to examine which acid-conjugate base pair to use. 3. Use the buffer equation to determine the buffer concentration: [HA] + [A-]. Optimal buffer on both sides of pH is when ratio of A:HA is 1:1

4. Dilute the solution and adjust the pH.

The same buffer can be made in different ways. The most practical method in the lab is encircled.

Weak acid + strong base

Weak acid + salt of acid

Salt of acid + strong acid

Weak base + strong acid

Weak base + salt of base

Salt of base + strong base

Kb =[NH4+][OH- ]

[NH3 ]= 1.8 X 10-5

WORKING pH: +/- 1 pKa = 1 + 9.26 = 10.3 and 9.26 - 1 = 8.3 OR works well over pH 8.3-10.3!

--Buffers work best when the concentrations of the acid and conjugate base are large and equal!

NH3(aq) + H2O <===> NH4+(aq) + OH-(aq)

Show using a chemical equation or use LeChatlier’s principle and explain why a solution of aqueous NH3 NH4Cl constitutes a buffer solution. Kb of NH3 is 1.8 X 10-5. Over what pH range should this buffer work?

pKb = -log(1.8 x 10-5) = 4.74, pKa = 14.00 - 4.74 = 9.26

You wish to prepare a 1.0L buffer with a pH of 3.90. A list of possible acids and their conjugate bases is shown below. Which combination should be selected? Suppose the total concentration of the buffer ([HA] + [A-]) is 0.5M. What is the concentration of each component?

2. Use the buffer equation and determine the ratio of base/acid. Rick will choose the acetate buffer (you could choose HF/F).

pH = pKa + log[HCOO-]initial

[HCOOH]initial

3.90 = 4.74 + log [HCOO-]initial

[HCOOH]initial-0.84 = log [HCOO-]initial

[HCOOH]initial

Acid Conjugate Base Ka pKa

HSO4- SO4- 1.2 x 10-2 1.92HCO2H HCO2- 1.8 x 10-5 4.74HCO3- CO3-2 4.8 x 10-11 10.32HF F- 6.6 x 10-4 3.18

1. Choose the buffer using pKa of the acid (+/- 1 unit away). CH3COOH -CH3COO- or HF/F are the only choice that fits this criteria.

Suppose the total concentration of the buffer is 0.5M. What is the concentration of each component?

total concentration = 0.5M = [HCOOH] + [HCOO-]

[HCOO-] = .144[HCOOH]

0.5M = [HCOOH] + 0.144[HCOOH]

0.5M = 1.144[HCOOH]

[HCOOH] = .5M/1.144 = 0.437M

[HCOO-] = .5 - 0.437 = 0.063M

10-0.84 = [HCOO-]initial

[HCOOH]initial= .144 HCO3

-(aq) + H2O(l) CO32-(aq) + H3O+(aq)

Identify the equilibrium and the acid-conjugate base pair and Ka

pH = pKa + log [conjugate base][acid]

pKaHCO3 = - log(KaHCO3) = - log(4.7 x 10-11) = 10.33

An environmental chemist requires a carbonate buffer of pH 10.00. How many grams of Na2CO3 must she add to 1.5 L of freshly prepared 0.20 M NaHCO3 to prepare the buffer? Ka of HCO3

- is 4.7 x 10-11.

CO32-

0.2010-0.33 =

10 = 10.33 + log [HCO3-][CO32-]

[HCO3-][CO32-]-0.33 = log

[CO32-] = 0.20 X 10-0.33 = 0.0935 M

mol CO32- = (1.5 L)(0.0935 mol CO32- /L) = 0.1402 moles CO32-

0.14 mole Na2CO3 X 105.99 gmol Na2CO3

= 15 g Na2CO3

[CO32-] = 0.20 X 10-0.33 = 0.0935 M [CO32-]

It’s now a molarity and mole problem from Chem 7. The problem said we need: 1.5L of 0.0935M CO32- .

mol Na2CO3 = 0.140 mol CO32- X = 0.14 mol Na2CO32-1 mol of CO32-1 mol Na2CO3

An environmental chemist needs a carbonate buffer of pH 10.00. How many grams of Na2CO3 must she add to 1.5 L of freshly prepared 0.20 M NaHCO3 to prepare the buffer? Ka of HCO3

- is 4.7 x 10-11.

What is the pH of a solution containing 0.30 M HCOOH (pKa = 3.77) and 0.52 M HCOOK (potassium formate)?

pH = pKa + log [conjugate base][acid]

pKa = 3.77

pH = 3.77 + log [0.52][0.30]

= 4.01

We can solve this using (1) buffer equation or (2) using an ice table and common ion effect

(1) Solve using the buffer equation

% ionization of an acid is the ratio of [H+] at equilibrium to [HA]initial X 100.

[H+]equilbrium

[HA]initalx 100%% ionization =

[OH-]equilbrium

[OH-]initalx 100%% ionizationb =

Calculate the percent ionization of a acetate buffer made up from .6M CH3COOH acetic acid and 0.1M CH3COOK. The Ka of acetic acid is 1.8 x 10-5

[H+]equilbrium

[HA]initalx 100%% ionization =

pH = pKa + log[A−]

[HA]Connect the dots!

pKa = -log Ka = -log (1.8 x 10-5) = 4.744

pH = 4.744 + log[0.1M ]

[0.6M ]= 3.965

[H+] = 10

−3.965= 1.08× 10

−4

% ionization =[H

+]

[HA]× 100 =

1.08× 10−4

0.6× 100 = 1.8× 10

−2

from data in problem

Buffer equation

[H+] concentration

A) Calculate the pH of the 0.30 M NH3/0.36 M NH4Cl buffer system. The pKb of NH3 is 4.75.

NH4+ (aq) + H2O H+ (aq) + NH3 (aq)

pH = pKa + log [NH3][NH4

+]

pKb = 4.75

pH = 9.25 + log [0.30][0.36]

= 9.17

pH = pKa + log [conjugate base][acid]

NH3 + H2O NH4+ (aq) + OH- (aq)

pKa = ?

pKw = pKa + pKb = 14

pKa = 14.00 - pKb pKa = 14.00 - 4.75 = 9.25

Calculating the impact of Added H3O+ on the pH

of a buffer solution (Sample Problem 19.1)

MAJORS ONLY

NON-MAJORS TO TITRATIONS

Calculating the effect of added H3O+ and OH- on buffer pH

Calculate the pH of the following solutions given that Ka of CH3COOH = 1.8 x 10-5 (assume the additions cause negligible volume changes).

(a) A buffer solution consisting of 0.50 M CH3COOH and 0.50 M CH3COONa

(b) After adding 0.020 mol of solid NaOH to 1.0 L of the buffer solution in part (a)(c) After adding 0.020 mol of HCl to 1.0 L of the buffer solution in part (a)

PLAN: We know the value of Ka and we can find initial the concentrations of conjugate acid and base. Use [Initial] >100 x 100 to eliminate x as variable. Proceed IN AN ORGANIZED FASHION stepwise through changes in the system slow

SOLUTION A: pH of a solution consisting of 0.50 M CH3COOH/0.50 M CH3COONa

Ka =[H3O+][CH3COO-]

[CH3COOH]

x = [H3O+] = Ka = 1.8 x 10-5 M

Simplify? 0.50 > 100 x 1.8 x 10-5 YES!

pH = -log (1.8 x 10-5 M) = 4.74

initialchangeequilibrium

0.50 + x

0.50 - x

---

0.50

0.50 + x

CH3COOH(aq) + H2O(l) CH3COO-(aq) + H3O+(aq)concentration

+ x+ x

Ka =[x][0.50]

[0.50]

=[x][0.50 + x]

[0.50 - x]

[OH-]added = 0.020 mol1.0 L soln

= 0.020 M NaOH

CH3COOH(aq) + OH-(aq) CH3COO-(aq) + H2O (l)concentration (M)

before additionaddition

after neutralization

0.50 - 0.50 -don’t care

-

- 0.020

0.48 0 0.52___________________________________________________________

Calculate the pH after adding 0.020 mol of solid NaOH to 1.0 L of the buffer solution in part (a) 0.50 M CH3COOH/0.50 M CH3COONa

Set up an ICE table and treat it like a stoichiometry problem. Note that the base reacts with the acetic acid to form the salt and water---all before equilbrium (conceptually)!

SOLUTION B:

Calculate mol of acid left:

___________________________________________________________-

change --.02 -0.020 +0.020

Treat this as two problems---OH- neutralizes some acid making some conjugate base in the process that then equilibrates.

Now set up a reaction our reaction table with acetic acid reacting in water the new starting values!

[H3O+] = 1.8 x 10-5 x0.48

0.52= 1.7 x 10-5

pH = -log 1.7 X 10-5 = 4.77

CH3COOH(aq) + H2O(l) CH3COO-(aq) + H3O+(aq)concentration (M)

initialchange

equilibrium

0.48 -- x

0.48 - x

-

-

0.52 0

x

+ x + x

0.52 + x___________________________________________________________

Ka =[H3O+][CH3COO-][CH3COOH]

1.8 X 10-5 =[H3O+]

[0.52][0.48]

SOLUTION B:

Calculate H+ [H3O+]added = 0.020 mol

1.0 L soln= 0.020 M H3O+

Calculate the pH after adding 0.020 mol of solid HCl to 1.0 L of the buffer solution in part (a) 0.50 M CH3COOH/0.50 M CH3COONa

SOLUTION C:

CH3COO-(aq) + H3O+(aq) CH3COOH(aq) + H2O (l)concentration (M)

before additionaddition

after addition

0.50 - 0.50 --

-

- 0.020

0.48 0 0.52___________________________________________________________

-change don’t care-0.02 -0.020 +0.020___________________________________________________________

Set up a reaction table with the new values for the equilibrium of the acetic acid/conjugate base in water!

Set up a reaction table with the new values.

CH3COOH(aq) + H2O(l) CH3COO-(aq) + H3O+(aq)concentration (M)

initialchange

equilibrium

0.52 -- x

0.52 - x

-

-

0.48+ x

0.48 + x

[H3O+] = 1.8 x 10-5 x 0.48

0.52= 2.0 x 10-5

___________________________________________________________

Ka =[H3O+][CH3COO-][CH3COOH]

1.8 X 10-5 =[H3O+]

[0.52][0.48]

pH = -log (2.0 x 10-5 M) = 4.70

pH = 4.70-acidpH = 4.77-basepH = 4.74-nothing

A) Calculate the pH of the 0.30 M NH3/0.36 M NH4Cl buffer system. B) What is the pH after the addition of 20.0 mL of 0.050 M NaOH to 80.0 mL of the buffer solution above (pKa = 9.25 of NH4+)

When NaOH is added it will neutralize the acid, 1 mole H+ for each mole of OH- added. Go slow be methodical and organized.

Treat the problem as two parts. The volume changes adding complexity. Just go slow and use reason.

1. First react the strong base NaOH with the acid in the system (can you identify the acid here?).

2. Then use the new values for the buffer concentrations to calculate the pH (using Henderson-Hasslebalch if you want).

A) Calculate the pH of the 0.30 M NH3/0.36 M NH4Cl buffer system. B) What is the pH after the addition of 20.0 mL of 0.050 M NaOH to 80.0 mL of the buffer solution above (pKa = 9.25 of NH4+)

NH4+ (aq) H+ (aq) + NH3 (aq)

pH = pKa + log [NH3][NH4

+]

pKa = 9.25

pH = 9.25 + log [0.30][0.36]

= 9.17

pH = pKa + log [conjugate base][acid]

A titration is the controlled addition of one substance (titrant) to a known quantity of another substance (titrate) until the stoichiometric requirements are met.

Titrant is added slowlyuntil color change

A buret holds the titrant while the solution to be titrated is prepared in the flask. A dye indicator flags the end point.

1. Strong Acid-Strong Base --- easy to handle--no hydrolysis of either ion. Just use simple stoichiometric principles from Chem 7.

2. Weak Acid-Strong Base --- neutralization followed hydrolysis of the anion (the conjugate base of the weak acid).

3. Strong Acid-Weak Base --- neutralization followed by hydrolysis of the cation (the conjugate acid).

4. Weak Acid-Weak Base ---not typical

5. Polyprotic Acid Neutralization---more complex

Acid-Bases titrations can be classified into roughly 5-types. We care about one.

1. Strong Acid - Strong Base Titration

The key is to recognize strong acids and strong bases and apply your 100% dissociation, solution chemistry and adding volumes.

2-moles OH per mole base

1-moles OH per mole base

2-moles H+ per mole acid

1-moles H+ per mole acid

Strong

1. Strong Acid-Strong Base Titration

NaOH (aq) + HCl (aq) H2O (l) + NaCl (aq)

OH- (aq) + H+ (aq) H2O (l) net-ionic reaction

Suppose we titrate 40.00 mL of 0.100 M HCl with 0.1000 M NaOH and monitor the pH of the solution as a function of volume of NaOH added.

VISUALIZE 1) H+ reacts with OH- from NaOH2) the total solution volume is changing as youadd NaOH: key to the calculation! 3) need to consider both together and arrive at equilibrium [H+]

1. HCl-NaOH Titration (40 mL of 0.100M HCl)

mL OH- mol OH- mol H+ left Total Volume pH = -log [H]0.00 0.00 0.0040 mol 40 mL 1.0010.00 0.0010 0.0030 mol 50 mL 1.2220.00 0.0020 0.0020 mol 60 mL 1.4830.00 0.0030 0.0010 mol 70 mL 1.8539.00 0.0039 0.0001 mol 79.0 mL 2.8939.50 0.00395 0.00005 mol 79.5 mL 3.2039.75 0.003975 0.000025 mol 79.75 mL 3.5039.95 0.003995 0.000005 mol 79.95 mL 3.9040.00 0.0040 0.00 mol 80.00 mL 7.0040.01 0.004001 0.000001mol 80.01 mL 9.4040.05 0.004005 0.000005 mol 80.05 mL 9.80

Initial Moles of H+ = Volume x Molarity HCl = 0.040 L X 0.100M = 0.00400 mol

1. Strong Acid-Strong Base Titrations (0.100M each)

NaOH (aq) + HCl (aq) H2O (l) + NaCl (aq)OH- (aq) + H+ (aq) H2O (l) net-ionic reaction

pH

mL of NaOH added

Equivalence point and end point are not the same.

Equivalence Point: stoichiometric amount of H+ reactswith OH-

End Point: dye indicator changes color.

Volume of 0.100M NaOH (mL)

pH 7.0 -

• Equivalence point:– The point in the reaction at which both acid and base have

been consumed stoichiometrically. Neither acid nor base is present in excess.

• End point:– The point at which the pH indicator changes color. It is not

identical to the equivalence point---but close.

• Titrant:– The solution of known concentration (often standardized)

and added to the solution of unknown concentration.

• Titration Curve:– Terminology for a graph or plot of Solution pH vs. volume

of titrant added to the unknown solution.

There are a number of terms used in acid-base titrations that we need to know.

A 0.100M NaOH solution is used to titrate 50.0mL of a 0.100M HCl. Calculate the pH of the solution after 0mL, 40mL and 50mL of titrant have been added. What indicator would you use?

Moles of HCl = 0.050L× 0.100M HCl = 0.005 mol HCl

Moles of NaOH = 0.040L× 0.100M NaOH = 0.004 mol NaOHMol HCl left = 0.005 mol HCl - 0.004 mol NaOH = 0.001 mol HClTotal Volume = 50.0 mL HCl + 50 mL NaOH = 100.0 mL total

[H+] =0.001 mol HCl

0.090 L= 0.0111M HCl pH = -log(0.0111) = 1.95

Moles of NaOH = 0.050L× 0.100M NaOH = 0.005 mol NaOH

Total Volume = 50.0 mL HCl + 50 mL NaOH = 100.0 mL totalMol HCl left = 0.005 mol HCl - 0.005 mol NaOH = 0

pH = -log(0) = Infinity ====now H+ is from water only [10-7]pH = 7

1. 0 mL of NaOH addition : pH = -log(0.100M) = 1

ACIDCOLOR

Acid-Base Indicators are weak acids (abbreviated HIn instead of HA) that have one color in its acid form and a different color in conjugate base form (In-).

HIn (aq) H+ (aq) + In- (aq)

BASICCOLOR

Ka =[H+][In-][HIn]

What color is an indicator at a specified pH? It depends on the pKa of the indicator! Use the buffer equation to understand pH = pKa + log [In-]/[HIn]

When [HIn] = [In-] then the pH = pKa. --If pH of solution is higher than pKa => In- color form dominates, --If pH lower than pKa => then acidic color form HIn dominates.

pH > pKapH < pKa

The color change of the indicator bromthymol blue (pKa = 7.1).

acidic (HI form)

Basic (I- form)

change occurs over ~2pH units

pH = pKa in between colors

pH < pKa pH > pKa

pH = pKa

Acid-Base Titration Curves

Solubility Equilibria (precipitates and equilibrium)Stalagmite (g--from the ground) Stalactite (c--from the ceiling) Both are precipitated CaCO3 (limestone caves and caverns)

Calcium Oxalate = Kidney Stones

Precipitation Reactions Can Hurt:As of April 16, 2003, Don Winfield of Caledonia, Ontario, Canada, passed 4,504 kidney stones, ranging in size from a grain of sand to a dried pea. Eventually, the kidney had to be removed.

--Large value of Ksp = very soluble (equilibrium is product-favored)

--Small value Ksp = very insoluble (equilibrium is reactant-favored)

Chemists use the equilibrium constant, Ksp to quantify the equilibrium between a slightly soluble solid and its saturated aqueous solution.

Al(OH)3 (s) 1Al3+ (aq) + 3OH- (aq) Ksp = [Al3+][OH-]3

Ksp = 1.3 x 10-33

--Heterogeneous equilibrium so we exclude the solid from Ksp.

Ksp is written like other equilibrium constants. The equilibrium is established when a solution is “saturated”---whereby no more solute will dissolve (a precipitate).

MgF2 (s) Mg2+(aq) + 2F-(aq) Ksp = [Mg2+][F-]2

Solid = precipitate not included in equilibrium constant

Molarity raised to stoichiometric coefficients don’t include solids in expression!

Balance equation keeping track of stoichiometry!

The Chem 7 “solubility rules” are a qualitative “summary” of Ksp values of compounds.

Large value of Ksp = very soluble (equilibrium is product-favored)

Small value Ksp = very insoluble (equilibrium is reactant-favored)

Saturated solutions have a solid precipitate in equilibrium with its ions (heterogeneous).

Al(OH)3 (s) 1Al3+ (aq) + 3OH- (aq) Ksp = [Al3+][OH-]3

When the value of Ksp exceeded a precipitate will form!

Ksp = 1.3 x 10-33

Write the ion-product expression for each of the following:

Ag2CO3 (s) 2Ag+(aq) + CO32-(aq) Ksp = [Ag+]2[CO3

2-]

Ca3(PO4)2 (s) 3Ca2+(aq) + 2PO43-(aq) Ksp = [Ca2+]3[PO3

3-]2

Ksp = [Mg2+][CO32-]

Ksp = [Fe2+][OH-] 2

Ksp = [Ca2+]3[PO43-]2

MgCO3(s) Mg2+(aq) + CO32-(aq)

(b) Fe(OH)2(s) Fe2+(aq) + 2OH- (aq)

Ca3(PO4)2(s) 3Ca2+(aq) + 2PO43-(aq)

Molar solubility (mol/L) is the solubility expressed as molarity hat produces a saturated solution--or a precipitate. It is the value at which Ksp is determined.

Solubility (g/L) is the number of grams of solute per liter of solution that will dissolve to produce a precipitate--a saturated solution. It is the concentration at which Ksp is determined.

Solubility Terms Know

Molar Mass

FormulaStoichiometry

Equilibrium Equation

Ion Concentrations

(mol/L)

Molar Solubility(mol/L)

Solubility(g/L) Ksp

Determining Ksp from solubility(a) Lead (II) sulfate is a key component in lead-acid car batteries. Its solubility in water at 250C is 4.25x10-3g/100mL solution. What is the Ksp of PbSO4?

Ksp = [Pb2+][SO42-] PbSO4(s) Pb2+(aq) + SO4

2-(aq)(a)

Ksp = [Pb2+][SO42-] = (1.40x10-4) (1.40x10-4) = 1.96x10-8

The stoichiometry for Pb2+ and SO4- are 1:1 with PbSO4

= 1.40x10-4M PbSO44.25x10-3g

100mL soln 303.3g PbSO4

1 mol PbSO41000mLL

Molar Solubility =

1. Determining Ksp from solubilityWhen lead (II) fluoride (PbF2) is shaken with pure water at 250C, the solubility is found to be 0.64g/L. Calculate the molar solubility and then find the Ksp of PbF2.

Ksp = (2.6x10-3)(5.2x10-3)2 = 7.0x10-8

PbF2(s) Pb2+(aq) + 2F-(aq) Ksp = [Pb2+][F-]2

= 2.6x10-3 M PbF20.64gL soln 245.2g PbF2

mol PbF2Molar Solubility =

[Pb+2] = 2.6x10-3 M PbF21 mol Pb2+

= 2.6x10-3 M Pb2+

1 mol PbF2

[F-] = 2.6x10-3 M PbF2 1 mol PbF2

2 mol F-= 5.2x10-3 M F-

Key part is to be careful with stoichiometry!

Determining solubility from Ksp

Calcium hydroxide is a major component of mortar, plaster and cement, and solutions of Ca(OH)2 are used in industry as a cheap, strong base. Calculate the molar solubility of Ca(OH)2 in water at 25 oC if its Ksp is 6.5 x 10-6.

Ksp = (S)(2S)2 = 4S3 = 6.5 x 10-6

Ca(OH)2(s) Ca2+ + 2OH- -initial

changeequilibrium

--

0 0+S + 2SS 2S

concentration (M)

_________________________________________

Ksp = [Ca2+][OH-]2

_________________________________________

S = 3

�6.5× 10−6

4= 1.2× 10−2MCa(OH)2

Calculate the solubility of silver chloride in g/L and mol/L given that Ksp = 1.6 X 10-10?

Ksp = [Ag+][Cl-] = s2 = 1.6 x 10-10

[Ag+] = 1.3 x 10-5 M [Cl-] = 1.3 x 10-5 M

Solubility of AgCl =

1.3 x 10-5 mol AgCl1 L soln

143.35 g AgCl1 mol AgClx = 1.9 x 10-3 g/L

AgCl (s) Ag+ (aq) + Cl- (aq)

Initial (M)Change (M)Equilibrium (M)

0.00+s

0.00+s

s s

--

s =�

Ksp =�

1.6× 10−10 = 1.3× 10−5 = [Ag+] = [Cl−]

No. of Ions

2

2

2

3

3

3

3

Within a family of Ksp related compounds solubility can be compared directly.

Formula Cation:Anion Ksp Solubility (M)

MgCO3 1:1 3.5 x 10-8 1.9 x 10-4

PbSO4 1:1 1.6 x 10-8 1.3 x 10-4

BaCrO4 1:1 2.1 x 10-10 1.4 x 10-5

Ca(OH)2 1:2 5.5 x 10-6 1.2 x 10-2

BaF2 1:2 1.5 x 10-6 7.2 x 10-3

CaF2 1:2 3.2 x 10-11 2.0 x 10-4

Ag2CrO4 2:1 2.6 x 10-12 8.7 x 10-5

Ksp to Solubility and Vis Versa

LeChatlier’s Principle: The presence of a stress on a system will cause the system to react to relieve the stress.

Solubility of Salts: A salt becomes less soluble if a common ion is added to an existing sparingly soluble solution (the common ion effect)

Ksp = [Pb2+][I-]2

Calculating the Effect of a Common Ion on Solubility

In Sample Problem 19.6, we calculated the solubility of Ca(OH)2 in water. What is its solubility in 0.10M Ca(NO3)2? Ksp of Ca(OH)2 is 6.5x10-6.

Ksp = 6.5x10-6 = (0.10 + S)(2S)2 = (0.10)(2S)2 S << 0.10

= 4.0x10-3 S =

!

6.5x10"64

Ca(OH)2(s) Ca2+(aq) + 2OH-(aq)Concentration(M)

InitialChangeEquilibrium

---

0.10 0+S +2S

0.10 + S 2S

The solubility of a insoluble ionic salt can be increased by the addition of a strong acid IF the anion of the salt is from a weak acid, or is OH-.

PbF2(s) Pb2+(aq) 2F-

Add HCl or HNO3

Anion of weak acid will be basic in water (hydrolysis) and react with acid if added

2F- + H2O <=> HF + OH-

H2O + Cl- or NO3-

equilibrium shifts

Adding H3O+ reacts with the anion of the weak acid CO32- the equilibrium to the right.

CaCO3(s) Ca2+(aq) + CO32-(aq)

CO32-(aq) + H3O+(aq) HCO3

- (aq) + H2O(l)

HCO3-(aq) + H3O+(aq) H2CO3(aq) + H2O(l)

H2CO3(aq) 2H2O(l) + CO2(g)

If the insoluble compound contains the anion of a weak acid, addition of H3O+ (from a strong acid) increases its solubility (LeChâtelier’s principle)

Predicting the Effect on Solubility of Adding Strong AcidWrite balanced equations to explain whether addition of H3O+ from a strong acid affects the solubility of these ionic compounds:

Br- is the anion of a strong acid.

No effect.(a) PbBr2(s) Pb2+(aq) + 2Br-(aq)

(b) Cu(OH)2(s) Cu2+(aq) + 2OH-(aq)H+ will react with OH- shift to the equilibrium right and increase solubility.

(c) FeS(s) Fe2+(aq) + S2-(aq)

S2- is the anion of a weak acid and will react with water to produce OH-.

Both weak acids serve to increase the solubility of FeS.

FeS(s) + H2O(l) Fe2+(aq) + HS-(aq) + OH-(aq)

Calculating the Effect of a Common Ion on Solubility

In Sample Problem 19.6, we calculated the solubility of Ca(OH)2 in water. What is its solubility in 0.10M Ca(NO3)2? Ksp of Ca(OH)2 is 6.5x10-6.

Ksp = 6.5x10-6 = (0.10 + S)(2S)2 = (0.10)(2S)2 S << 0.10

= 4.0x10-3 S =

!

6.5x10"64

Ca(OH)2(s) Ca2+(aq) + 2OH-(aq)Concentration(M)

InitialChangeEquilibrium

---

0.10 0+S +2S

0.10 + S 2S

Q = Ksp

Q < Ksp

Q > Ksp

Saturated solutionUnsaturated solution No precipitate

Supersaturated solution Precipitate will form

Ag2CO3 (s) 2Ag+ (aq) + CO32- (aq)

Ksp = [Ag+]2[CO32-]Qsp = [Ag+]2[CO3

2-]

before equilibrium after equilibrium

Ksp can be used to predict if a precipitate will form in a solution. (Remember Q vs Kc?).

Consider silver carbonate equilibrium:

What happen when [Ag+] = 0.1M and [CO32-] = 0.05M?

Predicting whether a precipitate will formA common laboratory method for preparing a precipitate is to mix solutions of the component ions. Does a precipitate form when 0.100 L of 0.30 M Ca(NO3)2 is mixed with 0.200 L of 0.060 M NaF?

CaF2(s) Ca2+(aq) + 2F-(aq) Ksp = 3.2 x 10-11

mol Ca2+ = 0.100 L(0.30 mol/L) = 0.030 mol Ca2+

[Ca2+] = 0.030 mol/0.300 L = 0.10 M

mol F- = 0.200 L(0.060 mol/L) = 0.012 mol F-

[F-] = 0.012 mol/0.300 L = 0.040 M

Q = [Ca2+][F-]2 = (0.10)(0.040)2 = 1.6 x 10-4

Q is >> Ksp, thus CaF2 will precipitate.

If 2.00 mL of 0.200 M NaOH are added to 1.00 L of 0.100 M CaCl2, will a precipitate form?

The ions present in solution are Na+, OH-, Ca2+, Cl-.The only possible precipitate is Ca(OH)2 (solubility rules).

Is Q > Ksp for Ca(OH)2?? If so.....a precipitate will occur

mol Ca2+ = 0.100 mol mol OH- = 4.0 x 10-4 mol

Ksp = [Ca2+][OH-]2 = 8.0 x 10-6

Q = [Ca2+]0[OH-]02 = 0.10 x (4.0 x 10-4)2 = 1.6 x 10-8

Q < Ksp No precipitate will form

Total volume = .002L + 1.00 L = 1.002 L