grade 12 - weebly

Grade 12 Chapter 15 Lesson Plan

402. The presence of a common ion decreases the dissociation. BQ1

Calculate the pH of 0.10M CH3COOH. Ka = 1.8 × 10-5

.

[H+] = √ = √( )( ) = 1.34 × 10

-3 M pH = 2.87

Calculate the pH of 0.10M CH3COOH dissolved in 0.10 M CH3COONa. Ka = 1.8 × 10-5

.

Given: [CH3COOH] =0.10M, [CH3COO-] = 0.10M, Ka = 1.8 × 10

-5 RTF: pH

CH3COOH(aq) CH3COO-(aq) + H

+(aq)

[initial] 0.10 0.10 -

[change] -x +x +x

[eqbm] 0.10 – x 0.10 + x x

Ka =

( )

= 1.8 × 10

-5 assume x << 0.1

[H+] = x = 1.8 × 10

-5 M (assumption true) pH = 4.75

BQ 1 Calculate the pH of 0.10M acetic acid solution in 0.050M sodium acetate(aq) solution. Ka = 1.8 × 10-5

.

Given: [CH3COOH] =0.10M, [CH3COONa] = 0.050M, Ka = 1.8 × 10-5

RTF: pH


+(aq)

[initial] 0.10 0.050 -

[change] -x +x +x

[eqbm] 0.10 – x 0.050+x x

Ka =

( )

= 1.8 × 10

-5 assume x << 0.050

x = ( )( )

= 6.0 × 10

-3 M = [H

+] pH = 2.22

403. A buffer: is a solution that resists changes in its pH upon small additions of acid or base.SQ1

SQ1. What is a buffer solution? A solution that resists change in its pH upon small addition of H+or OH

-.

404. A buffer consists of: a weak acid and its conjugate base or a weak base and its conjugate acid.

405. Buffer are made in one of three ways: SQ 2

mixing a weak acid (base) with its salt

mixing a weak acid (in excess) with a strong base

mixing a weak base (in excess) with a strong acid

SQ2 Which of the following procedures will produce a buffered solution?

i) Equal volumes of 0.3 M KOH and 1.0 M HCl solutions are mixed.

ii) Equal volumes of 0.5 M KOH and 1.0M C6H5COOH solutions are mixed.

iii) Equal volumes of 1.0 M C6H5COOH and 1.0 M C6H5COOK solutions are

mixed.

Which of the following are buffers? 0.10 M HCl and 0.10 M KCl

0.10 M HF and 0.10 M KF

0.10 M Na2HPO4 and 0.10 M NaH2PO4

Given 1:1 molar ratio of: NH3 / NH4Cl - which solution has the lowest pH?

H3PO4 / NaH2PO4 - which solution is almost neutral?

HCl / NaCl - which solution is a buffer at pH > 8?

NaOH / NH3 - which solution is a buffer at pH < 6?

NH3 / CH3COOH

406. Buffers work: because they have species that neutralized added H+ or OH

-. SQ3

HF/F- Buffer: If H

+ added: H

+ + F

- HF (H

+ neutralized)

If OH- added: HF + OH

- F

- + H2O (OH

- neutralized)

NH3/NH4+ Buffer: If H

+ added: H

+ + NH3 NH4

+ (H

+ neutralized)

If OH- added: NH4

+ + OH

- NH3 + H2O (OH

- neutralized)

SQ 3 Explain how a CH3COOH / CH3COO- buffer maintains pH relatively the same upon additions of a few

drops of HCl? Of NaOH?

When HCl(aq) is added, the added H+ will be neutralized by the acetate ions present in the buffer

solution. CH3COO-(aq) + H

+(aq) CH3COOH(aq)

When NaOH(aq) is added, the added OH- will be neutralized by the acetic acid molecules present in the

buffer solution. CH3COOH(aq) + OH-(aq) CH3COO

-(aq) + H2O(l)

407. pH of buffer solutions: can be found using the Henderson Hasselbalch equation. SQ4, BQ2, ex 31, 33

SQ 4 What is the Henderson - Hasselbalch equation? pH = pKa + log

BQ 2 Calculate the pH of a buffer solution which is 0.2M NaF and 0.3M HF.

Given: [NaF] = 0.2M, [HF] = 0.3M , Ka = 7.2 × 10-4

RTF: pH

It is a buffer solution, therefore pH = pKa + log

= -log7.2 × 10

-4 + log

= 3.14 +(– 0.18) = 2.96

How can we prepare a buffer with pH = 7 using only solutions of 0.10 M CH3COOH and CH3COONa?

Ka = 1.8 × 10-5

.

pH = pKa + log

7 = 4.75 + log

= 180

Use much more of the acetate ion than the acetic acid

408. If

is constant, pH of the buffer is unchanged: dilution has no effect on pH of buffer. BQ3, 4

BQ 3 Calculate the pH of a buffer solution which is 0.4M NaF and 0.6M HF. (Compare your result with the

one you got in question 2 above!)

pH = 2.96, the pH remains the same as the change in the concentrations did not change the ratio of

[base] to [acid].

BQ 4 A buffer solution is prepared which is 0.2M NaF and 0.3M HF. 100ml of water were added to 100 ml

of the above solution. Calculate the new pH.

pH = 2.96, the addition of water dilutes both the acid and the base. The ratio of [base] to [acid] does

not change, so the pH does not change.

409. Buffering capacity: measures the ability of a buffered solution to absorb H+ or OH

- without significant

change in its pH. The larger the concentrations of the weak acid (base) and its conjugate base (acid), the higher

the capacity. A ratio of

= 1, gives the best buffering capacity. (The ratio being equal to 1 is more

important than high concentrations). SQ5, 6

SQ5 What is meant by the buffering capacity of a buffer solution?

It is the ability of a buffered solution to absorb H+ or OH

- ions without a significant change in pH.

SQ6 Which ratio of [conjugate base] / [weak acid] would produce the most efficient buffer?

= 1

410. When choosing an acid to make a buffer of a given pH, the pKa of the acid must be as close as

possible to the required pH. SQ7, BQ6

SQ7 You wish to create a buffer solution with a pH of 5.00. Which of the following would be the best choice

to prepare the required buffer?

i) HCl

ii) H3AsO4 Ka =5.5 x 10-3

iii) CH3COOH Ka = 1.8 x 10-5

iv) HCN Ka = 4.9 x 10-10

i) HCl Ka very large pKa = very negative

ii) H3AsO4 Ka =5.5 x 10-3

pKa = 2.3

iii) CH3COOH Ka = 1.8 x 10-5

pKa = 4.7

iv) HCN Ka = 4.9 x 10-10

pKa = 9.3

Best acid to choose is CH3COOH Ka = 1.8 x 10-5

pKa = 4.7

BQ 6 A buffer solution of pH = 5 is to be prepared using CH3COONa and CH3COOH. What should be the

molar ratio [CH3COO- ] to [CH3COOH] to achieve this?

Given: pH = 5, Ka = 1.8 × 10-5

RTF:

pH = pKa + log

= - log1.8 × 10

-5 + log

= 5

log

= 5 – 4.74 = 0.26

= 1.82

A buffer made using equimolar amounts of NaHSO3 and Na2SO3 would be most effective between

the pH ranges (Ka H2SO3 = 1 × 10-2

, Ka HSO3- = 1 × 10

-7):

8 – 10 6 – 8 4 – 6 2 – 4 0.5 – 2

411. Steps to find pH of solution when a strong base/strong acid is added to a buffer. BQ 5

BQ 5 A buffer solution is prepared which is 0.10M CH3COONa and 0.20M CH3COOH. 10cm3of 0.20M

NaOH(aq) were added to 100cm3 of the above solution. Find the new pH.

Given: [CH3COOH] = 0.20M, [CH3COO-] = 0.10M, Vbuffer = 100cm

3, VNaOH = 10cm

3

[NaOH] = 0.20M, Ka = 1.8 × 10-5

RTF: pH

Step 1 Find number of moles of each species before mixing

n of OH- = (0.20)(10) = 2.0 mmoles

n of CH3COOH = (0.20)(100) = 20mmoles

n of CH3COO- = (0.10)(100) = 10mmoles

Step 2 Find number of moles left after mixing

Write the reaction occurring: CH3COOH(aq) + OH-(aq) CH3COO

- (aq) + H2O(l)

Initially 20 mmoles 2.0 mmoles 10 mmoles

Finally 18 mmoles ---- 12 mmoles

The resulting solution contains a weak acid with its conjugate base, a buffer.

Step 3 Use Hendersen Hasselbalch equation to find pH

pH = pKa + log

= -log1.8 × 10

-5 + log

= 4.56

412. Steps to find pH of a solution formed by mixing: Weak acid (in excess) with strong base. BQ7

BQ7 A solution is prepared by mixing 70ml, 0.2M HF and 30ml, 0.1M NaOH. Calculate the pH of the

solution. Given: VHF = 70ml, [HF] = 0.2M, VNaOH = 30ml, [NaOH] = 0.1M, Ka = 7.25 × 10

-4 RTF: pH

nHF = (70)(0.2) = 14mmoles

nNaOH = (30)(0.1) = 3mmoles

HF(aq) + OH-(aq) F

-(aq) + H2O(aq)

Initially 14 3 -

Finally 11 - 3

pH = pKa + log

= - log7.25 × 10

-4 + log

= 3.14 +(– 0.56) = 2.58

413. Steps to find pH of a solution formed by mixing: Weak basic salt (in excess) with strong acid. BQ 9

BQ 9 A solution is prepared by mixing 60ml, 0.2M NaF and 40ml, 0.1M HCl. Calculate the pH of the

solution. Given: VNaF = 60ml, [NaF] = 0.2M, VHCl = 40ml, [HF] = 0.1M, Ka HF = 7.25 × 10

-4 RTF: pH

n of F- = (60)(0.2) = 12mmoles

nHCl = (40)(0.1) = 4mmoles

F-(aq) + H

+(aq) HF(aq)

Initially 12 4 -

Finally 8 - 4

pH = pKa + log

= - log7.25 × 10

-4 + log

= 3.44

414. Steps to find pH of a solution formed by mixing: Weak base (in excess) with strong acid BQ10

BQ 10 A solution is prepared by mixing 20ml, 3M NH3 and 80ml, 0.2M HCl. Calculate the pH of the

solution. Given:

3NHV = 20ml, [NH3] = 3M, VHCl = 80ml, [HCl] = 0.2M, Kb NH3 = 1.8 × 10-5

RTF: pH

n of NH3 = (20)(3) = 60mmoles

nHCl = (80)(0.2) = 16mmoles

NH3(aq) + H+(aq) NH4

+(aq)

Initially 60 16 -

Finally 44 - 16

Ka =

= 5.56 × 10-10

pH = pKa + log

= - log5.56 × 10-10

+ log

=9.25 + 0.44 = 9.69

415. Steps to find pH of a solution formed by mixing: Acid salt (in excess) with strong base. BQ 8

BQ 8 A solution is prepared by mixing 80ml, 0.3M NH4Cl and 20ml, 0.2M NaOH. Calculate the pH of the

solution. Given:

4NH ClV = 80ml, [NH4Cl] = 0.3M, VNaOH = 20ml, [NaOH] = 0.2M, Kb of NH3 = 1.8 × 10-5

RTF: pH

Calculate the number of moles: n of NH4+ = (80)(0.3) = 24mmoles

nNaOH = (20)(0.2) = 4mmoles

NH4+(aq) + OH

-(aq) NH3(aq) + H2O(aq)

Initially 24 4 -

Finally 20 - 4

Ka =

= 5.56 × 10

-10 pH = pKa + log

= - log(5.56 × 10-10

) + log

=9.25 – 0.70 = 8.55

416. Titration: a technique in which one solution (titrant) is used to analyze another (analyte).

417. Equivalence point: is the point in the titration when enough titrant has been added to react exactly with

the substance being titrated, analyte. SQ 8

SQ8 What is meant by equivalence or stoichiometric point in titration?

Equivalence or stoichiometric point is is the point in the titration when enough titrant has been addded to

react exactly with the substance in solution being titrated.

418. End point: the point in the titration when the indicator changes colour. SQ 9

SQ9 What is meant by end point in titration?

End point is the point in the titration at which the indicator changes colour.

419. At the equivalence point of acid-base titrations: moles of H+ = moles of OH

-. BQ 11, 14

BQ 11

a) 25cm3 of a certain NaOH solution were needed to neutralize 20cm

3 of a 0.20M HCl solution.

Calculate the concentration of the base.

Given: [HCl] = 0.20M, VHCl = 20cm3, VNaOH = 25cm

3

RTF: [NaOH]

At the equivalence point: number of moles of H+ = number of moles of OH

-

[H+] = [OH

-]

(0.20)(20) = [OH-](25) [OH

-] = 0.16M

b) What volume of 0.5M HCl is needed to neutralise 40cm3 of a 0.02M Ca(OH)2 solution?

Given: [HCl] = 0.50M, ( ) = 40cm3, [Ca(OH)2] = 0.02M

RTF: [HCl]

At the equivalence point: number of moles of H+ = number of moles of OH

-

# of moles of OH- = 2 × # of moles of Ca(OH)2 = 2(0.02) = 0.04M

[H+] = [OH

-]

(0.50) = (0.04)(40) = 3.2cm3

Another way to solve this type of exercise is to write the reacting ratio:

Ca(OH)2(aq) + 2HCl(aq) CaCl2(aq) + 2H2O(l)

1 mole 2 moles

( ) nHCl

nHCl = 2 ( )

(0.50)( ) = 2(0.02)(40) = 3.2cm3

BQ 14 0.20g of a monoprotic acid was dissolved in 25 cm3 solution. The solution was titrated with 0.125M

NaOH solution. 20 cm3 of the base was needed to reach the equivalence point. What is the molar mass of the

acid?

Given: macid = 0.20g, Vacid = 25 cm3, [NaOH] = 0.125 M, Vbase = 20 cm

3 RTF: Molar mass of acid

Since monoprotic acid, at the equivalence point,

nacid = nbase

= [base]Vbase

= (20 × 10

-3)(0.125) Macid = 80 g/mol

420. Strong acid-strong base titration. BQ12, ex. 57

BQ 12

a. 50 cm3 of 0.50 M HNO3 solution were titrated with 0.50 M NaOH solution. Find the pH of the

solution when no base has been added yet.

Given: = 50 ml, [HNO3] = 0.50M, [NaOH] = 0.50M, VNaOH = 0ml

RTF: pH

HNO3 is a strong acid, it dissociates completely. [H+] = [HNO3] = 0.50M

pH = -log0.50 = 0.30

b. 50 cm3 of 0.50 M HNO3 solution were titrated with 0.50M NaOH solution. Find the pH of the solution

when 10cm3 of base are added.


RTF: pH

= (50)(0.50) = 25mmoles

nNaOH = (10)(0.50) = 5mmoles

n of acid in excess = 25 – 5 = 20 mmoles

[H+] = [HNO3]excess =

= 0.33 M

pH = -log0.33 = 0.48

c. 50 cm3 of 0.50 M HNO3 solution were titrated with 0.50 M NaOH solution. Find the pH of the

solution when 50cm3 of base are added.


RTF: pH

= (50)(0.50) = 25mmoles

nNaOH = (50)(0.50) = 25mmoles

n of acid in excess = 25 – 25 = 0 mmoles

Neither H+ nor OH

- is in excess, solution is neutral pH = 7

d. 50 cm3 of 0.50 M HNO3 solution were titrated with 0.50 M NaOH solution. Find the pH of the

solution when 60cm3 of base are added.


RTF: pH

= (50)(0.50) = 25mmoles

nNaOH = (60)(0.50) = 30mmoles

n of base in excess = 30 – 25 = 5 mmoles

NaOH is a strong base, it dissociates completely. nOH- = nNaOH = 5mmoles

[OH-]excess =

= 0.045 M

pOH = -log0.045= 1.34 pH = 14 – pOH = 14 – 1.34 = 12.7

e. 50 cm3 of 0.50 M HNO3 solution were titrated with 0.50 M NaOH solution. Draw a titration curve for

the change of pH versus volume of base added. In your curve the following points have to be clearly

indicated: The pH before the titration starts and at the equivalence point. The volume of base needed

for the equivalence point should be also shown.

VNaOH (cm3) pH

0 0.30

20 0.70

30 0.90

40 1.26

50 7.0

60 12.7

pH

13 –

12 –

11 –

10 –

9 –

8 –

7 – • equivalence point pH = 7 6 – 5 –

4 –

3 –

2 –

1 –

10 20 30 40 50 60 70 80 90 100

Volume of NaOH (cm3)

421. Strong base - weak acid titration. BQ13, ex. 61

BQ 13

a. 50 cm3 of 0.10 M CH3COOH solution were titrated with 0.10 M KOH solution. Find the pH of the

solution when no base has been added yet. Ka = 1.80 × 10-5

Given: = 50 ml, [CH3COOH] = 0.10M, [KOH] = 0.10M, VKOH = 0ml,

Ka = 1.80 × 10-5

RTF: pH

CH3COOH is a weak acid, it dissociates partially:


+(aq)

[initially] 0.10M - -

[cange] -x +x +x

[eqbm] 0.10 – x x x

Ka =

= 1.80 × 10

-5 assume x << 0.10

[H+] = x = √( )( = 1.34 × 10

-3 M (assumption valid)

pH = -log1.34 × 10-3

= 2.87

b. 50 cm3 of 0.10 M CH3COOH solution were titrated with 0.10 M KOH solution. Find the pH of the

solution when 10 ml of base has been added. Ka = 1.80 × 10-5


Ka = 1.80 × 10-5

RTF: pH

= (50)(0.10) = 5mmoles nKOH = (10)(0.10) = 1mmole

CH3COOH(aq) + OH-(aq) CH3COO

-(aq) + H2O(l)

Initially 5mmoles 1mmole -

After

reaction

4mmoles - 1mmole

After the reaction we end up with a solution containing a weak acid and its conjugate base buffer

solution.

pH = pKa + log

= 4.74 – 0.60 = 4.14

c. 50 cm3 of 0.1 M CH3COOH solution were titrated with 0.1 M KOH solution. Find the pH of the

solution when 25cm3 base are added. (What is the significance of this stage of titration?)


Ka = 1.80 × 10-5

RTF: pH

= (50)(0.10) = 5mmoles

nKOH = (25)(0.10) = 2.5mmoles


-(aq) + H2O(l)

Initially 5mmoles 2.5mmole -

After

reaction

2.5mmoles - 2.5mmoles

After the reaction we end up with a solution containing a weak acid and its conjugate base buffer

solution.

pH = pKa + log

= 4.74

This point is called the half-way equivalence point. The buffer is at its best capacity and the pH = pKa

d. 50 cm3 of 0.1 M CH3COOH solution were titrated with 0.1 M KOH solution. Find the pH of the

solution when 50cm3 base are added. (What is the significance of this stage of titration?)


Ka = 1.80 × 10-5

RTF: pH

= (50)(0.10) = 5 mmoles

nKOH = (50)(0.10) = 5 mmoles


-(aq) + H2O(l)


After

reaction

- - 5mmoles

[CH3COO-] =

= 0.050M

After the reaction we end up with a solution containing a weak base CH3COO-(aq).

CH3COO

-(aq) + H2O(l) CH3COOH(aq)+ OH

-(aq)

[initially] 0.050M - - [change] -x +x +x [eqbm 0.050 – x x x

Kb =

=

= 5.56 × 10

-10 assume x << 0.050

[OH-] = x = √( )( ) = 5.27 × 10

-6M (assumption valid)

pOH = 5.28

pH = 14 – pOH = 14 – 5.28 = 8.72

e. 50 cm3 of 0.1 M CH3COOH solution were titrated with 0.1 M KOH solution. Find the pH of the

solution when 60cm3 base are added.


Ka = 1.80 × 10-5

RTF: pH

= (50)(0.10) = 5mmoles

nKOH = (60)(0.10) = 6mmoles

n of base in excess = 6 – 5 = 1 mmole

KOH is a strong base, it dissociates completely. nOH- = nKOH = 1mmole

[OH-]excess =

= 9.09 × 10

-3 M

pOH = -log 9.09 × 10-3

= 2.05 pH = 14 – pOH = 14 – 2.04 = 11.95

f. 50 cm3 of 0.1 M CH3COOH solution were titrated with 0.1 M KOH solution. Draw a titration

curve for the change of pH versus volume of base added. In your curve the following points

have to be clearly indicated: The pH before the titration starts and at the equivalence point. The

volume of base needed for the equivalence point should be also shown.

Volume of base added pH

0 cm3 2.87

10 cm3 4.14

20 cm3 4.56

30 cm3 4.92

40 cm3 5.34

49.95 cm3 6.73

50 cm3 8.72

50.05 cm3 9.70

60 cm3 11.95

70 cm3 12.7

pH

13 –

12 –

11 –

10 –

9 –

8 –

7 –

6 – 5 –

4 –

3 –

2 –

1 –

10 20 30 40 50 60 70

Volume of KOH (cm3)

422. Half equivalence point: is when half the acid/base is neutralized. At that point, nHA = pH = pKa

Buffer is at its best capacity. SQ 12

SQ12 In the titration of weak acid with strong base what is meant by "the half way point"?

What can you say about the buffering action of the resulting solution?

How does the pH at the half-equivalence point compare with the value of Ka?

The half equivalence point is the point in the titration when half the starting material has been titrated.

At the half-equivalence point, pH = pKa, and the buffer is at its best buffering capacity.

423. Comparison between titration curves. SQ11

SQ 11 How is the titration curve for a weak acid-strong base different from that of a strong acid-strong

base?

Strong Acid – strong base Titration

Weak Acid – strong base Titration

No small initial rise A small initial rise

Its vertical section is longer Its vertical section is shorter

Its rise is sharp Its rise is not sharp

Before equivalence point the pH steadily

increases

Before the equivalence point, there is an almost

horizontal section where the solution behaves as a

buffer

pH at equivalence point = 7 pH at equivalence point is greater than 7

424. The effect of acid strength and quantity on its titration with a strong base. BQ15

Acid strength Acid quantity

Volume of base needed No effect The greater the quantity

pH at the equivalence point The stronger the acid the closer the pH to 7 No effect

Shape of pH curve The stronger the acid the sharper the curves

and the longer the vertical section

Same general shape

List, other than water, all the species present in a solution of weak acid HA(aq) in decreasing concentration.

HA > H+ = A

- > OH

-

Correct procedures for a titration include:

Draining the pipette by touching the tip to the side of the container used for titration.

Rinsing the buret with distilled water just before filling it with the liquid to be titrated.

Swirling the solution frequently during the titration. A, C

A student titrated a weighed sample of a solid acid (HA) with standard aqueous base to find its molar mass.

What effect will each of the following have on the determined molar mass?

This graph was obtained when 0.10M acid was

titrated with NaOH. How much is pKa?

pH = pKa at half equivalvence point.

Equivalence point is at V = 50 cm3 half -

equivalence point is at V = 25 cm3. pH at that

point is around 4.

Adding more water than is needed to dissolve the acid. No effect

Adding some base beyond the equivalence point. Obtained molar mass too small

Failure to rinse all the acid from the weighing paper into the titration vessel. Obtained molar mass

too large

.

BQ 15 25 ml of 0.20M HA was titrated with 0.20M NaOH. The pH at the equivalence point was 8.00. What is

Ka of HA? (Are the volumes necessary in this problem? Why?)

Given: Vacid = 25 cm3, [acid] = 0.20M, [NaOH] = 0.20 M RTF: Ka

At equivalence point, we have a solution of A-.

HA(aq) + OH-(aq) A

-(aq) + H2O(l)


After reaction - - 5mmoles

A 3.00g sample of a weak acid HA was

titrated with 1.0 M NaOH. The pH

titration curve obtained is shown below.

Find pKa and the molar mass of the acid.

pKa around 6

nacid = nbase 𝒎

𝑴 = CV M =

𝒎

𝑪𝑽

𝟑 𝟎𝟎

(𝟏 𝟎)(𝟐𝟎 𝟏𝟎 𝟑)

M = 150 g/mole

An unknown weak acid HA was dissolved

and titrated with NaOH. Titration curve

obtained shown, find pKa.

pH = pKa = around 4

[A-] =

( )( )

= 0.10M A

- is a base, it reacts with water setting up the following equilibrium:

A

-(aq) + H2O(l) HA(aq) + OH

-(aq)


pH is given to be 8.00 at equivalence point pOH = 6

At equivalence point, [OH-] = 1.0 × 10

-6 M [HA] = 1.0 × 10

-6 M

Kb =

= ( )

= 1.0 × 10

-11 Ka =

= 1.0 × 10

-3

425. Acid base indicator: is a substance, usually a weak organic acid, that changes color within a fairly

narrow range of pH values because the acid and its conjugate base have different colors. SQ 14

SQ 14. What is an acid-base indicator ?

An acid base indicator is a substance, usually an organic acid,that changes color within a fairly narrow

range of pH values because the acid and its conjugate base are of two different colors.

426. Indicators you must know:

Indicator Color in Acid Color in Base Changes color at pH

Phenolphthalein Colorless Pink 9

Methyl orange Red Yellow 4

427. Indicator changes color over the range: its pKa 1

Suppose an indicator HIn has a Ka = 1.0 × 10-5

. HIn is yellow and In- is red.

If placed in an acidic solution, it will exist as HIn solution’s color is yellow. As a base is added, HIn In-.

A color change occurs when [In-] =

[HIn]

.

Since both HIn and In- are present, then the indicator acts as a buffer: pH = pKa + log

.

Solution becomes orange when pH = -log 1.0 × 10-5

+ log

= 5.0 - 1 = 4.0

If placed in a basic solution, it will exist as In- solution’s color is red. As a acid is added, In

- HIn. A

color change occurs when [HIn] =

[In

-]

.

Solution becomes orange when pH = -log 1.0 × 10-5

+ log = 5.0 + 1 = 6.0

Summary: When

= 0.1 we see color of HIn

When

we see color of In

-

When 0.1 =

<

we see intermediate color

Useful pH of indicator is pKa 1

When pH > pKa + 1 we see color of In-

When pH < pKa – 1 we see color of HIn

When pKa + 1 > pH > pKa – 1 we see color of intermediate

428. In choosing an indicator for a titration: we want the color change to occur approximately at the pH

of the equivalence point. Since pH changes rapidly around the equivalence point, we need not be exact.

This is especially true for strong acid – strong base titrations.

When a strong acid is titrated with a strong base using phenolphthalein, the color changes suddenly at the end

point. The color can be switched back and forth by addition of only a single drop of acid or base. What is the

reason for the sudden color change? A large change in pH occurs near the enpoint of a titration

429. In choosing an indicator for a titration:its useful pH range must lie completely within the vertical

part of the pH titration curve.

430. For a strong acid – strong base titration there is a wider choice of indicators than for a strong

acid – weak base or weak base – strong acid titration. SQ15, MCBQ 1- 8, BQ 16, 18

SQ 15 In a particular titration experiment the pH of the solution changed from 4.0, one drop before the

equivalence point to 10, one drop after the equivalence point. Will any of these two indicators, with their range

of colour change be suitable for the above titration? HIn' : 3.8 to 5.4; HIn" : 8 to 9.8

The best indicator is one whose pH range change lies completely within the pH rise. HIn" has a pH color

change between 8 to 9.8, which lies completely between 4.0 and 10. Therefore, HIn’’ is the better

indicator to choose for this titration.

BQ 16 A solution of 0.100 M HCl and a solution of 0.100 M NaOH are prepared. A 40.0 mL sample of one of

the solutions is added to a beaker and then titrated with the other solution. A pH electrode is used to obtain the

data that are plotted in the titration curve shown below. (b)

(a) Identify the solution that was initially added to the beaker. Explain your reasoning.

The solution in the beaker was the 0.100 M HCl because the initial pH was 1 (the pH of 0.100 M HCl).

(b) On the titration curve above, circle the point that corresponds to the equivalence point.

The point with coordinates (40.0, 7) is circled.

(c) At the equivalence point, how many moles of titrant have been added?

Given: [NaOH] = 0.100M RTF: n

nNaOH = (0.100)(40.0 × 10-3

) = 0.00400 mol NaOH

(d) The same titration is to be performed again, this time using an indicator. Use the information in the table

below to select the best indicator for the titration. Explain your choice.

Indicator pH range of color change

Crysatl violet 0 – 1.8

Bromocresol Purple 5.2 – 6.9

Thymophthalein 9.5 – 10.5

Bromocresol purple would be best because its color change will occur closest to the equivalence

point (at equivalence point pH changes from about 4 to 10).

(e) What is the difference between the equivalence point of a titration and the end point of a titration?

The equivalence point in a titration occurs when the number of moles of titrant added is exactly

sufficient to react completely with the number of moles of the titrated species present in the

sample being titrated.

The end point of a titration is the point in a titration at which the indicator undergoes its color

change.

(f) On the grid provided below, sketch the titration curve that would result if the solutions in the beaker and

buret were reversed (i.e., if 40.0 mL of the solution used in the buret in the previous titration were

titrated with the solution that was in the beaker).

pH

E

pH

8.7 D

C

B

A

Volume of base added

8. Which of the following indicator is the best choice for the titration?

[-A-] Methyl orange 3.2 – 4.4 [-D-] Phenol phthalein 8.2 – 10.0

[-B-] Methyl red 4.8 – 6.6 [-E-] Alizarin 11.0 – 12.4

[-C-] Bromothymol blue 6.1 – 7.6

BQ 18 CH3COOH(aq) + H2O(l) CH3COO-(aq) + H3O

+(aq)

Acetic acid dissociates in water as shown in the equation above. A 25.0ml sample of aqueous solution of

0.100M acetic acid is titrated using standardized 0.100M NaOH.

a) After addition of 12.5 ml of 0.100M NaOH, the pH of the resulting solution is 4.74. Calculate each of

the following.

(i) [H+] in solution

Given: = 25 ml, [CH3COOH] = 0.100M, [NaOH] = 0.100M,

VNaOH = 12.5ml, Ka = 1.80 × 10-5

, pH = 4.74

RTF: [H+]

This point is called the half-way equivalence point. At this point the pH = pKa.

[H+] = Ka = 1.80 × 10

-5M

(ii) [OH-] in the solution

Given: [H+] = 1.80 × 10

-5M RTF: [OH

-]

[OH-] =

= 5.56 × 10-10

M

The diagram represents the pH titration curve

of a weak monoprotic acid with a 0.100 M

NaOH. Use it to answer the following

questions.

1. The pH at this point is less than 3. A

2. The pH at this point is greater than 8

and less than 10. D

3. The pH at this point could be used to

determine the acid dissociation constant. B

4. What part of the curve corresponds to

the optimum buffer action? B

5. At this point, the solution is buffered.

B

6. Of the points shown on the graph, this

is the point when the solution is most basic.

E 7. Of the points shown on the graph, this

is the point is the equivalence point of the

titration. D

(iii) The number of moles of NaOH added

Given: [NaOH] = 0.100M, VNaOH = 12.5ml RTF: nNaOH

nNaOH = (0.100)(12.5 × 10-3

) = 1.25 × 10-3

mole

(iv) The number of mole of CH3COO- in the solution.

Given: = 25 ml, [CH3COOH] = 0.100M, [NaOH] = 0.100M, VNaOH = 12.5ml, Ka = 1.80 × 10-5

RTF: [CH3COO-]

= (25)(0.10) = 2.5mmoles

nNaOH = (12.5)(0.10) = 1.25mmoles


-(aq) + H2O(l)

Initially 2.5mmoles 1.25mmol

e

-

After reaction 1.25mmoles - 1.25mmoles

v) The number of moles of CH3COOH present the solution

Given: = 25 ml, [CH3COOH] = 0.100M, [NaOH] = 0.100M, VNaOH = 12.5ml, Ka = 1.80 × 10-5

RTF: [CH3COOH]

= (25)(0.10) = 2.5mmoles

n NaOH = (12.5)(0.10) = 1.25mmoles


-(aq) + H2O(l)

Initially 2.5mmoles 1.25mmole -

After reaction 1.25mmoles - 1.25mmoles

b) State whether the solution at the equilibrium of the titration is acidic, basic, or neutral. Explain your

reasoning.

At the equivalence point, we have a solution of CH3COO-. This ion dissociates in water as follows:

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

-(aq)

Since OH- ion are formed then the solution is basic.

In a different experiment, 0.118 g sample of a mixture of solid CH3COOH and solid NaCl is dissolved in water

and titrated with 0.100M NaOH. The equivalence point was reached when 10.00 ml of the base solution is

added.

c) Calculate the following:

(i) The mass in grams of acetic acid solid in the mixture

Given: [NaOH] = 0.100M, VNaOH = 10.00ml RTF: m of CH3COOH

At the equivalence point, nNaOH = nacid since the base is monobasic and the acid is monoprotic

n of acid = (0.100)(10.00) = 1.00 mmole m = (1.00 × 10-3

)(60) = 0.0600g

(ii) The mass percentage of acetic acid in the solid

Given: m of acid = 0.0600g, mtotal = 0.118g RTF: mass % of acid

m % =

= 50.8%

d) Calculate the pH at the equivalence point of the titration of 25.0 ml 0.10M CH3COOH with 0.10 NaOH

solutions.

Given: = 25.0 ml, [CH3COOH] = 0.10M, [NaOH] = 0.10M, VKOH = 25.0ml, Ka = 1.80 × 10-5

RTF: pH

= (25.0)(0.10) = 2.5 mmoles nKOH = (25.0)(0.10) = 2.5 mmoles


-(aq) + H2O(l)

Initially 2.5mmoles 2.5mmole -

After reaction - - 2.5mmoles

[CH3COO-] =

= 0.05M

After the reaction we end up with a solution containing a weak base CH3COO-(aq).

CH3COO

-(aq) + H2O(l) CH3COOH(aq)+ OH

-(aq)


Kb =

=

= 5.56 × 10

-10 assume x << 0.050

[OH-] = x = √( )( ) = 5.27 × 10

-6M (assumption valid)

pOH = 5.28

pH = 14 – pOH = 14 – 5.28 = 8.72

e) The pKa values for several indicators are given below. Which of the indicators listed is most suitable

for this titration? Justify your answer.

Indicator pKa

Erythrosine 3

Litmus 7

O-cresolphtalein 9.2

Thymolphtalein 10

Indicators change color over a pH range = pKa 1

Erythrosine 2 – 4

Litmus 6 – 8

o-Cresolphthalein 8.2 – 10.2

Thymolphthalein 9 – 11

Since this a titration of a weak acid with a strong base the color will change at the pH > 7, (pH at

the equivalence point = 8.57).

Therefore, o-Cresolphthalein is the best indicator because 8.57 lies in the range of color change of

o-Cresolphthalein.

431. The titration curves of polyprotic acid titrations: have as many rises as there are acidic ions. SQ13

SQ13 Sketch the titration curve of H3PO4 with NaOH.

This is titration curve for 100mL, 0.025M acetic

acid with 0.10M NaOH.

Choose the best indicator from the following, given

their pH range of color change:

a. Methyl orange 3.2 – 4.4

b. Methyl red 4.8 – 6.0

c. Bromothymol blue 6.1 – 7.6

d. Phenolphthalein 8.2 – 10.0

e. Alizarin 11.0 – 12.4

What part of the curve corresponds to optimum

buffer action? Point V

An unknown acid was titrated with a solution of NaOH. The pH titration curve is shown above.

Which of the following could the acid be?

Acetic acid, CH3COOH Hydrochloric acid, HCl Formic acid, HCOOH

Nitric acid, HNO3 Oxalic acid, H2C2O4 Boric acid, H3BO3

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