Expt 6-Colorimetric Dtermination of pH

Experiment 6:Experiment 6:Colorimetric Colorimetric

Determination of Determination of pHpH

Report by:Tricia Ojon and

Katrina Mae Lee

Objectives:Objectives:

• To be able to determine the pH of an unknown solution colorimetrically

• To be able to calculate the ionization constant of a weak acid

INTRODUCTIINTRODUCTIONON

ColorimetryColorimetry

• Colorimetry refers to any technique by which an unknown color is evaluated in terms of standard colors.

• It is the determination of the spectral absorbance of a solution

pH indicatorpH indicator

• It can be used as a tool in colorimetry.• It is a halochromic (a material which

changes color when a change in pH occurs) chemical compound that, when added in solutions, can help detect changes in pH and present the result visually through color change.

• It is also called acid-base indicator because it is a chemical indicator of hydronium ions which dictates if the solution is acidic or basic.

• The following is a table of examples of pH indicators with their respective pH range for color transition to occur and respective colors in pH levels higher and lower than this pH range:

pH indicatorpH indicator

pH indicatorpH indicator

Indicator Lower pH color

pH Range(transition

interval)

Higher pH

color

Thymol blue Red 1.2 – 2.8 Yellow

Bromophenol blue

Yellow 3.0 – 4.6 Purple

Chlorophenol red

Yellow 4.8 – 6.4 Violet

Bromothymol blue

Yellow 6.0 – 7.6 Blue

Phenol red Yellow 6.8 – 8.4 Red

McIlvaine buffer system McIlvaine buffer system

•Buffer system which makes use of a citrate and a phosphate, particularly citric acid and Na2HPO4, to volumetrically set for pH in a wide range(2.2 to 8).

BufferBuffer solutionsolution• Solutions that contain a conjugate

acid-base pair in reasonable concentrations.

• They can resist changes in pH because they contain both an acidic species to react with the OH- ion, and a basic species to react with the H3O+ ion.

• Exhibit the common ion effect.

CommonCommon ionion effecteffect

• A special case of LeChatelier’s Principle.

• This occurs when a given ion is added to an equilibrium mixture that already contains that ion, and the position of equilibrium shifts away from forming more of it.

Common ion effectCommon ion effect• For example, a solution of a weak

electrolyte is altered by adding one of its ions from another source, causing a shift in equilibrium position away from the added component, thereby suppressing the ionization of the weak electrolyte.

• CH3COOH plus NaCH3COO

Common ion effectCommon ion effect

CH3COOH + H2O ↔ H3O+ + CH3COO-

NaCH3COO → Na+ + CH3COO-

The addition of CH3COO- by NaCH3COO causes a shift in equilibrium position away from the added component, thereby suppressing the ionization of CH3COOH.

pHpH

• Measure of acidity or basicity of a substance

• Calculated by:• pH = -log [H+]

• pH = -log [H3O+]

Acid dissociation constant, Acid dissociation constant, KKaa

•An equilibrium constant for the dissociation of an acid (HA) in H2O to yield the conjugate base (A-) and H3O+

• Ka = [H3O+][A-]

[HA]

Henderson-Hasselbalch Henderson-Hasselbalch equationequation

• Equation used for measuring the pH of buffered solutions

• pH = pKa + log [conjugate base]

[acid]

• Assumption: the concentration of the acid and its conjugate base at equilibrium will remain the same as the formal concentration.

• This neglects the dissociation of the acid and the hydrolysis of the base. The dissociation of water itself is neglected as well.

• These approximations will fail when dealing with relatively strong acids or bases (pKa more than a couple units away from 7), dilute or very concentrated solutions (less than 1 mM or greater than 1M), or heavily skewed acid/base ratios (more than 100 to 1).

Henderson-Hasselbalch Henderson-Hasselbalch equationequation

EXPERIMENTEXPERIMENTALAL

Preparation of Buffer Preparation of Buffer SolutionsSolutions

• A set of McIlvaine buffers were accurately prepared in test tubes of uniform sizes labeled according to their respective pH levels.

• Five drops of the appropriate indicators to use for each pH level were added to each of the buffer solutions.

• The following tables summarize the amounts (in mL) of 0.2M Na2HPO4 and 0.1M Citric Acid needed to make a buffer solution with a certain pH level and the indicator appropriate for each:

Preparation of Buffer Preparation of Buffer SolutionsSolutions

Indicator: THYMOL BLUE (1.2 – 2.8)

pH 0.2M Na2HPO4 (mL) 0.1M Citric Acid (mL)

2.2 0.20 9.80

2.4 0.62 9.38

2.6 1.06 8.91

2.8 1.58 8.42

Preparation of Buffer Preparation of Buffer SolutionsSolutions

Indicator: BROMOPHENOL BLUE (3.0 – 4.6)

pH 0.2M Na2HPO4 (mL) 0.1M Citric Acid (mL)

3.0 2.05 7.95

3.2 2.47 7.53

3.4 2.85 7.15

3.6 3.22 6.78

3.8 3.55 6.45

4.0 3.25 6.15

4.2 4.14 5.86

4.4 4.41 5.59

4.6 4.67 5.33

Preparation of Buffer Preparation of Buffer SolutionsSolutions

Indicator: CHLOROPHENOL RED (4.8 – 6.4)

pH 0.2M Na2HPO4 (mL) 0.1M Citric Acid (mL)

4.8 4.93 5.07

5.0 5.15 4.85

5.2 5.20 4.80

5.4 5.58 4.42

5.6 5.80 4.20

5.8 6.05 3.95

6.0 6.31 3.69

6.2 6.61 3.39

6.4 6.92 3.08

Preparation of Buffer Preparation of Buffer SolutionsSolutions

Indicator: BROMOTHYMOL BLUE (6.0 – 7.6)

pH 0.2M Na2HPO4 (mL) 0.1M Citric Acid (mL)

6.0 6.31 3.69

6.2 6.61 3.39

6.4 6.92 3.08

6.6 7.34 2.66

6.8 7.72 2.28

7.0 8.24 1.76

7.2 8.69 1.31

7.4 9.08 0.92

7.6 9.37 0.63

Preparation of Buffer Preparation of Buffer SolutionsSolutions

Indicator: PHENOL RED (6.8 – 8.0)

pH 0.2M Na2HPO4 (mL) 0.1M Citric Acid (mL)

6.8 7.72 2.28

7.0 8.24 1.76

7.2 8.69 1.31

7.4 9.08 0.92

7.6 9.37 0.63

7.8 9.57 0.43

8.0 9.72 0.28

Preparation of Buffer Preparation of Buffer SolutionsSolutions

Colorimetric Determination Colorimetric Determination of pHof pH

• Five test tubes were labeled with the 5 indicators, namely, thymol blue, bromophenol blue, chlorophenol red, bromothymol blue, and phenol red.

• Equal amounts of 0.01M HOAc were placed in each of the test tubes.

• Two drops of the proper indicator was added only to the test tube with the correct label for the indicator being used.

Colorimetric Determination Colorimetric Determination of pHof pH

• Each of the 5 resulting colors was then compared to the previously prepared buffer solutions in Part A while limiting the comparison to those included in the pH range of the indicator used.

Colorimetric Determination Colorimetric Determination of pHof pH

• For example, if thymol blue is used, the buffers that will be used for comparison will be those labeled 2.2, 2.4, 2.6 and 2.8.

• If the color of the solution matches that of a buffer’s, the pH of the solution is taken as the same as the pH of that buffer.

Colorimetric Determination Colorimetric Determination of pHof pH

• If the color of the solution does not match that of any of the buffers’ applicable, other test tubes of the remaining indicators will be used for comparison until the colors match.

Colorimetric Determination Colorimetric Determination of pHof pH

• The above procedure was repeated for each of the following solutions replacing 0.01M HOAc:

• 1 mL 0.1 M HOAc + 1 mL 0.1 M NaOAc+ 8 mL H2O

• 1 mL 0.1 M HOAc + 0.1 mL 0.1 M NaOAc+ 8.9 mL H2O

• 0.1 mL 0.1 M HOAc + 1 mL 0.1 M NaOAc+ 8.9 mL H2O

Colorimetric Determination Colorimetric Determination of pHof pH

RESULTSRESULTS

ResultsResultsSolution Observed

pH

A 0.01M HOAc 4.8

B 1mL 0.1M HOAc + 1mL 0.1M NaOAc + 8mL H20

4.8

C 1mL 0.1M HOAc + 0.1mL 0.1M NaOAc + 8.9mL H2O

3.8

D 0.1mL 0.1M HOAc + 1mL 0.1M NaOAc + 8.9mL H2O

8.0

DISCUSSIONDISCUSSION

• Preparation of Buffer Solutions• It is practical to use only the

appropriate indicators for certain pH ranges because there will be no color transition in pH levels higher or lower than the respective range of the indicator.

• If there is no color transition, the resulting indication of the pH of solutions will be very vague and there will be no firm basis for comparison.

DiscussionDiscussion

• From the buffer solutions B to D, it can be inferred that as the ratio of the molarity of NaOAc to that of HOAc in the mixture ([NaOAc]/[HOAc]) increases, the pH level also increases and the solution becomes less acidic.

DiscussionDiscussion

• Solution D is the least acidic because it has the highest [NaOAc] to [HOAc] ratio in the mixture while solution C is the most acidic because it has the lowest [NaOAc] to [HOAc] ratio in the mixture.

DiscussionDiscussion

Solution Observed pH

A 0.01M HOAc 4.8

B 1mL 0.1M HOAc + 1mL 0.1M NaOAc + 8mL H20

4.8

C 1mL 0.1M HOAc + 0.1mL 0.1M NaOAc + 8.9mL H2O

3.8

D 0.1mL 0.1M HOAc + 1mL 0.1M NaOAc + 8.9mL H2O

8.0

Recall:

• CH3COOH + H2O ↔ H3O+ + CH3COO-

• NaCH3COO → Na+ + CH3COO-

HOAc + H2O ↔ H3O+ + OAc-

NaOAc → Na+ + OAc-

• From this equation, it is known that NaCH3COO is a strong electrolyte and CH3COOH is a weak electrolyte.

• Because it dissociates completely, NaCH3COO produces much of the CH3COO-

• The produced CH3COO- is a common ion of the dissociation of CH3COOH, thus, a common ion effect occurs on the solution.

CH3COOH + H2O ↔ H3O+ + CHCH33COOCOO--

NaCH3COO → Na+ + CHCH33COOCOO--

• This production of additional CH3COO- causes the ionization of CH3COOH to be suppressed based on LeChatelier’s Principle of chemical equilibrium.

• The system consumes the additional CH3COO- by forming more CH3COOH. This means a backward reaction.

• This also means the reduction of formation of H3O+ , which determines the level of acidity of the solution.

• In this case, it is reduced, therefore, the solution becomes less acidic.

GUIDE GUIDE QUESTIONS AND QUESTIONS AND

ANSWERSANSWERS

1.) Calculate the ionization constant of acetic acid using colorimetric data.

pH of 0.01 M HOAc 4.8

[H3O+] of 0.01 M HOAc 1.58x10-5 M

Calculated Ka of HOAc 2.50x10-8

To get [H3O+]:pH = - log [H+]pH = - log [H3O+][H3O+] = 10 -pH

[H3O+] = 10 -4.8

[H3O+] = 1.58x10-5 M

pH of 0.01 M HOAc 4.8

[H3O+] of 0.01 M HOAc 1.58x10-5 M

Calculated Ka of HOAc 2.50x10-8

To get Ka:HOAc → H+ + OAc-

Initial 0.01 0 0Change - 1.58x10-5 + 1.58x10-5 + 1.58x10-5

Equilibrium 9.98x10-3 1.58x10-5 1.58x10-5

Ka = [H+][OAc-] = (1.58x10-5 )( 1.58x10-5 )= 2.50x10-8

[HOAc] 9.98x10-3

2.) Calculate the pH of three mixtures of HOAc and NaOAc (solutions A, B and C) using the Henderson-Hasselbalch equation and compare with the observed pH. (Use the value of ionization constant of HOAc at 25°C.) Support your answers with computations.

A 1 mL 0.1 M HOAc + 1 mL 0.1 M NaOAc+ 8 mL H2O

B 1 mL 0.1 M HOAc + 0.1 mL 0.1 M NaOAc+ 8.9 mL H2O

C 0.1 mL 0.1 M HOAc + 1 mL 0.1 M NaOAc+ 8.9 mL H2O

Ka

• Ka of HOAc at 25°C = 1.8x10-5

pKa

• pKa = -log Ka = - log (1.8x10-5) = 4.74

Henderson-Hasselbalch equation• pH = pKa + log [conjugate base]

[acid]

Recall:

• CH3COOH + H2O ↔ H3O+ + CH3COO-

• NaCH3COO → Na+ + CH3COO-

HOAc + H2O ↔ H3O+ + OAc-

NaOAc → Na+ + OAc-

HOAc + H2O ↔ H3O+ + OAc-

NaOAc → Na+ + OAc-

The acid is HOAc.It’s conjugate base is OAc- .Because HOAc has very little

dissociation, it is assumed that most of the OAc- comes from the ionization of NaOAc, therefore,[OAc-] = [NaOAc].

For solution A:1 mL 0.1 M HOAc +1 mL 0.1 M NaOAc + 8 mL H2O

• Let X = [acid] = [HOAc] M1 V1 = M2 V2

(0.1M) (1mL) = (X) (10mL) X = 0.01MX = 0.01M

HOAc + H2O ↔ H3O+ + OAc-

NaOAc → Na+ + OAc-

• Let Y = [conjugate base] = [OAc-] = [NaOAc]

M1 V1 = M2 V2

(0.1M) (1mL) = (Y) (10mL) Y = 0.01MY = 0.01M

For solution A:1 mL 0.1 M HOAc +1 mL 0.1 M NaOAc + 8 mL H2O

HOAc + H2O ↔ H3O+ + OAc-

NaOAc → Na+ + OAc-

• pH = pKa + log [conjugate base]

[acid]= 4.74 + log 0.01

0.01= 4.74 + 0= 4.744.74

For solution A:1 mL 0.1 M HOAc +1 mL 0.1 M NaOAc + 8 mL H2O

• Let X = [acid] = [HOAc] M1 V1 = M2 V2

(0.1M) (1mL) = (X) (10mL) X = 0.01MX = 0.01M

HOAc + H2O ↔ H3O+ + OAc-

NaOAc → Na+ + OAc-

For solution B:1 mL 0.1 M HOAc +0.1 mL 0.1 M NaOAc + 8.9 mL H2O

• Let Y = [conjugate base] = [OAc-] = [NaOAc]

M1 V1 = M2 V2

(0.1M) (0.1mL) = (Y) (10mL)

Y = 0.001MY = 0.001MHOAc + H2O ↔ H3O+ + OAc-

NaOAc → Na+ + OAc-

For solution B:1 mL 0.1 M HOAc +0.1 mL 0.1 M NaOAc + 8.9 mL H2O

• pH = pKa + log [conjugate base]

[acid]= 4.74 + log 0.001

0.01= 4.74 - 1= 3.743.74

For solution B:1 mL 0.1 M HOAc +0.1 mL 0.1 M NaOAc + 8.9 mL H2O

• Let X = [acid] = [HOAc] M1 V1 = M2 V2

(0.1M) (0.1mL) = (X) (10mL) X = 0.001MX = 0.001M

HOAc + H2O ↔ H3O+ + OAc-

NaOAc → Na+ + OAc-

For solution C:0.1 mL 0.1 M HOAc +1 mL 0.1 M NaOAc + 8.9 mL H2O

• Let Y = [conjugate base] = [OAc-] = [NaOAc]

M1 V1 = M2 V2

(0.1M) (1mL) = (Y) (10mL) Y = 0.01MY = 0.01M

HOAc + H2O ↔ H3O+ + OAc-

NaOAc → Na+ + OAc-

For solution C:0.1 mL 0.1 M HOAc +1 mL 0.1 M NaOAc + 8.9 mL H2O

• pH = pKa + log [conjugate base]

[acid]= 4.74 + log 0.01

0.001= 4.74 + 1= 5.745.74

For solution C:0.1 mL 0.1 M HOAc +1 mL 0.1 M NaOAc + 8.9 mL H2O

SolutionpH

observed calculated

0.01 M HOAc 4.8 xxxxx

1 mL 0.1 M HOAc + 1 mL 0.1 M NaOAc + 8 mL H2O (A) 4.8 4.74

1 mL 0.1 M HOAc + 0.1 mL NaOAc + 8.9 mL H2O (B) 3.8 3.74

0.1 mL 0.1 M HOAc + 1 mL 0.1 M NaOAc + 8.9 mL H2O (C) 8.0 5.74

Comparison:Comparison:

CONCLUSION AND CONCLUSION AND RECOMMENDATIONSRECOMMENDATIONS

• Although it is less precise than the use of a pH meter, the use of colorimetry and pH indicators can be used to determine the unknown pH of a solution in approximation. But first, one must understand and acquire knowledge on which indicator to use on which pH range and their respective color changes to be able to successfully perform colorimetric analysis.

• Buffers can resist dramatic changes in pH but, because they exhibit the common ion effect, a slight change is still experienced by the solution.

• The effects of the common ion effect on the pH of a weak electrolyte such as CH3COOH is a decrease in the ionization of the electrolyte, which reduces the production of H3O+, thereby causing an increase in the pH of the solution.

• To compute for the theoretical pH of a buffered solution and to compare it to the observed pH by colorimetry, use the Henderson-Hasselbalch equation but, take note that this is limited to only certain strengths of acids and bases.

• The solutions needed should be correctly, carefully and accurately prepared using the proper tools to achieve the accurate pH using the colorimetric method.

Thank you for listening!Thank you for listening!