CHEM 322EXPERIMENT #5

POTENTIOMETRIC TITRATION OF A WEAK BASE

INTRODUCTION

In this experiment, your unknown Soda Ash sample containing sodium carbonate, a weak dibase from

experiment 4, will be titrated with your standardized HCl (aq) solution using a pH electrode to determine

the equivalence point, and thus the percent sodium carbonate in the unknown sample. As seen below, the

base reacts with the acid in steps:

CO32- (aq) + H3O+ (aq) → H2O (l) + HCO3- (aq)

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

Figure 5- 1 Titration Curve.

In figure 5-1 titration curve, the point labeled ep1 represents the 1st equivalence point, at which the

number of millimoles of weak conjugate base1 originally present equals the number of millimoles of the

strong acid added. Point A represents the 1st halfway point in the titration. The point labeled ep2

represents the 2nd equivalence point, at which the number of millimoles of the weak conjugate base2

originally present equals the number of millimoles of the strong acid added. Point B represents the 2nd

halfway point in the titration. Point C is the excess acid that has been added. Note: that the number of

millimoles of acid required to reach ep2 should be exactly twice that required to reach ep1.

OBJECTIVES

1.

2.

To learn about potentiometric titrations.

To construe titration curves for strong and weak acid-base titration.

To titrate an unknown Soda Ash sample containing sodium carbonate, a weak dibase with

standardized HCl (aq) solution using a pH electrode.

4. Using a titration curve compute the equivalence points, pKa, and the percent sodium carbonate in

an unknown sample.

3.

pH

Chemists use an operator called “p,” which is used to describe the negative logarithm of a given number

(e.g., pX= -log (x)). The use of this operator with hydrogen ion concentrations gives rise to a term known

as pH. A notational definition of pH is pH= -log(aH+) ≈ -log ([H+]) where aH+ and [H+] are the activity or

concentration of “hydrogen ions” in a sample. The content of hydrogen ions or hydronium ions (H3O+) in

a sample is express as activity or concentration. The same type of operation that is used to obtain pH

from aH+ or [H+] can be used to describe other numbers that have a broad range of values. For example,

the term “pOH” can be used to represent the term -log (aOH-) or pKw could be used in place of -log (Kw).

Kw is related to [H3O+], or [H+] and [OH-] through Kw= [H+] [OH-] and by taking the logarithm of both

sides then it is pKw = pH + pOH. When the latter equation is combined with Kw = 1.0 x 10-14 at 25℃,

then it results in 14.00 = pH + pOH. This is relationship is the basis for the pH scale that is often used

with water and aqueous samples. In using this scale, an aqueous solution with a pH less than 7.0 is said

to be acidic because it contains greater mounts of hydrogen ions than hydroxide ions. An aqueous

solution with a pH greater than 7.0 is called basic (or alkaline), because it contains a greater mount of

hydroxide ions. An aqueous solution that is at or near a pH of 7.0 is said to be neutral, because it has

approximately the same amounts of hydrogen ions and hydroxide ions, making the solution neither acidic

nor basic.

Monoprotic acid (HA) is an acid that can donate only one H+ to a base.

Monoprotic base (B) will react with an acid to accept essentially one mole of hydrogen ion (or form one

mole of hydroxide ions in water) per mole of the original base.

Polyprotic acid/base is an acid or a base that gives up or accepts more than one hydrogen ion, for

example carbonic acid (H2CO3).

Amphiprotic is a substance that can act as either an acid or a base, which means they can accept or donate

hydrogen ions, example is water, H2O, and sulfuric acid, HSO4-.

Potentiometry

Potentiometry is a technique that is based on the measurement of a cell potential with essentially zero

current passing through the system. The measured potential is related to the chemical composition of the

two electrodes and the solutions in which they are placed. Potentiometry is the type of electrochemical

method that forms the basis for use of the fluoride electrode and the pH electrode. In this experiment the

combined used of a pH electrode along with titration will help you achieve the purpose of this

experiment, this is called potentiometric titration, which measures the potential of an acid- base reaction

through the course of a titration as various amounts of titrant is combine with the analyte.

A pH electrode is an ion selective electrode (ISE) that responds to individual types of anions and cations.

The pH electrode and its read-out device give an indication of the amount of H3O+ (aq) ion present in

aqueous solutions. So, the pH meter is a common type of ISE, an indicator electrode that is selective for

the detection of hydrogen ions only. The pH meter has a thin glass membrane that records the potential

of the glass electrode, E, that depends upon the two hydronium ion activities (concentrations): those

outside the glass electrode and those inside the glass electrode.

Titration Curves

A plot of the measured response versus the amount of added titrant during a titration is called a titration

curve. The point of this curve at which exactly enough titrant has been added to react with all the analyte

is known as the “equivalence point” and the point at which we can detect that enough titrant has been

added represents the “end point.”

Figure 5- 2 Titration curve of a weak base and a strong acid.

Figure 5-2 is a titration curve of a reaction of a weak base and a strong acid, HCl (aq). Where curve a

(blue) represents a titration of 10.0mL of 0.100M base (pKb1 =4.00, pKb2 =9.00) with 0.100M HCl (aq).

The two equivalence points are C and E. Points B and D are half-neutralization points, whose pH values

equals pKa2 and pKa1, respectively. Curve b (gray) represents a titration of 10.0 mL of 0.100M nicotine

(pKb1=6.15, pKb2=10.85) with 0.100M HCl (aq). There is no sharp break at the second equivalence point,

J, because the pH is too low.

Figure 5-3 of several titration curves, (a) is the

traditional titration curve of pH vs volume of

titrant, this is the experimental points in the

titration of 1.430 mg of xylenol orange, a

hexaprotic acid, dissolved in 1.000 mL of

aqueous 0.10M NaNO3. The titrant was

0.06592M NaOH. Plot (b) is the first (1st)

derivative,

∆𝑝𝑝𝑝𝑝

,

∆𝑉𝑉

of the titration curve. Plot (c) is

the second (2nd) derivative

∆�

∆𝑝𝑝𝑝𝑝

�

∆𝑚𝑚𝑚𝑚

∆𝑉𝑉

, which is the

derivative of the curve in panel (b). The end

points are taken as maxima in the 1st derivative

curve, and zero crossing of the 2nd derivative

curve.

Figure 5- 3 Several titration curves.

PROCEDURES

1. Use your experiment 3 unknown Soda Ash. The sample should be dry.

2. Rinse beakers, electrode, and stirring bar with deionized water.

3. Determine how much to weigh for each trial:

Sample weight of unknown = Molarity of HCl (aq) x

~18 𝑚𝑚𝑚𝑚 𝑥𝑥 105.99 𝑔𝑔/𝑚𝑚𝑚𝑚𝑚𝑚

10 𝑥𝑥 2 𝑥𝑥 % 𝑁𝑁𝑎𝑎2 𝐶𝐶𝐶𝐶3

Note: If 18 mL is the buret region you desire the second equivalence point to occur, your molarity is

0.1103 M and your unknown was found to contain 25% Na2CO3, (% calculation determined from

experiment 4), then the equation would be:

Sample weight of unknown = 0.1103M of HCl (aq) x

~18 𝑚𝑚𝑚𝑚 𝑥𝑥 105.99 𝑔𝑔/𝑚𝑚𝑚𝑚𝑚𝑚

10 𝑥𝑥 2 𝑥𝑥 25%

4. Weigh the sample by difference using a weighing boat and transfer into a 250 mL beaker. As usual

record the weight to 0.1 mg.

5. Dissolve the sample in approximately 70 mL of water.

6. Standardize the pH meter and electrode with the pH 7 and pH 4 buffers.

7. Trial 1 (Test Run)

a. Titrate the base potentiometrically with the standardized HCl (aq) solution, prepared in

Experiment 4.

b. Use a magnetic stirring bar and stirring plate to mix the solution, being sure to allow adequate

stirring between additions of titrant and be careful to avoid breaking the fragile electrodes with the stirring

bar (use ring stand and clamp).

c. Mount burette on ring stand, fill with HCl titrant.

d. Dispense 1.0 mL increments of HCl until ~pH = 2.0, finally add 3 mL more using 1.0 mL

increments.

e. Record mL titrant added vs. pH in tabular form in notebook.

f. Based on the

∆𝑝𝑝𝑝𝑝

∆𝑚𝑚𝑚𝑚

change, decide approximate 1st and 2nd equivalence point regions. Adequate

data points in these regions are essential in order to locate both equivalence points precisely in the

subsequent trials.

8. Trials 2-4

a. Repeat steps 1-5, and step 7.

b. Dispense 1.0 mL increments of HCl (aq) until 2 mL before equivalence point 1 then dropwise

until 2 mL past Equivalence Point 1 (Ep1). Continue with 1.0 mL increments until 2 mL before

Equivalence Point 2 (Ep2), then drop wise until 2 mL past Ep2. Add an additional 3.0 mL in 1.0 mL

increments past Ep2.

c. Record mL titrant used vs. pH in tabular form in notebook.

d. When finished, rinse and store pH electrodes in pH 7 buffer.

Figure 5- 4 pH meter titration set up.

REVIEW OF pH EQUILIBRIA CALCULATIONS

You may want to read the electrochemistry section of your analytical text and review the theory of ion

selective electrodes.

For H2O (l) + H2O (l) → OH- (aq) + H3O+ (aq) at 25℃ the equilibrium constant is

Kw= Kw= [H3O+] [OH-] = 1.0 x 10-14 therefore [H3O+] = 1.0 x 10-7, [OH-] = 1.0 x 10-7.

If [H3O+] > [OH-], the solution is acidic. If [H3O+] < [OH-], the solution is basic (alkaline).
We say that pH is the negative pH= -log [H3O+], pOH= -log [OH-], and its relationship is 14.00 = pH +
pOH. For example, HCl (aq) concentration is 8.7 x 10-4 M, the pH = -log (8.7 x 10-4) = 3.06, this is an
acidic solution.
For a NaOH solution whose concentration is 5.8 x 10-5 M, the pOH = -log (5.8 x 10-5) = 4.24 so, pH=
14.00 – 4.24 = 9.76, thus it is a basic solution.
Acid
⇌ Conjugate Base
For the reaction of an Acid (HA), HA (aq) + H2O (l) ⇌ A- (aq) + H3O+ (aq)
Base
⇌ Conjugate Acid
For the reaction of a base (A-), A- (aq) + H2O (l) ⇌ OH- (aq) + HA (aq)
The equilibrium constant for an acid is Ka – acid dissociation constant.
Ka =
[𝐻𝐻3 𝑂𝑂 + ][𝐴𝐴− ]
[𝐻𝐻𝐻𝐻]
[H3O+] =
A stronger acid has a larger ionization constant than a weak acid.
𝐾𝐾𝑎𝑎 𝑥𝑥 [𝐻𝐻𝐻𝐻]
[𝐴𝐴− ]
pH = pKa + log
pH = pKa + log
if you take the log ⇒ -log [H3O+] = -log Ka – log(
[𝐴𝐴− ]
[𝐻𝐻𝐻𝐻]
[𝐻𝐻𝐻𝐻]
)
[𝐴𝐴− ]
thus this becomes
(Henderson-Hasselbach equation), where pKa = -log Ka, and Ka = 10-pKa.
[𝐶𝐶𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜 𝐵𝐵𝑎𝑎𝑎𝑎𝑎𝑎]
[𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴]
[𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴]
[𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴]
[𝐻𝐻𝐻𝐻]
= 1, when
[𝐴𝐴− ]
= pH = pKa + log
When the condition becomes [HA] = [A-], when
Ka = [H3O+], then pH=pKa.
The pH before any acid is added to the analyte.
Principle species:
CO32- (aq) + H3O+ (aq) → H2O (l) + HCO3- (aq)
HCO3- (aq) + H3O+ (aq) → H2O (l) + H2CO3 (aq)
CO32- (aq) + H2O (l) → OH- (aq) + HCO3- (aq) this is Kb1.
𝑥𝑥 2
𝐾𝐾
Kb = 𝐹𝐹−𝑥𝑥 ⇒ x = �(𝐹𝐹)𝑥𝑥 � 𝐾𝐾𝑤𝑤 � if F – x ≈ F.
𝑎𝑎
Note: F = formality of CO32- (aq) and x = [OH-]
pH = 14.00 - pOH
The pH after HCl (aq) added but before the first equivalence point.
Principle species:
CO32- (aq) + H3O+ (aq) → H2O (l) + HCO3- (aq)
Treat as a buffer:
𝐾𝐾
Ka2 = 𝐾𝐾 𝑤𝑤
𝑏𝑏1
�𝐶𝐶𝐶𝐶−2 �
pH = pKa2 + log �[𝐻𝐻𝐶𝐶𝑂𝑂3 −]�
3
The pH at the first equivalence point.
Principle species for HCO3- (aq):
Bicarbonate (HCO3- (aq)) acts as an amphiprotic substance!
Let CO32- (aq) = A-2; HCO3- (aq) = HAThese two reactions compete with one another.
HA- (aq) + H2O (l) ↔ A- (aq) + H3O+ (aq) this is Ka2.
HA- (aq) + H2O (l) ↔ H2A + OH- (aq)
Use this equation: [H3O ] = �
+
Note: when the terms:
1
2
Ka1
(𝐾𝐾𝑎𝑎2 𝑪𝑪𝐻𝐻𝐻𝐻− + 𝐾𝐾𝑤𝑤 )
𝑪𝑪𝐻𝐻𝐻𝐻−
>1

𝐾𝐾𝑎𝑎2

𝑪𝑪

1+ 𝐻𝐻𝐻𝐻−

𝐾𝐾𝑎𝑎1

1/2

�

and Ka1CHA-> Kw and this simplified to [H3O+] = (Ka1Ka2)1/2

and pH = (pKa1 + pKa2).

The pH after the first equivalence point but before the second equivalence point.

Principle species:

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

Treat as a buffer:

𝐾𝐾

Ka1 = 𝐾𝐾 𝑤𝑤

𝑏𝑏2

[𝐻𝐻𝐻𝐻𝐻𝐻3 ]

�

2 𝐶𝐶𝑂𝑂3 ]

pH= pKa1 + log �[𝐻𝐻

The pH at the second equivalence point.

Principle species:

H2CO3 (aq) + H2O (l) → HCO3- (aq) + H3O+ (aq) this is Ka1.

Treat as weak acid; where x = H3O+ (aq); and F = formality of H2CO3 (aq)

𝑥𝑥 2

Ka = 𝐹𝐹−𝑥𝑥 ⇒ x = �(𝐹𝐹)𝑥𝑥 [𝐾𝐾𝑎𝑎 ]

The pH past second equivalence point.

Treat as strong acid, remember dilution effects!

pH = -log [H3O+]

Figure 5-1 depicts every principle described in pH equilibria calculations, and by monitoring the change

in pH and the amount titrant added for any acid-base titration the data can be used to prepare a titration

graph and derivative plots to determine the end point (equivalence point), as is shown in figure 5-3.

CONSTRUCTION AND COMPUTATION OF TITRATION CURVES

Titration curves should be included and plotted with the volume of HCl (aq) added along the x-axis

(abscissa) and the pH along the y-axis (ordinate). Draw a smooth curve through the data points.

1. Use a graphing program such as excel to plot your titration curves as described below.

a. Plot pH (y-axis) vs volume of titrant, mL (x-axis) (label these columns as B and A,

respectively) to obtain the Titration Curve.

b. To obtain the ∆𝑚𝑚𝑚𝑚, subtract mL2 – mL1, mL3 – mL2, mL4 – mL3, and so on (label this

column as C), whose formula is = A2(cell) – A1(cell).

c. To obtain the ∆𝑝𝑝𝑝𝑝, subtract pH2 – pH1, pH3 – pH2, pH4 – pH3, and so on (label this

column as D), whose formula is = B2(cell) – B1(cell).

d. To obtain the average volume of titrant, mL, calculate the average volume for two of the

𝑚𝑚𝑚𝑚 +𝑚𝑚𝑚𝑚1 𝑚𝑚𝑚𝑚3 +𝑚𝑚𝑚𝑚2 𝑚𝑚𝑚𝑚4 +𝑚𝑚𝑚𝑚3

volume added at a time, for example, 2

,

,

, and so on (label this

2

2

2

column as E) whose formula is = (A2 (cell) + A1 (cell))/2).

e. To obtain

∆𝑝𝑝𝑝𝑝

,

∆𝑚𝑚𝑚𝑚

divide ∆𝑝𝑝𝑝𝑝1/ ∆𝑚𝑚𝑚𝑚1, ∆𝑝𝑝𝑝𝑝2/ ∆𝑚𝑚𝑚𝑚2, ∆𝑝𝑝𝑝𝑝3/ ∆𝑚𝑚𝑚𝑚3, and so on (label this

column as F), whose formula is = D1(cell)/C1(cell).

f.

Plot

∆𝑝𝑝𝑝𝑝

∆𝑚𝑚𝑚𝑚

(y-axis) vs average volume of titrant, mL (x-axis) (columns E and F) to obtain

the 1 derivative curve.

g. To obtain Average of the average volume of titrant, mL (x-axis), this is the average of

𝐴𝐴𝐴𝐴𝐴𝐴.𝑚𝑚𝑚𝑚2 +𝐴𝐴𝐴𝐴𝐴𝐴.𝑚𝑚𝑚𝑚1 𝐴𝐴𝐴𝐴𝐴𝐴.𝑚𝑚𝑚𝑚3 +𝐴𝐴𝐴𝐴𝐴𝐴.𝑚𝑚𝑚𝑚2 𝐴𝐴𝐴𝐴𝐴𝐴.𝑚𝑚𝑚𝑚4 +𝐴𝐴𝐴𝐴𝐴𝐴.𝑚𝑚𝐿𝐿3

column F,

,

,

(label this column as G)

st

2

2

2

whose formula is = ((E2(cell) + E1(cell))/2).

h. To obtain the change of the average of the average volume of titrant ∆𝐴𝐴𝐴𝐴𝑔𝑔. 𝑚𝑚𝑚𝑚(1) subtract

Avg.mL2 – Avg.mL1, Avg.mL3 – Avg.mL2, Avg.mL4 – Avg.mL3, and so on from column

G (label this column as H), whose formula is = G2 – G1.

i.

F1(cell)/I1).

j.

2.

3.

4.

5.

6.

∆�

∆𝑝𝑝𝑝𝑝

�

∆𝑚𝑚𝑚𝑚

To obtain ∆𝐴𝐴𝐴𝐴𝑔𝑔.𝑚𝑚𝑚𝑚(2

(label this column as I) whose formula is = ((F2 (cell) –

)

∆�

∆𝑝𝑝𝑝𝑝

�

∆𝑚𝑚𝑚𝑚

Plot ∆𝐴𝐴𝐴𝐴𝑔𝑔.𝑚𝑚𝑚𝑚(2) (y-axis) vs Average of the average volume of titrant, mL

𝐴𝐴𝐴𝐴𝑔𝑔 𝑜𝑜𝑜𝑜𝐴𝐴𝐴𝐴𝑔𝑔. 𝑚𝑚𝑚𝑚(2) (x-axis) (columns I and G) to obtain the 2nd derivative curve. (The

endpoints should be similar to those found for the 1st derivative).

Calculate the weight of the unknown.

Plot: Titration Curve, 1st derivative curve, and 2nd derivative curve.

Identify the Equivalence point 1, Equivalence point 2, pKa1, pKa2, pKas halfway, and the pH, and

their suggested volume for each trial’s 1st and 2nd derivative plots. Remember pH = pKa.

Calculate the mass of Soda Ash in unknown sample for each trial.

Calculate the mass % Sodium Carbonate in Soda Ash unknown sample for each trial. From the

endpoint determination using the 1st and 2nd derivative and the mass of the unknown sample,

calculate the percent sodium carbonate of the unknown.

% Na2CO3 (s) =mL of HCl (aq) 𝑥𝑥 Molarity of HCl (aq) 𝑥𝑥

1 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 𝑁𝑁𝑁𝑁2 𝐶𝐶𝑂𝑂3

2 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 𝐻𝐻𝐻𝐻𝐻𝐻 (𝑎𝑎𝑎𝑎)

𝑥𝑥

0.10599 𝑁𝑁𝑁𝑁2 𝐶𝐶𝑂𝑂3

1 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 𝑁𝑁𝑁𝑁2 𝐶𝐶𝑂𝑂3

𝑥𝑥

7. Compare this mass % of Sodium Carbonate with that from experiment #4.

1

𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤ℎ𝑡𝑡

𝑥𝑥 100.

To determine the EQUIVALENCE POINT, find the midpoint of the steep portion in the graph, and there

is a fast and abrupt change of pH found at this point in the titration curve. The midpoint is the first point

at with the curve has zero slope, it is the most vertical point in the titration curve. Its corresponding

volume on the abscissa is the volume of HCl (aq) required to reach the equivalence point (Ep). The

equivalence point is the point at which the pH of the solution is equal to the dissociation constant of the

acid, pKa. Determine the pKa of the acid by graphically finding the pH corresponding to half the volume

of base necessary to reach the equivalence point. At this halfway point, the solution is a buffer with equal

concentrations of HA and A- and its pH is equal to the pKa of the unknown base. Computing the

Equivalence points in the 1st derivative plot is shown by the highest point/maximum slope on the curve

(peaks). As for the 2nd derivative curve, the equivalence points are located where the curve passes through

zero at the same point on the x-axis. Y-axis (pH) is zero at x-intercept volume, this point coincides with

the inflection point (point at which the curves changes direction) of the original data plot and the peak of

the first derivative. When the titration curve, 1st and 2nd derivative curves are overlaid it is noticeable that

the equivalence points coincide in the same place (volume) for all three curves, if it does not, then the

calculations of the derivatives must be checked or the titration has to be repeated, to get accurate pH and

volume readings.

This excel spreadsheet represents the titration of a weak base. Note: If a dibasic specie is used, then there

would be two end points.

mL

Formulas:

1

2

3

4

5

6

7

8

9

10

11

12

A

notebook

0

5

7.5

10

12.5

15

18

19

19.2

19.4

19.6

20

pH

B

notebook

11.70

10.84

10.58

10.36

10.14

9.88

9.41

9.08

8.98

8.85

8.67

8.35

C

∆𝑚𝑚𝑚𝑚

=A2-A1

5.0

2.5

2.5

2.5

2.5

3.0

1.0

0.2

0.2

0.2

0.4

0.5

D

∆𝑝𝑝𝑝𝑝

=B2-B1

-0.86

-0.26

-0.22

-0.22

-0.26

-0.47

-0.33

-0.10

-0.13

-0.18

-0.32

-0.42

E

Avg. Vol.

mL1

=(A2+A1)/2

2.5

6.25

8.75

11.25

13.75

16.5

18.5

19.1

19.3

19.5

19.8

20.25

F

∆𝑝𝑝𝑝𝑝

∆𝑚𝑚𝑚𝑚

=D1/C1

-0.172

-0.104

-0.088

-0.088

-0.104

-0.157

-0.330

-0.500

-0.650

-0.900

-0.800

-0.840

G

Avg of Vol

(E) Avg.mL1

(Avg.mL2)

=(E2+E1)/2

4.38

7.50

10.00

12.50

15.13

17.50

18.80

19.20

19.40

19.65

20.03

20.50

H

∆ 𝐴𝐴𝐴𝐴𝑔𝑔. 𝑚𝑚𝑚𝑚(

2)

=G2-G1

3.13

2.50

2.50

2.63

2.38

1.30

0.40

0.20

0.25

0.38

0.48

0.88

I

∆𝑝𝑝𝑝𝑝

∆�

�

∆𝑚𝑚𝑚𝑚

∆𝐴𝐴𝐴𝐴𝑔𝑔. 𝑚𝑚𝑚𝑚(2)

=(F2-F1)/H1

0.022

0.006

0.000

-0.006

-0.022

-0.133

-0.425

-0.750

-1.000

0.267

-0.084

0.251

T1 pH vs mL (B & A)

14.00

12.00

pH

10.00

8.00

6.00

4.00

2.00

0.00

0

10

20

30

40

50

40

50

mL HCl (aq)

T2 ∆𝑝𝑝𝐻𝐻/∆𝑚𝑚𝐿𝐿 vs Ave.mL (F & E)

0

∆𝑝𝑝𝐻𝐻/∆𝑚𝑚𝐿𝐿

-0.5

0

10

20

30

-1

-1.5

-2

-2.5

-3

Ave. mL

T3 ∆(∆𝑝𝑝𝐻𝐻/∆𝑚𝑚𝐿𝐿)/(∆𝐴𝐴𝑣𝑣𝑒𝑒.𝑚𝑚𝐿𝐿(2 )) vs Avg mL2

∆(∆𝑝𝑝𝐻𝐻/∆𝑚𝑚𝐿𝐿)/(∆𝐴𝐴𝑣𝑣𝑒𝑒.𝑚𝑚𝐿𝐿(2 ))

3.000

2.000

1.000

0.000

0.00

-1.000

10.00

20.00

30.00

-2.000

-3.000

-4.000

-5.000

Avg mL 2

40.00

50.00

Potentiometric Titration Curve

14.00

12.00

10.00

Measured response – pH

8.00

6.00

4.00

2.00

0.00

0

5

10

15

20

25

30

35

-2.00

-4.00

-6.00

Added volume of titrant

Titration Curve

1st Derivative

2nd Derivative

40

45

50

REPORT DATA SHEET

Calculations for mass of unknown, and % of unknown.

Avg M

HCl (aq)

Initial Calc. Eq.

Final

Calc.

Trial 1

M HCl

(aq)

Vol

(mL)

Vol (mL)

MW

Na2CO3

Millimoles

Na2CO3

g, soda ash

% Na2CO3

Avg

% of

Unknown

Mass of

Unknown

% Unknown

Avg.

Sd

Trial 2

Trial 3

The tables from excel for each trial needs to be included in the Data and Result, using the format below.

mL

Formulas:

1

2

A

notebook

pH

B

notebook

C

∆𝑚𝑚𝑚𝑚

=A2-A1

D

∆𝑝𝑝𝑝𝑝

=B2-B1

E

Avg. Vol.

mL1

=(A2+A1)/2

F

∆𝑝𝑝𝑝𝑝

∆𝑚𝑚𝑚𝑚

=D1/C1

G

Avg of Vol

(E) Avg.mL1

(Avg.mL2)

=(E2+E1)/2

H

∆ 𝐴𝐴𝐴𝐴𝑔𝑔. 𝑚𝑚𝑚𝑚(

2)

=G2-G1

I

∆𝑝𝑝𝑝𝑝

�

∆𝑚𝑚𝑚𝑚

∆𝐴𝐴𝐴𝐴𝑔𝑔. 𝑚𝑚𝑚𝑚(2)

=(F2-F1)/H1

∆�

Equivalence point information for each trial.

mL vs pH

mL

pH

1st derivative curve

pKa

mL

pH

2nd derivative curve

pKa

mL

pH

pKa

Eq1

Eq2

Average and Standard Deviation of Equivalence points.

Trials

Eq1 pH

Average

Sd

Trials

1

1

2

2

3

3

Eq2 pH

Average

Sd

Do not forget to include the Titration Curve, 1st, and 2nd derivative curves, along with the overlaid plots

for all three curves.

REFERENCES

1. Hage, D.S., & Carr, J.D. Analytical Chemistry and Quantitative Analysis. Pearson Education,

Inc. New Jersey, USA; 2011; pp 171, 173, 174, 176, 185, 283, 284, 321, 351, 357.

2. Harris, D. Quantitative Chemical Analysis. W.H. Freeman & Company. New York, USA; 2007;

8th Ed.; pp 213, 215.

3. Penrose, W.R. Chemistry 247 Quantitative Chemical Analysis Laboratory Manual. Illinois

Institute of Technology. Illinois, USA; January 2003 Ed.; pp 25, 26.

4. http://measurenet-tech.com/the-components/probeware/ph/index.html

5. https://opentextbc.ca/chemistry/chapter/14-7-acid-base-titrations/

6. Potentiometric Titration of Soda Ash.

https://people.wou.edu/~postonp/ch312/pdf/Lab3_Soda_Ash_w11.pdf

7. Potentiometric Acid-Base Titration. Brooklyn College, New York, USA.

http://academic.brooklyn.cuny.edu/chem/maggie/teach/chem41/files/ph.pdf

8. http://www.glimme.net/apchem/Ch14-15/AqueousEquilibrium-4-Handout.pdf

EXPERIMENT #5 POTENTIOMETRIC TITRATION OF A WEAK BASE

CHEM 322 QUANTITATIVE ANALYSIS LABORATORY

DATA SET A

Avg M HCl (aq)

Vol (mL)

0.0855

Initial Calc Eq

Final Calc

Trial 1

Trial 2

Trial 3

M HCl

Vol (mL)

0.0855

0.0855

0.0855

MW Na2CO3

18

105.99

MiliM Na2CO3 g soda ash

0.10599

0.10599

0.10599

TRIAL 1 TEST RUN

A

B

VOL (HCl), mL

C

A2-A1

pH

0

3

6.4

9

13

15.5

17.6

19

19.2

19.4

19.6

20

21.2

22

24

27

30

32.8

36

39

39.2

39.4

39.6

39.8

40

ΔmL

13.8

11.15

10.65

10.36

10.14

9.78

9.41

9.18

8.98

8.85

8.67

8.48

7.77

7.42

7.1

6.87

6.54

5.97

5.4

5.11

4.96

4.83

4.65

4.34

3.85

D

B2-B1

ΔpH

E

(A2+A1)/2

AvgmL1

40.8

41

42

44

45

49

3.45

3.12

2.85

2.49

2.39

2.25

mL vs pH

mL

pH

pKa

mL

1st derivative

pH

pKa

mL

2nd derivative

pH

pKa

Eq pt 1

Eq pt 2

Eq pt 1

Eq pt 2

Eq pt 1

Eq pt 2

% Na2CO3 Avg

Mass of Unknown

52

% of Unknown

Average

Sd

This is calculated for each trial using VOLUME of ep

TRIAL 1 TEST RUN

F

D1/C1

ΔpH/ΔmL

G

(E2+E1)/2

Avg of AvgmL1

H

G2-G1

Δ Avg of AvgmL1

I

(F2-F1)/H1

(Δ(ΔpH/ΔmL))/ Δave.mL (2)

Trials Eqpt 1 pH

1

2

3

Average

Sd

Trials Eqpt 2 pH

1

2

3

Average

Sd

Trials %Na2CO3

1

2

3

Average

Sd

Plot 1 ph vs vhlc

plot 2 ΔpH/ΔmL vs Ave.mL (1)

plot 3 ∆(∆pH/∆mL)/∆ml vs mlavg2

Y vs X

(B vs A)

(F vs E)

(I vs G)

Titration curve

1st Derivative

2nd Derivative

h trial using VOLUME of ep1 of the 2nd derivative pH.

A

B

VOL (HCl), mL

C

A2-A1

ΔmL

pH

0

6

8.5

11

13.5

16

19

20

20.2

20.4

20.6

21

21.5

22

24

27

31

35

39

40

40.2

40.4

40.6

40.8

41

12.7

11.84

11.58

11.36

11.14

10.88

10.41

10.08

9.98

9.85

9.67

9.35

8.93

8.62

8.09

7.71

7.34

6.97

6.39

6.06

5.96

5.83

5.65

5.34

4.88

41.2

41.4

42

43

44

48

4.42

4.12

3.72

3.43

3.37

2.91

mL

mL vs pH

pH

mL

1st derivative

pH

mL

2nd derivative

pH

Eq pt 1

Eq pt 2

Eq pt 1

Eq pt 2

Eq pt 1

Eq pt 2

TRIAL 2

D

B2-B1

ΔpH

E

F

(A2+A1)/2

D1/C1

AvgmL1

ΔpH/ΔmL

G

(E2+E1)/2

Avg of AvgmL1

H

G2-G1

Δ Avg of AvgmL1

L vs pH

pKa

derivative

pKa

derivative

pKa

Trials Eqpt 1 pH

1

2

3

Average

Trials Eqpt 2 pH

1

2

3

Average

Trials %Na2CO3

Average

1

2

3

I

(F2-F1)/H1

(Δ(ΔpH/ΔmL))/ Δave.mL (2)

A

B

VOL (HCl), mL

0

5

7.5

10

12.5

15

18

19

19.2

19.4

19.6

20

20.5

21

23

26

30

34

38

39

39.2

39.4

39.6

39.8

40

pH

11.7

10.84

10.58

10.36

10.14

9.88

9.41

9.08

8.98

8.85

8.67

8.35

7.93

7.62

7.09

6.71

6.34

5.97

5.39

5.06

4.96

4.83

4.65

4.34

3.88

40.2

40.4

41

42

43

47

3.42

3.12

2.72

2.43

2.37

2.23

mL vs pH

Sd

mL

Eq pt 1

Eq pt 2

1st derivat

Sd

mL

Eq pt 1

Eq pt 2

Sd

2nd derivat

mL

Eq pt 1

Eq pt 2

C

A2-A1

ΔmL

D

B2-B1

ΔpH

E

F

(A2+A1)/2

D1/C1

AvgmL1

ΔpH/ΔmL

TRIAL 3

G

(E2+E1)/2

Avg of AvgmL1

mL vs pH

pH

1st derivative

pH

pKa

pKa

Trials Eqpt 1 pH

1

2

3

Trials Eqpt 2 pH

1

2

3

Trials %Na2CO3

2nd derivative

pH

pKa

1

2

3

H

G2-G1

Δ Avg of AvgmL1

I

(F2-F1)/H1

(Δ(ΔpH/ΔmL))/ Δave.mL (2)

Average

Sd

Average

Sd

Average

Sd

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