Titration of Unknown Acid Chemistry Lab Questions

Acid-Base ReactionsIntroduction
You know that vinegar, lemon juice and many other foods taste sour. However, it was not until
a few hundred years ago that it was discovered why – they are all acids. The term acid, in fact,
comes from the Latin term acere, which means “sour”. Although there are many slightly
different definitions of acids and bases, in this lesson we will introduce the fundamentals of
acid/base chemistry.
Definitions of Acid and Bases
In the seventeenth century, the Irish writer and amateur chemist Robert Boyle first labeled
substances as either acids or bases by the following characteristics:
• Acids taste sour, are corrosive to metals, change litmus red, and become less acidic
when mixed with bases.
• Bases feel slippery, change litmus blue, and become less basic when mixed with acids.
Arrhenius
While different chemists tried to explain why acids and bases behave the way they do, the first
practical definition of acids and bases was proposed in the late 1800’s by Svante Arrhenius. He
proposed that water can dissolve many compounds by separating them into their individual
ions. Arrhenius suggested that acids are compounds that contain hydrogen and can be dissolved
in water to release hydrogen ions into the solution. For example, hydrochloric acid (HCl)
dissolves in water as follows:
Equation 1
HCl(aq) + H2O (l)→ H3O+(aq) + Cl-(aq)
You can see that the acid (HCl) separates into the ions: H+ (H3O+) and Cl-.
Arrhenius then defined bases as substances that dissolve in water to release hydroxide ions
(OH-) into solution. For example, a typical base such as NaOH (sodium hydroxide) will
dissolve in water and produce HO- and Na+ ions in solution.
Equation 2
NaOH(aq) → HO-(aq) + Na+(aq)
Neutralization
As you can see from the above equations, acids release H+ and bases release OH- into solution.
If we were to mix an acid and base together, the H+ ion would combine with the OH- ion to
make the molecule H2O.
Equation 3
H+(aq) + OH- (aq) → H2O(l)
The neutralization reaction of an acid with a base will produce water and a salt, as shown in the
examples below:
Equation 4
HCl(aq) + NaOH(aq) → H2O(l) + NaCl(aq)
Acid
Base
Water
Salt
Equation 5
HBr(aq) + KOH(aq) → H2O(l) + KBr(aq)
Acid
Base
Water
Salt
When we place 20 mL of an acid in a flask and begin to add a base like NaOH from a buret, the
following neutralization reaction will occur:
HCl(aq) + NaOH(aq) → H2O(l) + NaCl(aq)
Acid
Base
Water
Salt
Figure 1
(from http://wps.prenhall.com/wps/media/objects/3311/3390507/blb0406.html)
How does an acid-base reaction work?
Let us assume that we have 100 molecules of acid (H+) in the flask as shown in the image above
(Figure 1). We also assume that each drop from the buret adds one molecule of base (OH-) into
the flask. Before we add any base there are 100 molecules of acid (H+) in solution.
When we add one drop of base (OH-), it reacts with one of the acid molecules (H+) in solution,
leaving 99 molecules of acid (H+) remaining in the flask.
When we add 50 drops of base (OH-), 50 acid molecules (H+) react with the added molecules of
base. We would then have 50 molecules of acid (H+) remaining in the flask.
Finally, we add 100 drops of base (OH-) into the flask and all of the
acid molecules (H+) react with the base. There is no more acid or
base in solution, only salt and water. When this happens, we say that
the acid has been neutralized and refer to this as the equivalence
point. Since we cannot see the reaction taking place, we use organic
compounds called acid-base indicators to tell us when all the acid has
been consumed. The most common indicator used is phenolphthalein.
It is clear in an acidic solution and pink in a basic solution (Figure 2).
Figure 2
(from http://www.titrations.info/acid-base-titration-sulfuric-acid)
Name:
Last, First
General Chemistry 1: Week 11 – Titration
Standardization of NaOH
Total Volume NaOH
Moles NaOH
Molarity of NaOH
Mean NaOH Molarity
Percent Error
Standardization of Unknown Acid
Selected Trials
Initial buret reading of NaOH (mL)
Final buret reading of NaOH(mL)
Total Volume of NaOH (mL)
Molarity of NaOH
Volume of Unknown Acid (mL)
Molarity of Unknown Acid
Avg Molarity
Percent Error
10,00
Titration
Lab time:
(Copy exactly from right)
North or South
(Full word)
Acid Name:
Do NOT enter units into the
cells
Enter values exactly as you
have them written on your
paper
Options:
T8
T11
T2
W11
W2
R8
Standardization of NaOH Part A
Preparation of KHP Tuesday 11 South
KHP
Trial 1
Trial 2
Trial 3
Trial 4
Empty Container Mass (Grams)
135.000
118.360
159.230
195.630
Container with KHP (Grams)
135.568
118.925
159.951
196.430
Mass of KHP (grams)
Moles KHP
Standardization of NaOH
NaOH
Trial 1
Trial 2
Trial 3
Trial 4
Initial Buret Reading
(mL)
3.00
6.98
3.75
10.36
Final Buret Reading
(mL)
30.96
34.96
39.13
14.18
Total Volume of NaOH
delivered
(mL)
Moles of NaOH
Molarity of NaOH
Mean Molarity Value
of Best 3 Trials
% Error
Titration of an Unknown Acid Part B
Tues 11 South
Acid Trial 1
Acid Trial 2
Acid Trial 3
Acid Trial 4
Initial Buret Reading
of NaOH
5.00
6.00
8.25
27.36
Final Buret Reading
of NaOH
40.00
46.12
43.02
62.59
10.00 mL
10.00 mL
10.00 mL
10.00 mL
Total Volume of NaOH
used in Titration
Mean Molarity from the
Standardization of NaOH from Part A
Moles of NaOH
Volume of Unknown Acid Titrated
Moles of Unknown Acid
Molarity of Unknown Acid
Average Unknown Acid Molarity for
selected Trials
Percent Error for each titration
selected
Standardization of NaOH Part A
Preparation of KHP Tuesday 11 North
KHP
Trial 1
Trial 2
Trial 3
Trial 4
Empty Container Mass
(Grams)
135.000
118.360
159.230
195.630
Container with KHP (Grams)
135.568
118.925
159.951
196.430
Mass of KHP (grams)
Moles KHP
Standardization of NaOH
NaOH
Trial 1
Trial 2
Trial 3
Trial 4
Initial Buret Reading (mL)
3.00
6.98
3.75
10.36
Final Buret Reading
(mL)
30.96
34.96
39.13
14.18
Total Volume of NaOH
delivered
(mL)
Moles of NaOH
Molarity of NaOH
Mean Molarity Value of
Best 3 Trials
% Error
Titration of an Unknown Acid Part B
Tues 11 North
Acid Trial 1
Acid Trial 2
Acid Trial 3
Acid Trial 4
Initial Buret Reading
of NaOH (mL)
2.30
6.00
8.25
23.25
Final Buret Reading
of NaOH (mL)
32.48
43.73
38.33
53.63
10.00
10.00
10.00
10.00
Total Volume of NaOH
used in Titration (mL)
Mean Molarity from the
Standardization of NaOH from Part A
Moles of NaOH
Volume of Unknown Acid Titrated
(mL)
Moles of HCl
Molarity of HCl
Mean HCl Molarity for selected Trials
Percent Error for each titration
selected
Titration of an Unknown Acid Part B
Acid Trial 1
Acid Trial 2
Acid Trial 3
Acid Trial 4
10
10
10
10
Initial Buret Reading
of NaOH (mL)
Final Buret Reading
of NaOH (mL)
Total Volume of NaOH
used in Titration (mL)
Enter Mean Molarity from the
Standardization of NaOH from Part A
Moles of NaOH
Volume of Unknown Acid Titrated
(mL)
Moles of HCl
Molarity of HCl
Average HCl Molarity for selected
Trials
Percent Error for each titration
selected
Acid-Base Reactions
Introduction
You know that vinegar, lemon juice and many other foods taste sour. However, it was not until
a few hundred years ago that it was discovered why – they are all acids. The term acid, in fact,
comes from the Latin term acere, which means “sour”. Although there are many slightly
different definitions of acids and bases, in this lesson we will introduce the fundamentals of
acid/base chemistry.
Definitions of Acid and Bases
In the seventeenth century, the Irish writer and amateur chemist Robert Boyle first labeled
substances as either acids or bases by the following characteristics:
• Acids taste sour, are corrosive to metals, change litmus red, and become less acidic
when mixed with bases.
• Bases feel slippery, change litmus blue, and become less basic when mixed with acids.
Arrhenius
While different chemists tried to explain why acids and bases behave the way they do, the first
practical definition of acids and bases was proposed in the late 1800’s by Svante Arrhenius. He
proposed that water can dissolve many compounds by separating them into their individual
ions. Arrhenius suggested that acids are compounds that contain hydrogen and can be dissolved
in water to release hydrogen ions into the solution. For example, hydrochloric acid (HCl)
dissolves in water as follows:
Equation 1
HCl(aq) + H2O (l)→ H3O+(aq) + Cl-(aq)
You can see that the acid (HCl) separates into the ions: H+ (H3O+) and Cl-.
Arrhenius then defined bases as substances that dissolve in water to release hydroxide ions
(OH-) into solution. For example, a typical base such as NaOH (sodium hydroxide) will
dissolve in water and produce HO- and Na+ ions in solution.
Equation 2
NaOH(aq) → HO-(aq) + Na+(aq)
Neutralization
As you can see from the above equations, acids release H+ and bases release OH- into solution.
If we were to mix an acid and base together, the H+ ion would combine with the OH- ion to
make the molecule H2O.
Equation 3
H+(aq) + OH- (aq) → H2O(l)
The neutralization reaction of an acid with a base will produce water and a salt, as shown in the
examples below:
Equation 4
HCl(aq) + NaOH(aq) → H2O(l) + NaCl(aq)
Acid
Base
Water
Salt
Equation 5
HBr(aq) + KOH(aq) → H2O(l) + KBr(aq)
Acid
Base
Water
Salt
When we place 20 mL of an acid in a flask and begin to add a base like NaOH from a buret, the
following neutralization reaction will occur:
HCl(aq) + NaOH(aq) → H2O(l) + NaCl(aq)
Acid
Base
Water
Salt
Figure 1
(from http://wps.prenhall.com/wps/media/objects/3311/3390507/blb0406.html)
How does an acid-base reaction work?
Let us assume that we have 100 molecules of acid (H+) in the flask as shown in the image above
(Figure 1). We also assume that each drop from the buret adds one molecule of base (OH-) into
the flask. Before we add any base there are 100 molecules of acid (H+) in solution.
When we add one drop of base (OH-), it reacts with one of the acid molecules (H+) in solution,
leaving 99 molecules of acid (H+) remaining in the flask.
When we add 50 drops of base (OH-), 50 acid molecules (H+) react with the added molecules of
base. We would then have 50 molecules of acid (H+) remaining in the flask.
Finally, we add 100 drops of base (OH-) into the flask and all of the
acid molecules (H+) react with the base. There is no more acid or
base in solution, only salt and water. When this happens, we say that
the acid has been neutralized and refer to this as the equivalence
point. Since we cannot see the reaction taking place, we use organic
compounds called acid-base indicators to tell us when all the acid has
been consumed. The most common indicator used is phenolphthalein.
It is clear in an acidic solution and pink in a basic solution (Figure 2).
Figure 2
(from http://www.titrations.info/acid-base-titration-sulfuric-acid)
Name:
Last, First
General Chemistry 1: Week 11 – Titration
Standardization of NaOH
Total Volume NaOH
Moles NaOH
Molarity of NaOH
Mean NaOH Molarity
Percent Error
Standardization of Unknown Acid
Selected Trials
Initial buret reading of NaOH (mL)
Final buret reading of NaOH(mL)
Total Volume of NaOH (mL)
Molarity of NaOH
Volume of Unknown Acid (mL)
Molarity of Unknown Acid
Avg Molarity
Percent Error
10.00
Titration
Lab time:
(Copy exactly from right)
North or South
(Full word)
Acid Name:
Do NOT enter units into the
cells
Enter values exactly as you
have them written on your
paper
Options:
T8
T11
T2
W11
W2
R8
Determination of the Concentration of the NaOH Solution:
Part A
1. Balanced Equation for the reaction of KHP and NaOH:
NaOH (aq) + KHC8H4O4 (aq) →
Acid
Base
HOH (l) + KNaC8H4O4 (aq)
Water
Salt
2. The Grams of KHP
 On the data table provided, we have provided the mass of the
An empty container and same container w/ KHP.
 Use mass by difference to calculate the grams of KHP
3. Convert Grams to Moles
 Use the following molar mass for KHP in this exercise
 Molar mass of KHP 204.2g = 1 mole KHP
𝑀𝑜𝑙𝑒𝑠 𝐾𝐻𝑃 = 𝐺𝑟𝑎𝑚𝑠 𝐾𝐻𝑃 𝑥
1 𝑀𝑜𝑙𝑒 𝐾𝐻𝑃
204.2 𝐺 𝐾𝐻𝑃
 3 significant figures are required: 0.00356 (Yes)
0.003(No)
4. Moles of NaOH
You have completed the Titration and you stopped at the pale pink endpoint.
o What does this mean? The reaction is over. All the acid has been reacted.
Question: How many moles of NaOH where used in the reaction?
Moles of KHP = Moles of NaOH
Why?
# 𝑀𝑜𝑙𝑒𝑠 𝑜𝑓 𝐾𝐻𝑃 𝑥
1 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑁𝑎𝑂𝐻
= # 𝑀𝑜𝑙𝑒𝑠 𝑜𝑓 𝑁𝑎𝑂𝐻
1 𝑀𝑜𝑙𝑒 𝑜𝑓 𝐾𝐻𝑃
Therefore, if you calculated that you had 0.00325 moles of KHP then
you would have the same numbers of Moles of NaOH: 0.00325 .
5. Calculate the Volume of NaOH Delivered
a. We have provided you with the initial and final buret reading for each
trial. Use the Volume by difference calculation to determine the
volume delivered:
Final Buret Reading – Initial Burt Reading = Volume Delivered (mL)
6. To Calculate Concentration of NaOH (Molarity)
# 𝑚𝑜𝑙𝑒𝑠 𝑁𝑎𝑂𝐻
1000 𝑚𝐿 𝑀𝑜𝑙𝑒𝑠
𝑥
=
𝑫𝒆𝒍𝒊𝒗𝒆𝒓 𝑽𝒐𝒍𝒖𝒎𝒆 𝒐𝒇 𝑵𝒂𝑶𝑯 (𝒎𝑳)
1 𝐿𝑖𝑡𝑒𝑟
𝐿𝑖𝑡𝑒𝑟
7. Calculate the Mean Value for your best three trials
8. Calculate the percent error for your best three trials
𝑇𝑟𝑖𝑎𝑙 𝑀𝑜𝑙𝑎𝑖𝑟𝑦
𝑀𝑒𝑎𝑛 𝑀𝑜𝑙𝑎𝑙𝑟𝑖𝑡𝑦
Proceed to Part B
𝑥 100
Determination of Unknown Acid Concentration:
Part B
1. Balanced Equation for the reaction of HCL and NaOH:
HCl(aq) + NaOH(aq) → H2O(l) + NaCl(aq)
Acid
Base
Water
Salt
2. The moles of NaOH is calculated as follows
𝑀𝑜𝑙𝑒𝑠 𝑁𝑎𝑂𝐻 = 𝑀𝑜𝑙𝑎𝑟𝑖𝑡𝑦 𝑜𝑓 𝑁𝑎𝑂𝐻 𝑋 𝑉𝑜𝑙𝑢𝑚𝑒 𝑁𝑎𝑂𝐻 𝐿𝑖𝑡𝑒𝑟𝑠
i.
The 𝑀𝑜𝑙𝑎𝑟𝑖𝑡𝑦 𝑜𝑓 𝑁𝑎𝑂𝐻 comes from the mean Molarity value
determined in the standardization of NaOH Part A.
ii.
To calculate the volume of NaOH Deliver use the technique:
Volume by difference:
Final Buret Reading – Initial Burt Reading Volume Delivered
iii.
At the pale pink end point the reaction is complete
𝑀𝑜𝑙𝑒𝑠 𝑜𝑓 𝑁𝑎𝑂𝐻 𝑥
1 𝑚𝑜𝑙𝑒𝑠 𝐻𝐶𝐿
= 𝑀𝑜𝑙𝑒𝑠 𝑜𝑓 𝐻𝐶𝑙
1 𝑀𝑜𝑙𝑒 𝑜𝑓 𝑁𝑎𝑂𝐻
Moles of HCl = Moles of NaOH
iv.
Calculate Concentration of HCL (Molarity = Moles/Liter)
? 𝑚𝑜𝑙𝑒𝑠 𝐻𝐶𝐿
1000 𝑚𝐿 𝑀𝑜𝑙𝑒𝑠
𝑥
=
𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝐴𝑐𝑖𝑑
1 𝐿𝑖𝑡𝑒𝑟
𝐿𝑖𝑡𝑒𝑟
The volume of the acid was 10.00 mL
? 𝑚𝑜𝑙𝑒𝑠 𝐻𝐶𝐿 1000 𝑚𝐿 𝑀𝑜𝑙𝑒𝑠
𝑥
=
10.00
1 𝐿𝑖𝑡𝑒𝑟
𝐿𝑖𝑡𝑒𝑟
Standardization of NaOH Part A
Preparation of KHP
Trial 1
KHP
Trial 2
Trial 3
Trial 4
Empty Container Mass
(Grams)
Container with KHP (Grams)
Mass of KHP (grams)
Moles KHP
Standardization of NaOH
NaOH
Initial Buret
Reading (mL)
Final Buret Reading
(mL)
Total Volume of
NaOH delivered
(mL)
Moles of NaOH
Molarity of NaOH
Mean Molarity
Value of Best 3
Trials
% Error od Best 3
trials
Trial 1
Trial 2
Trial 3
Trial 4
Titration of an Unknown Acid Part B
Acid Trial 1
Acid Trial 2
Acid Trial 3
Acid Trial 4
10
10
10
10
Initial Buret Reading
of NaOH (mL)
Final Buret Reading
of NaOH (mL)
Total Volume of NaOH
used in Titration (mL)
Enter Mean Molarity from the
Standardization of NaOH from Part A
Moles of NaOH
Volume of Unknown Acid Titrated
(mL)
Moles of HCl
Molarity of HCl
Average HCl Molarity for selected
Trials
Percent Error for each titration
selected