32
Neutral or Not?
Neutral or Not?
Exploring Salt Hydrolysis
OBJECTIVE
Students will prepare a salt solution of assigned molarity and determine the pH of the solution using a
pH meter. They will then be required to explain via equations and calculations why some salts are
neutral and others are not.
T E A C H E R
P A G E S
LEVEL
Chemistry
NATIONAL STANDARDS
UCP.2, UCP.3, A.1, A.2, B.2, B.3, E.1, E.2, F.4
CONNECTIONS TO AP
AP Chemistry:
III. Reactions C. Equilibrium 2. Quantitative treatment b. Equilibrium constants for reactions in
solution 3. Common ion effect; buffers; hydrolysis
TIME FRAME
45 minutes class time; 45 minutes homework
MATERIALS
(For a class of 28 working in pairs)
7 pH meters or probes (2 lab groups per meter)
7 wash bottles of distilled water
several balances
weighing paper
14 50 mL graduated cylinders or
50 mL volumetric flasks
14 100 mL beakers
14 stirring rods
Select at least five salt samples such as these:
(1 known sample per lab pair of ≈ 5 g each)
NaHCO3 sodium bicarbonate (basic)
KHCO3 potassium bicarbonate (basic)
CaC2H3O2 calcium acetate (basic)
NH4Cl ammonium chloride (acidic)
(NH4)2SO4 ammonium sulfate (acidic)
NaCl sodium chloride (neutral)
NH4C2H3O2 ammonium acetate (neutral)
K2SO4 potassium sulfate (neutral)
NH4Cl ammonium chloride (acidic)
*If pH meters are not readily available, wide-range indicator paper (0–14) may also be used. In fact, it
is often interesting to have students use both pH meters and pH paper. Students can then compare the
two results and even calculate the percent of difference between them.
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TEACHER NOTES: PRIOR TO DATA COLLECTION
This experiment should follow a study of acids and bases so that students will already be familiar with
the proper set up and use of the pH meter.
The first part of this exercise requires that students prepare a 0.25 M solution of an assigned salt.
You will need to provide them with the chemical formula (including waters of hydration) so that the
appropriate molar mass is calculated. For example, calcium acetate is commonly sold as
Ca(C2H3O2)2 γ H2O. If the water of hydration is not considered, the molarity will not be calculated
correctly.
The second part of the lesson requires that you guide the students through the chemical logic that
explains their observations. The term for this phenomenon is salt hydrolysis. Teacher notes are
provided for you to use in your discussion, however you may wish to share them with the students at the
conclusion of the laboratory exercise. In order for students to be successful with this portion of the
activity, they must have experience with Ka, Kb, and pH calculations prior to this lesson. By introducing
salt hydrolysis and their pH calculation, students will have a good background when attempting buffer
calculations in AP* Chemistry.
Laboratory comments:
It is always a good idea to run the experiment ahead of time to ensure that the results are as you expect.
The pH of the distilled water must be checked in advance. The pH meters must be calibrated for
accuracy.
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P A G E S
The students are asked to gather information from four lab groups with salts different than their own.
Therefore, you must have at least five different salts available or modify the number of samples needed.
A list of suggested salts is found in the materials list, but many other soluble salts could be used. If
using other salts, make sure that you check the pH of the salts before distributing. Hopefully, this lab
will remove the misconception that all salts are neutral and will pique students’ curiosity as to why this
is so.
T E A C H E R
Before beginning, review with students the proper technique for preparing the solution. The sample
should be massed and then placed into a 50 mL volumetric flask or 50 mL graduated cylinder.
If students use weighing paper, the solid is washed from the paper into the volumetric flask with distilled
water prior to filling to the appropriate mark on the glassware. It is always a good idea to add about half
of the water, mix, and allow the salt to completely dissolve before filling to final volume. Remind
students to mix the final solution well before measuring the pH. Students should pour their solution into
the 100 mL beaker before trying to measure the pH.
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Neutral or Not?
TEACHER NOTES: AFTER COLLECTION OF DATA
Part I: Determining the relative pH of salts and an introduction to writing hydrolysis reactions.
The products of any acid-base neutralization reaction are a salt and water. The hydrogen ion from the
acid reacts with the hydroxide ion from the base to give water. The anion from the acid reacts with the
cation from the base to give a salt. This salt can be neutral, acidic or basic. Determining the qualitative
pH of the aqueous salt solution is easy as long as you remember the strong acids and the strong bases.
T E A C H E R
P A G E S
Consider KHCO3. First, determine which acid reacted with which base to form this salt. This salt could
have been formed from potassium hydroxide and carbonic acid. Next, determine the relative strength of
the acid and the base. KOH is a strong base and H2CO3 is a weak acid. “Strong wins” and determines
the pH of the resulting salt. K2HCO3 is therefore a basic or alkaline salt. Furthermore, the cation or
anion of the “winner” is also a spectator ion. When this salt reacts with water, a hydrolysis reaction
occurs that produces a basic solution.
By considering the strength of the acid and base that reacted to produce the salt, the following
combinations are possible:
● strong base + strong acid = neutral salt (no hydrolysis)
● strong base + weak acid = basic salt (therefore, OH− is a product in the hydrolysis reaction)
● weak base + strong acid = acidic salt (therefore, H+ or H3O+ is a product in the hydrolysis reaction)
● weak base + weak acid = ? The determination depends on the Ka and Kb values.
Whichever is greater wins!
This makes the process of determining relative pH fairly straightforward. But how can we determine the
true pH of a salt solution quantitatively? We can calculate it!
Soluble salts are electrolytes and therefore ionize in solution. For the purpose of this lab exercise, all of
the salts used will be soluble, so we will assume that they all ionize 100%. If we know the concentration
of the original salt then we know the concentration of the components of that salt.
Example 1: What is the relative pH of a 0.50 M NaCl solution? Justify your answer.
Solution: The salt ionizes 100% and all species are in a one to one mole ratio as shown below.
NaCl(s) → Na+(aq) + Cl−(aq)
Since the salt is neutral, the relative pH of the solution is seven.
A strong acid reacted with a strong base to produce neutral salt. Sodium ion comes from the strong
base, sodium hydroxide. Chloride ion originates from the strong acid, hydrochloric acid. Neither ion
reacts with water so the solution is neutral and no hydrolysis occurs.
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Example 2: What is the relative pH of potassium bicarbonate, KHCO3? Justify your answer. The salt
will ionize 100% as shown below.
KHCO3(s) → K+(aq) + HCO3−(aq)
Solution: Potassium bicarbonate has a relative pH greater than seven. Writing a correct hydrolysis
reaction serves as justification.
To write a correct hydrolysis reaction, students should first ask themselves, “Which acid and which
base reacted to form this salt?” Once KOH and H2CO3 are identified, they should next ask themselves
“Is potassium hydroxide a strong or weak base? Is carbonic acid a strong or weak acid?”
⎯⎯
→ OH−
+ HOH ←⎯
⎯
(It may be helpful for students to write water as HOH until they get the hang of this.) The ion that was
contributed by the strong is also a spectator ion (K+ in this case). That means the ion from the weak
component, is the participant in the hydrolysis reaction (HCO3− in this case). The bicarbonate ion is the
other reactant in the hydrolysis reaction. Students now write:
In this format, it is easier for the student to see that the hydroxide ion remains in solution and is
responsible for the alkaline pH, and that the remaining hydrogen ion will bind to the bicarbonate ion to
produce the weak acid carbonic acid. Remind students that weak acids dissociate less than 10% as a
general rule. The final hydrolysis reaction is represented below.
⎯⎯
→ OH− + H2CO3
HCO3− + HOH ←⎯
⎯
Students should re-write the equation as shown below:
⎯⎯
→ H2CO3 + OH−
HCO3− + H2O ←⎯
⎯
Example 3: Will a solution of ammonium bromide be acidic, basic or neutral? Justify your answer.
NH4Br(s) → NH4+(aq) + Br−(aq)
Solution: Acidic. The base that reacted to form the salt is ammonia, a weak base. The acid that reacted
is HBr, a strong acid and the bromide ion is a spectator ion. Remember, “strong wins” therefore the
solution will be acidic. Students begin by writing
⎯⎯
→ H+
HOH ←⎯
⎯
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P A G E S
⎯⎯
→ OH−
HCO3− + HOH ←⎯
⎯
T E A C H E R
Once they have determined that a strong base reacted with a weak acid, it should be obvious that
“strong wins” and the resulting salt solution will be basic. Furthermore, we know OH− must be a
product if a basic salt solution results. Have them write the following on their paper once they
determine a basic salt is formed.
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Neutral or Not?
Since we established bromide ion is a spectator, the only other reactant possibility is the ammonium ion,
so the hydrolysis equation now becomes:
⎯⎯
→ H+
NH4+ + HOH ←⎯
⎯
Now students can see that ammonium ion is reacting with hydroxide ion to produce ammonium
hydroxide, a weak base.
⎯⎯
→ H+ + NH4OH
NH4+ + HOH ←⎯
⎯
⎯⎯
→ H3O+ + NH4OH
Students should re-write the equation as: NH4+ + H2O ←⎯
⎯
T E A C H E R
P A G E S
Note that credit is given on the AP* Chemistry exam for either NH3, H+ or H3O+ as products.
Part II: Calculating the pH of salts.
Can we calculate the exact pH of a salt? Yes! In order to do so, we must know the following:
• the concentration of the salt solution
• the hydrolysis reaction for the salt unless it is a neutral salt in which case its pH = 7.0
• the appropriate Ka or Kb value from a list of ionization constants (a sample list is found at the
conclusion of the notes)
• ionization constants can be calculated using this relationship: Kw = Ka × Kb = 1.0 × 10−14
• how to set up the equilibrium expression, substitute values, and solve for the concentration of
hydronium ion or hydroxide ion, whichever is appropriate
• how to calculate pH and relate it to pOH and Kw
Example 4: Calculate the pH of a 0.50 M solution of HCO3. The Ka for carbonic acid is 4.2 × 10−7.
The salt will ionize 100% as follows:
KHCO3(s) → K+(aq) + HCO3−(aq)
Solution: The pH = 10.04. Note there were 2 sig. figs. on the initial concentration which requires 2
decimal places on the pH value.
A strong base reacted with a weak acid to form the potassium bicarbonate salt which hydrolyzes.
Recall that “strong wins” thus the resulting solution will be basic (OH− will be a product). The ion that
was contributed by the strong is also a spectator ion (K+ in this case). That means the bicarbonate ion
participates as a reactant in the hydrolysis reaction.
⎯⎯
→ OH− + H2CO3 which is re-written as HCO3− + H2O ←⎯
⎯⎯
→ H2CO3 + OH−
HCO3− + HOH ←⎯
⎯
⎯
Since OH− is present as a product, the equilibrium is alkaline and we must determine the Kb value for
bicarbonate ion. The Ka for carbonic acid is 4.2 × 10−7. Use the relationship for Kw as follows:
K w 1.0 × 10−14
Kb =
=
= 2.4 × 10−8
−7
K a 4.2 × 10
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The next step is to set up the Kb expression and solve for pH:
⎯⎯
→ H 2CO3 (aq) + OH − (aq)
R: HCO3− (aq) + H 2O(aq ) ←⎯
⎯
I:
0.50 M
0
0
C:
−x
+x
+x
x
x
E: 0.50 − x
“neglect
Neglect x since 0.50 is much larger than 100K a .
Instruct students to complete their ANALYSIS and CONCLUSION questions for the next class session.
HYPOTHESIS
Student answers will vary. One possible hypothesis might be worded as follows:
• I expect that my salt, potassium bicarbonate, will be basic since it was formed from a strong base
and a weak acid.
[ H 2CO3 ] ⎡⎣OH − ⎤⎦
⎡⎣ HCO3− ⎤⎦
x 2 = 1.2 × 10−8
= 2.4 × 10−8 =
x2
0.50
P A G E S
Kb =
x = ⎡⎣OH − ⎤⎦ = 1.2 × 10−8 = 1.1 × 10−4
∴ pOH = − log (1.1 × 10−4 ) = 3.96
∴ pH=14 − pOH = 14 − 3.96 = 10.04
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T E A C H E R
POSSIBLE ANSWERS TO THE CONCLUSION QUESTIONS AND SAMPLE DATA
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Neutral or Not?
DATA AND OBSERVATIONS
Chemical Formula for Salt
pH Observed
KHCO3
10.02
NH4Cl
5.04
NH4C2H3O2
7.02
NaHCO3
9.98
NaCl
7.04
T E A C H E R
P A G E S
ANALYSIS
1. Show the calculations for the preparation of your salt solution:
50. mL ×
•
1 L
0.25 mol 100.12 g KHCO3
×
= 1.25 g KHCO3
×
1000 mL
L
1 mol
Measure out the calculated mass of the solid and place it into the 50. mL volumetric flask or the
50. mL graduated cylinder. Be sure to rinse the solid from the weighing paper into the flask, add
distilled water until about ½ filled. Mix well until the salt dissolves and continue adding water
until the 50 mL mark is reached. Mix again.
2. Calculate the pH for each of the salts listed in your data table.
a. Formula for salt: KHCO3
•
Calculation of pH:
KHCO3 → K+ + HCO3− (100% dissociation)
⎯⎯
→ H2CO3 + OH−
HCO3− + H2O ←⎯
⎯
K w 1.0 × 10−14
=
= 2.4 × 10−8
−7
K a 4.2 × 10
[ H 2CO3 ] ⎡⎣OH − ⎤⎦
x2
−8
2.4
10
=
×
=
Kb =
0.25
⎡⎣ HCO3− ⎤⎦
x 2 = 6.0 × 10−9
Kb =
x = ⎡⎣OH − ⎤⎦ = 6.0 × 10−9 = 7.7 × 10−5 M
∴ pOH = − log ⎡⎣7.7 × 10−5 ⎤⎦ = 4.11
∴ pH=14 − pOH = 14 − 4.11 = 9.89 (basic)
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b. Formula for salt: NH4Cl
•
Calculation of pH:
NH4Cl → NH4+ + Cl−
⎯⎯
→ NH3 + H3O+
NH4+ + H2O ←⎯
⎯
K w 1.0 × 10−14
Ka =
=
= 5.6 × 10−10
−5
K b 1.8 × 10
Ka =
[ NH 3 ] ⎡⎣ H 3O+ ⎤⎦
⎡⎣ NH 4 + ⎤⎦
x 2 = 1.4 × 10−10
= 5.6 × 10−10 =
x2
0.25
T E A C H E R
x = ⎡⎣ H 3O + ⎤⎦ = 1.4 × 10−10 = 1.2 × 10−5 M
∴ pH = − log ⎡⎣1.2 × 10−5 ⎤⎦ = 4.93(acidic)
c. Formula for salt: NH4C2H3O2
•
Calculation of pH:
NH4C2H3O2 → NH4+ + C2H3O2−
P A G E S
Ka for acetic acid = 1.8 × 10−5
Kb for ammonia = 1.8 × 10−5
Both ions present have the same affinity to react with water and the salt is therefore neutral.
d. Formula for salt: NaHCO3
•
Calculation of pH:
NaHCO3 → Na+ + HCO3−
⎯⎯
→ H2CO3 + OH−
HCO3− + H2O ←⎯
⎯
Kb =
K w 1.0 × 10−14
=
= 2.4 × 10−8
K a 4.2 × 10−7
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Neutral or Not?
Kb =
[ H 2CO3 ] ⎡⎣OH − ⎤⎦
⎡⎣ HCO3− ⎤⎦
x 2 = 6.0 × 10−9
= 2.4 × 10−8 =
x2
0.25
x = ⎡⎣ OH − ⎤⎦ = 6.0 × 10−9 = 7.7 × 10−5 M
∴ pOH = − log ⎡⎣7.7 × 10−5 ⎤⎦ = 4.11
∴ pH=14 − pOH = 14 − 4.11 = 9.89 (basic)
e. Formula for salt: NaCl
•
Calculation of pH:
T E A C H E R
P A G E S
NaCl → Na+ + Cl−
Neither ion in solution has a strong attraction to hydrolyze with the water, therefore the solution
is neutral.
CONCLUSION QUESTIONS
a.
•
•
Which of the salts listed in your data table are neutral? Explain why this must be true.
NH4C2H3O2 and NaCl are both neutral. Ammonium acetate is formed from a weak acid and
weak base having the same numerical value for their respective ionization constants. The salt
formed is therefore neutral.
Sodium chloride produces ions in solution that do not have an affinity to react with water and
therefore they remain in solution yielding a neutral pH.
3. Which of the salts listed in your data table are basic? Explain.
• KHCO3 and NaHCO3 are both basic. The bicarbonate ion is a fairly strong conjugate base and
reacts with water to produce the hydroxide ion. The production of hydroxide yields a basic
solution.
4. Which of the salts listed in your data table are acidic? Explain.
• NH4Cl is acidic. The ammonium ion is a fairly strong conjugate acid and reacts with water to
produce the hydronium ion. The production of the hydronium ion yields an acidic solution.
5. Compare the pH obtained experimentally to the calculated pH for each of the salts listed in your data
table. Explain any discrepancies.
• Student answers will vary. Students should include some of the following as explanations:
– If the solutions were not accurately prepared, the molarities of the original salt solution may
differ from 0.25 M and therefore have yielded a pH that was incorrect.
– The pH of the distilled water may have been too low to yield accurate results.
– The pH meters may not have been properly calibrated to yield accurate measurements.
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6. A student performed this same experiment but accidentally spilled some of the measured acidic salt
while transferring it to the volumetric flask. The student did not think that a small quantity of spilled
salt would matter much, so he continued with the rest of the experiment. To his surprise, the
measured pH and the calculated pH were not the same. Using your knowledge of salt hydrolysis,
explain whether the experimental pH was too low or too high as a result of his error.
• The experimental pH was too high. The true molarity would be smaller yielding a smaller value
for the hydronium ion concentration. When the (–log of the hydronium ion concentration) is
calculated, the result is a larger pH.
T E A C H E R
P A G E S
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Neutral or Not?
Neutral or Not?
Exploring Salt Hydrolysis
Salts are produced as a result of a neutralization reaction between an acid and a base. The name for this
type of chemical reaction seems a misnomer in some respects since the salts that are formed from these
reactions are not always neutral. Salts often hydrolyze with water to yield acidic or basic solutions.
PURPOSE
In this lab you will prepare a salt solution of assigned molarity, measure the pH of the solution and then
compare with others in your class. You will then use calculations to verify the accuracy of your pH.
MATERIALS
pH meter
wash bottle filled with distilled water
100 mL beakers
50 mL graduated cylinder or
50 mL volumetric flask
balance
stirring rod
weighing paper
variety of solid salts
Safety Alert
1. Wear safety goggles at all times in the laboratory.
2. Some salts may be skin irritants. Flush with copious amounts of water if a sample comes
in contact with your skin.
PROCEDURE
1. Obtain a salt sample from your teacher. Record the correct chemical formula on your student
answer page. Use your knowledge of hydrolysis reactions to formulate a hypothesis as to whether
your salt will be acidic, basic or neutral and record your hypothesis on your student answer page.
2. You need to prepare exactly 50. mL of a 0.25 M solution of the salt that you were assigned.
Show all of the calculations necessary to properly prepare this solution in the space provided on your
student answer page. Pay particular attention to significant figures and units, but record your mass
to the same degree of accuracy provided by the balance you will be using. In other words, if your
balance measures to 0.01 gram accuracy, round your answer to the nearest hundredth of a gram.
3. Have your teacher check the accuracy of your calculation before preparing your solution.
4. Prepare your sample using a clean, dry 50 mL volumetric flask or a clean, dry 50 mL graduated
cylinder. Mix your solution thoroughly following proper laboratory techniques. Transfer the
solution into a clean 100 mL beaker.
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5. Measure the pH of your salt solution and record the value in your data table.
6. Clean the lab area and dispose of your salt sample as instructed by your teacher. Be sure to turn the
pH meter or other data collection device off after rinsing with distilled water.
7. Gather the data for four different salts from other lab groups. Be sure to record both their chemical
formulas and their pHs in your data table.
8. Listen and take good notes as your teacher explains why salts are not always neutral. Follow along
as the pH for a salt is calculated.
9. Complete the ANALYSIS and CONCLUSION questions on your student answer page.
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Neutral or Not?
Name______________________________________
Period _____________________________________
Neutral or Not?
Exploring Salt Hydrolysis
HYPOTHESIS
DATA AND OBSERVATIONS
Chemical Formula for Salt
pH Observed
ANALYSIS
678
Substance
Ionization Constant
H2SO4
Ka1 = very large!
HSO4−
Ka2 = 1.2 × 10−2
H3PO4
Ka1 = 7.5 × 10−3
H2PO4−
Ka2 = 6.2 × 10−8
HPO42−
Ka3 = 3.6 × 10−13
HC2H3O2
Ka = 1.8 × 10−5
H2CO3
Ka1 = 4.2 × 10−7
HCO3−
Ka2 = 4.8 × 10−11
NH3
Kb = 1.8 × 10−5
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b.
Show the calculations for the preparation of your salt solution:
c.
Calculate the pH for each of the salts listed in your data table.
32
d. Formula for salt: ________________________
•
Calculation of pH:
e. Formula for salt: ________________________
•
Calculation of pH:
f. Formula for salt: ________________________
•
Calculation of pH:
g. Formula for salt: ________________________
•
Calculation of pH:
h. Formula for salt: ________________________
•
Calculation of pH:
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Neutral or Not?
CONCLUSION QUESTIONS
1. Which of the salts listed in your data table are neutral? Explain why this must be true.
2. Which of the salts listed in your data table are basic? Explain.
3. Which of the salts listed in your data table are acidic? Explain.
4. Compare the pH obtained experimentally to the calculated pH for each of the salts listed in your data
table. Explain any discrepancies.
5. A student performed this same experiment but accidentally spilled some of the measured acidic salt
while transferring it to the volumetric flask. The student did not think that a small quantity of spilled
salt would matter much, so he continued with the rest of the experiment. To his surprise, the
measured pH and the calculated pH were not the same. Using your knowledge of salt hydrolysis,
explain whether the experimental pH was too low or too high as a result of his error.
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