CHE 311 Exam 2. Review (Chapters 7-9)
Test Date: TBD
Exam 2 will be held right after Chapter 9. You are allowed to have a 3”×5” “cheating sheet”, and
you must have a calculator. All answers must be placed in the “Answer Sheet” for grading.
There will be 7-8 types of questions on exam 2 with an overall score of 100 + 10 (bonus) = 110
points. Exam 2 covers chapters 7-9. You should focus on the following concepts/topics:
(1) weak/strong acids/bases and their conjugates; pH calculations from these solutions. For
example, what would be the pH of 0.10 M acetic acid, ammonia, HCl, NaOH, NH4+, sodium
acetate, or some of their mixtures?
(2) Features of buffer solutions, how to calculate and prepare a buffer solution? At what conditions
the buffer would have the highest buffering capacity? What happens if a strong acid/base is added
to a buffer? How to calculate the pH changes?
(3) Solubility calculations for ALL kinds of sparingly soluble compounds in pure water, added
electrolytes (common ions effect as well as ionic strength effect on activity coefficients).
Alternatively, for a given solubility, how to calculate the Ksp? Why added electrolyte, acid/base,
common ions, complexing agents, dilutions would (or would not) affect the solubility? Would the
Ksp also change?
(4) Mass balance (MB) and charge balance (CB) expressions. Make sure to list ALL pertinent
equations including water dissociation, weak acid/base-water dissociations, and precipitate
dissociations. NO charge balance can be given if not all of the solution components are given.
What happens if a strong electrolyte is added to a saturated solution, and how does this addition
affect the MB and CB equations?
(5) Understand how to use charge or mass balance equation to calculate the pH of extremely diluted
strong acid/base aqueous solutions.
CHE 311 Exam 2 Review Questions
1. (2×2×4 = 16 pts) Identify the acid on the left and its conjugate base on the right in the
following reactions:
(a) HClO4 + H 2O
H3O + + ClO −4
CH3 NH 3+ + OH −
(b) CH3 NH 2 + H 2 O
(c)
2H 2 PO4-
H3PO4 + HPO4 2-
( Note : only one on the left is acid)
(d) H 2 PO 4− + H 2 O
HPO 24− + H 3O +
2. (4+5+5 = 14 pts) Calculate the molar solubility of CaF2 (Ksp = 3.9 10−11 ):
•
(a) CaF2 in water.
•
(b) CaF2 in 0.0200 M NaCl (consider ionic strength effect only).
•
(c) The solution that results when you mix 50.00 mL of 0.150 M NaF with 50.00 mL of
0.0250 M CaCl2 (consider common ions effect only).
3. (3×6 = 18 pts) Indicate the effect on the solubility of AgI upon addition of the following
chemicals:
•
(a) KI
•
(b) HNO₃
•
(c) NH₃ (ammonia)
•
(d) water
•
(e) NaCl
•
(f) Sucrose
4. (3×3×2 = 18 pts) Write the charge-balance (CB) and mass-balance (MB) expressions of the
following solutions:
•
(a) Saturated PbI₂ (Ksp[PbI₂] = 1.4 10−8 )
•
(b) 0.10 M NaI mixed with saturated PbI₂
•
(c) Saturated PbS (Ksp = 3e-38) mixed with 0.20 M Na2S {Ka1[H2S] = 1.0e-7 and Ka2[H2S] =
1.0e-13}
5. (3×3 = 9 pts) Calculate the pH of the following solutions:
•
(a) 5.0 10−8 M Ba(OH)₂
•
(b) Mixture of 0.10 M HCl (30.00 mL) + 0.10 M NH₃ (30.00 mL)
•
(c) Mixture of 0.025 M HCl (25.00 mL) + 0.050 M NaHCO3 (25.00 mL)
6. (4×4 = 16 pts) Multiple choices:
(i) Which salt would precipitate first if a 0.100 M solution of Na₂CO₃ is added to a solution
containing 0.200 M Ca²⁺, 0.200 M Sr²⁺, and 0.200 M Mg²⁺?
K sp for CaCO3 = 4.8 10−9 ,
K sp for SrCO3 = 1.0 10−10 ,
K sp for MgCO3 = 6.8 10−6
(a) CaCO 3
(b) SrCO3
(c) MgCO3
(ii) What final concentration of sulfide ion S2− must be present in solution to precipitate 99.99%
of 0.10 M Cd²⁺ (Ksp[CdS] = 8.0 10−27 )?
(a) 8.0 10−22 M
(b) 8.0 10−15 M
(c) 8.0 10−10 M
(iii) A 0.100 M solution of formic acid (HCOOH) had a pH of 2.38. What is the estimated K a
value for formic acid?
(a) 1.8 10−4
(b) 1.2 10−5
(c) 4.6 10−2
(iv) The activity coefficient:
(a) decreases as ionic strength increases.
(b) increases with the addition of a non-ionic solute.
(c) is the same for all ions in a solution.
7. (3×3 = 9 pts) Oxalic acid (HOOC-COOH) is a dicarboxylic acid with K a 1 = 5.4 10−2 and
K a 2 = 5.4 10−5 . To prepare a buffer at pH 3.5, the buffer solution should contain component (a)
_____________ and (b) ____________. If the solution has a mixture of 0.100 M sodium
hydrogen oxalate (NaHC₂O₄) and 0.050 M sodium oxalate (Na₂C₂O₄), then the solution's pH
should be approximately (c) __________.
8. Bonus. (10 pts)
A weak acid HA (pKa = 4.8) and a strong base MOH are available. How would you prepare 250
mL of a pH 5.5 buffer solution using these components?
CHE 311 Mock Exam 2, Q3 Solution
Q3. Indicate the effect on the solubility of AgI upon addition of the following chemicals:
•
(a) KI
•
(b) HNO₃
•
(c) NH₃ (ammonia)
•
(d) water
•
(e) NaCl
•
(f) Sucrose
Let's analyze the effect of each chemical on the solubility of silver iodide (AgI), which is
sparingly soluble in water:
(a) KI (Potassium Iodide):
•
Effect: The solubility of AgI will Significantly decrease.
•
Explanation: The addition of iodide ions (I⁻) from KI will shift the equilibrium toward
the solid AgI, according to Le Chatelier's principle, reducing the solubility due to the
common ion effect.
(b) HNO₃ (Nitric Acid):
•
Effect: The solubility of AgI will slightly increases.
•
Explanation: HNO₃ is a strong acid, and the formed H⁺and NO3- will increase the ionic
strength. The activities of Ag+ and I- ions therefore decrease, leading the solubility of
AgI increases slightly.
(c) NH₃ (Ammonia):
•
Effect: The solubility of AgI will significantly increase.
•
Explanation: Dissociated Ag+ will react favorably with NH₃ to form very stable and
soluble complexes Ag(NH₃)+ and Ag(NH₃)2+, resulting in remarkably increasing of the
solubility.
(d) Water:
•
Effect: No effect on the molar solubility (molar solubility remains the same).
•
Explanation: [Ag+] and [I-] concentrations will remain the same as Ksp is a constant.
However, the total mass of AgI dissolved will increase with the addition of water.
•
(e) NaCl (Sodium Chloride):
•
Effect: The solubility of AgI will increase.
•
Explanation: (1) Ionic strength increase results in increasing of solubility, (2) the
addition of chloride ions (Cl⁻) introduces a competing ion (AgCl is also sparingly
soluble), and this shifts the equilibrium solid AgI toward AgCl when a relatively high
concentration of Cl- is used, leading to the increase of AgI solubility.
(f) Sucrose:
•
Effect: The solubility of AgI will remain unchanged.
•
Explanation: Sucrose is a non-electrolyte and doesn't dissociate into ions. It won't
affect the ionic equilibrium of AgI in solution.
CHE 311 Mock Exam 2 Solution for Q5
5. (3×3 = 9 pts) Calculate the pH of the following solutions:
(a) 5.0e-8 M Ba(OH)₂
(b) Mixture of 0.10 M HCl (30.00 mL) + 0.10 M NH₃ (30.00 mL)
(c) Mixture of 0.025 M HCl (25.00 mL) + 0.050 M (25.00 mL)
Solution:
(a) Simplified approach:
𝟓 × 𝟏𝟎−𝟖 M BaOH𝟐
Dissociation of Ba(OH)₂: BaOH2 → Ba2+ + 2OH− Each Ba(OH)₂ molecule releases
2 OH⁻ ions, so the concentration of OH⁻ in solution will be 2 × 5 × 10−8 M =
1 × 10−7 M.
However, this OH⁻ concentration is similar to the concentration of OH⁻ in pure water
due to autoionization, which is also [OH− ] = 1 × 10−7 M.
We need to consider both sources of OH⁻ (from Ba(OH)₂ and from water). The total
OH⁻ concentration is slightly more than 1 × 10−7 M, but for simplicity, let's assume:
[OH− ]total ≈ 1 × 10−7 + 1 × 10−7 = 2 × 10−7 M
Find pOH: 𝑝𝑂𝐻 = −log(2 × 10−7 ) = 6.70
Find pH: 𝑝𝐻 = 14 − 𝑝𝑂𝐻 = 14 − 6.70 = 7.30
So, the pH of 5 × 10−8 M BaOH2 is approximately 7.30.
(b) Precise approach—using the change balance expression
1. Dissociation of Ba(OH)₂
Ba(OH)₂ dissociates fully in water: Ba(OH)2 → Ba2+ + 2OH− Thus, for every mole of
Ba(OH)2 , we get:
•
[Ba2+ ] = 5 × 10−8 M
•
[OH− ] = 2 × 5 × 10−8 = 1 × 10−7 M
2. Charge Balance Equation
In this solution, the charge neutrality condition is given by: 2[Ba2+ ] + [H+ ] = [OH− ] Here:
•
[Ba2+ ] = 5 × 10−8 M
1
•
[H+ ] is related to [OH− ] by the water dissociation constant 𝐾𝑤 = 1 × 10−14 , which
𝐾
means: [H+ ] = [OH𝑤− ]
1×10−14
Substitute this into the charge balance equation: 2(5 × 10−8 ) + [OH−] = [OH− ]
1×10−14
This simplifies to: 1 × 10−7 + [OH−] = [OH− ]
3. Solve for [OH− ]
Now we solve this equation for [OH− ]. Let 𝑥 = [OH− ]: 1 × 10−7 +
Rearranging:
1×10−14
𝑥
1×10−14
𝑥
=𝑥
= 𝑥 − 1 × 10−7
Multiplying both sides by 𝑥: 1 × 10−14 = 𝑥(𝑥 − 1 × 10−7 )
Expanding: 1 × 10−14 = 𝑥 2 − 1 × 10−7 𝑥
Solve this quadratic equation for 𝑥 (which is [OH− ]) using numerical methods.
4. Calculate pH
𝐾
Once we find [OH− ], we can calculate [H+ ] from: [H+ ] = [OH𝑤−]
Finally, calculate the pH: 𝑝𝐻 = −log[H+ ]
Results:
From the numerical solution: [OH− ]total = 1.62 × 10−7 M
𝑝𝐻 = 7.21
(b) Mixture of 0.10 M HCl (30.00 mL) and 0.10 M NH₃ (30.00 mL)
Step 1: Neutralization reaction between HCl and NH₃:
HCl is a strong acid, and NH₃ is a weak base. The reaction is: HCl + NH₃ → NH₄+ + Cl−
Both HCl and NH₃ have equal concentrations and volumes, so they will fully neutralize
each other:
•
Moles of HCl = 0.10 M × 0.0300 L = 0.0030 mol
•
Moles of NH₃ = 0.10 M × 0.0300 L = 0.0030 mol
Thus, the reaction will produce 0.0030 mol of NH+
4 , with no HCl or NH₃ remaining.
2
Step 2: Calculate concentration of NH+
4:
The total volume of the solution is: 𝑉total = 30.00 mL + 30.00 mL = 60.00 mL = 0.0600 L
0.0030 mol
+
The concentration of NH+
4 is: [NH4 ] = 0.0600 L = 0.0500 M
Step 3: Find the pH (using the Henderson-Hasselbalch equation):
+
+
−5
NH+
4 is a weak acid, and it dissociates as: NH4 ⇌ NH3 + H The 𝐾𝑏 for NH₃ is 1.8 × 10 ,
1×10−14
𝐾
𝑤
−10
and the 𝐾𝑎 for NH+
4 can be calculated using: 𝐾𝑎 = 𝐾 = 1.8×10−5 = 5.56 × 10
𝑏
Using the approximation for weak acids: [H+ ] = √𝐾𝑎 · [NH+
4] =
√(5.56 × 10−10 ) × (0.0500) = 5.27 × 10−6 M
Find pH: 𝑝𝐻 = −log(5.27 × 10−6 ) = 5.28
So, the pH of the mixture is approximately 5.28.
(c) Mixture of 0.025 M HCl (25.00 mL) and 0.050 M NaHCO₃ (25.00 mL)
Step 1: Neutralization reaction between HCl and NaHCO₃:
The reaction is: HCl + NaHCO3 → NaCl + H2 CO3
The moles of each species are:
•
Moles of HCl = 0.025 M × 0.025 L = 6.25 × 10−4 mol
•
Moles of NaHCO3 = 0.050 M × 0.025 L = 1.25 × 10−3 mol
Since HCl is the limiting reagent, it will fully react with NaHCO₃, leaving excess NaHCO₃:
Remaining NaHCO3 = 1.25 × 10−3 − 6.25 × 10−4 = 6.25 × 10−4 mol
+
The resulting product is H2 CO3 , which partially dissociates into HCO−
3 and H .
Step 2: Calculate the concentrations:
The total volume is: 𝑉total = 25.00 mL + 25.00 mL = 50.00 mL = 0.0500 L
The concentration of remaining NaHCO3 is: [NaHCO3 ] =
The concentration of produced H2CO3 is: [H2CO3] =
6.25×10−4
0.0500
6.25×10−4
0.0500
= 0.0125 M
= 0.0125 M
Step 3: Find the pH of the buffer solution:
H2 CO3 and HCO−
3 form a weak acid-conjugate base couple, or a buffer solution.
3
Using the Henderson-Hasselbalch equation:
[HCO−
3]
𝑝𝐻 = 𝑝𝐾𝑎1 + log (
) = 𝑝𝐾𝑎1 = −log(4.3 × 10−7 ) = 6.37
[H2 CO3 ]
Thus, the pH of the mixture is approximately 6.37.
4
CHE 311 Exam 2 Review Questions Q6 Key (in red)
6. (4×4 = 16 pts) Multiple choices:
(i) Which salt would precipitate first if a 0.100 M solution of Na₂CO₃ is added to a solution
containing 0.200 M Ca²⁺, 0.200 M Sr²⁺, and 0.200 M Mg²⁺?
K sp for CaCO3 = 4.8 10−9 ,
K sp for SrCO3 = 1.0 10−10 ,
K sp for MgCO3 = 6.8 10−6
(a) CaCO 3
(b) SrCO3
(c) MgCO3
(ii) What final concentration of sulfide ion S2− must be present in solution to precipitate 99.99%
of 0.10 M Cd²⁺ (Ksp[CdS] = 8.0 10−27 )?
(a) 8.0 10−21 M
(b) 8.0 10−15 M
(c) 8.0 10−10 M
(iii) A 0.100 M solution of formic acid (HCOOH) had a pH of 2.38. What is the estimated K a
value for formic acid?
(a) 1.8 10−4
(b) 1.2 10−5
(c) 4.6 10−2
(iv) The activity coefficient:
(a) decreases as ionic strength increases.
(b) increases with the addition of a non-ionic solute.
(c) is the same for all ions in a solution.