UNIVERSITY OF NEW ENGLAND CHEM 1011 GENERAL CHEMISTRY II FINAL EXAM QUESTIONS & ANSWERS 202… EXAM
P R O F E S S I O N A L P R A C T I C E M AT E R I A L S
University of New England CHEM 1011
General Chemistry II Final Exam
Questions & Answers 2026-2027 |
Updated Verified Questions
(Rationales)
Verified Answers Exam Ready With Rationales
121 QUESTIONS
DOCUMENT OVERVIEW
This document provides 121 verified questions with correct answers and detailed rationales focused on
general chemistry concepts. It serves as a comprehensive resource for students seeking to enhance their
understanding of chemistry. Students can use it effectively for study, review, and preparation for
certifications or exams in the subject area.
CONTENTS
01 • Solubility and Ksp 02 03 • Thermodynamics
• Equilibrium and Reaction Rates
04 • Electrochemistry 05 • Nuclear Chemistry 06 • Radiation and Health
07 • Acid-Base Chemistry
E XA M Q U EST I O N S
Q1 QUESTION 1 OF 121
A solid ionic compound, such as PbCl2 in water, that is classified as insoluble
CORRECT ANSWER
will dissociate very slightly
Page 1
, RATIONALE
Ionic compounds like PbCl2 exhibit limited solubility due to strong ionic bonds that resist dissociation, resulting in only a minimal amount of the compound
separating into ions in solution. This slight dissociation leads to a low concentration of Pb²⁺ and Cl⁻ ions, which characterizes its insolubility.
Q2 QUESTION 2 OF 121
A solid ionic compound, such as KCl in water, that is classified as soluble
CORRECT ANSWER
will not dissociate if added to a saturated solution
RATIONALE
Ionic compounds like KCl dissociate into their constituent ions when dissolved in water; however, in a saturated solution, the concentration of ions already
reaches equilibrium, preventing further dissociation of additional solid KCl. This principle of solubility equilibrium explains why adding more solid does not
change the concentration of ions in a saturated solution.
Q3 QUESTION 3 OF 121
The solubility product expression for the dissolution of BaS is
CORRECT ANSWER
Ksp = [Ba2+][S2-]
RATIONALE
The solubility product expression quantifies the equilibrium between the solid salt and its ions in solution, reflecting the concentrations of the dissociated ions
raised to their stoichiometric coefficients. For barium sulfide, BaS dissociates into Ba²⁺ and S²⁻ ions, forming the expression Ksp = [Ba²⁺][S²⁻].
Q4 QUESTION 4 OF 121
The solubility product expression for the dissolution of BaF2 is
CORRECT ANSWER
Ksp = [Ba2+][F-]2
RATIONALE
The solubility product constant (Ksp) quantifies the equilibrium between a solid and its ions in solution, with the concentrations of the ions raised to the power
of their coefficients in the balanced dissolution equation. For BaF2, this results in the expression Ksp = [Ba2+][F-]^2, reflecting the stoichiometry of one barium
ion and two fluoride ions produced upon dissolution.
Q5 QUESTION 5 OF 121
The solubility product expression for the dissolution of Ca3(PO4)2 is
CORRECT ANSWER
[Ca^2+]^3[PO4^3-]^2
Page 2
, RATIONALE
The solubility product expression quantifies the equilibrium concentrations of ions in a saturated solution, derived from the dissociation of Ca3(PO4)2 into
three calcium ions and two phosphate ions. This relationship reflects the stoichiometry of the dissolution reaction, where the concentrations of the ions are
raised to the power of their respective coefficients in the balanced equation.
Q6 QUESTION 6 OF 121
The solubility of AgBr will be greatest in
CORRECT ANSWER
0.1 M NaCl
RATIONALE
In the presence of NaCl, the common ion effect reduces the solubility of AgBr due to the increased concentration of Br⁻ ions, while the ionic strength of the
solution enhances AgBr dissociation. This dynamic allows for greater solubility of AgBr compared to pure water or other dilute solutions.
Q7 QUESTION 7 OF 121
The solubility of CaS will be greatest in
CORRECT ANSWER
0.1 M Mg(NO3)2
RATIONALE
Calcium sulfide (CaS) exhibits increased solubility in solutions containing Mg(NO3)2 due to the common ion effect, where the presence of magnesium ions
reduces the concentration of sulfide ions through selective precipitation. This dynamic shifts the dissolution equilibrium of CaS to favor greater solubility in the
presence of magnesium ions.
Q8 QUESTION 8 OF 121
What is the effect of adding NaOH to a saturated solution of Ca(OH)2?
CORRECT ANSWER
[Ca2+] will decrease, [OH-] will increase, there will be more Ca(OH)2(s)
RATIONALE
Adding NaOH to a saturated Ca(OH)2 solution increases the hydroxide ion concentration, driving the equilibrium to the left according to Le Chatelier's
principle, which reduces the solubility of Ca(OH)2 and precipitates more solid. this results in decreased calcium ion concentration while hydroxide ion
concentration rises.
Q9 QUESTION 9 OF 121
Page 3
, The Ksp of NiCO3 is 1.4 x 10-7 at 25 °C. What is the molar solubility of NiCO3 at 25 °C?
CORRECT ANSWER
Ksp = (S)(S)
Ni+ = S
CO3- = S
Ksp = (S)^2
1.47 * 10^-7 = (S)^2
square root (1.47 * 10^-7)
S = 3.7 * 10^-4
RATIONALE
The solubility product constant (Ksp) quantifies the equilibrium between solid NiCO3 and its ions in solution, allowing for the calculation of molar solubility
through the relationship Ksp = [Ni²⁺][CO₃²⁻]. By setting both ion concentrations equal to S and solving the equation Ksp = S², the molar solubility is determined
as 3.7 x 10⁻⁴ M.
Q10 QUESTION 10 OF 121
The Ksp of AgBr is 5.0 x 10-13 at 25 °C. What is the molar solubility of AgBr at 25 °C?
CORRECT ANSWER
Ksp = (S)(S)
Ag+ = S
Br- = S
Ksp = (S)^2
5.0 * 10^-13 = (S)^2
square root (5.0 * 10^-13)
S = 7.1 * 10^-7
RATIONALE
The molar solubility of AgBr can be calculated using the solubility product constant (Ksp), which defines the relationship between the ion concentrations at
equilibrium; since AgBr dissociates into Ag+ and Br- in a 1:1 ratio, the Ksp expression simplifies to S^2, allowing for the determination of S by taking the square
root of Ksp. Thus, substituting Ksp into the equation yields a molar solubility of approximately 7.1 x 10^-7 mol/L.
Q11 QUESTION 11 OF 121
The Ksp for CaF2 is 4.0 x 10-11. What is the molar solubility of CaF2?
CORRECT ANSWER
Ksp = (S)(2S)^2
Ca2+ = S
F2- = 2S
Ksp = (4S)^3
4.0 * 10^- = (4S)^
1.0 * 10^-11 = S^3
cube root (1.0 * 10 ^-11)
S = 2.2 * 10^-4
RATIONALE
The molar solubility of CaF2 is derived from its solubility product constant (Ksp), where the relationship between ion concentrations in a saturated solution
allows for the calculation of solubility using the formula Ksp = [Ca²⁺][F⁻]². By substituting the variable S for Ca²⁺ and 2S for F⁻, and solving the resulting equation,
the molar solubility is determined.
Page 4
P R O F E S S I O N A L P R A C T I C E M AT E R I A L S
University of New England CHEM 1011
General Chemistry II Final Exam
Questions & Answers 2026-2027 |
Updated Verified Questions
(Rationales)
Verified Answers Exam Ready With Rationales
121 QUESTIONS
DOCUMENT OVERVIEW
This document provides 121 verified questions with correct answers and detailed rationales focused on
general chemistry concepts. It serves as a comprehensive resource for students seeking to enhance their
understanding of chemistry. Students can use it effectively for study, review, and preparation for
certifications or exams in the subject area.
CONTENTS
01 • Solubility and Ksp 02 03 • Thermodynamics
• Equilibrium and Reaction Rates
04 • Electrochemistry 05 • Nuclear Chemistry 06 • Radiation and Health
07 • Acid-Base Chemistry
E XA M Q U EST I O N S
Q1 QUESTION 1 OF 121
A solid ionic compound, such as PbCl2 in water, that is classified as insoluble
CORRECT ANSWER
will dissociate very slightly
Page 1
, RATIONALE
Ionic compounds like PbCl2 exhibit limited solubility due to strong ionic bonds that resist dissociation, resulting in only a minimal amount of the compound
separating into ions in solution. This slight dissociation leads to a low concentration of Pb²⁺ and Cl⁻ ions, which characterizes its insolubility.
Q2 QUESTION 2 OF 121
A solid ionic compound, such as KCl in water, that is classified as soluble
CORRECT ANSWER
will not dissociate if added to a saturated solution
RATIONALE
Ionic compounds like KCl dissociate into their constituent ions when dissolved in water; however, in a saturated solution, the concentration of ions already
reaches equilibrium, preventing further dissociation of additional solid KCl. This principle of solubility equilibrium explains why adding more solid does not
change the concentration of ions in a saturated solution.
Q3 QUESTION 3 OF 121
The solubility product expression for the dissolution of BaS is
CORRECT ANSWER
Ksp = [Ba2+][S2-]
RATIONALE
The solubility product expression quantifies the equilibrium between the solid salt and its ions in solution, reflecting the concentrations of the dissociated ions
raised to their stoichiometric coefficients. For barium sulfide, BaS dissociates into Ba²⁺ and S²⁻ ions, forming the expression Ksp = [Ba²⁺][S²⁻].
Q4 QUESTION 4 OF 121
The solubility product expression for the dissolution of BaF2 is
CORRECT ANSWER
Ksp = [Ba2+][F-]2
RATIONALE
The solubility product constant (Ksp) quantifies the equilibrium between a solid and its ions in solution, with the concentrations of the ions raised to the power
of their coefficients in the balanced dissolution equation. For BaF2, this results in the expression Ksp = [Ba2+][F-]^2, reflecting the stoichiometry of one barium
ion and two fluoride ions produced upon dissolution.
Q5 QUESTION 5 OF 121
The solubility product expression for the dissolution of Ca3(PO4)2 is
CORRECT ANSWER
[Ca^2+]^3[PO4^3-]^2
Page 2
, RATIONALE
The solubility product expression quantifies the equilibrium concentrations of ions in a saturated solution, derived from the dissociation of Ca3(PO4)2 into
three calcium ions and two phosphate ions. This relationship reflects the stoichiometry of the dissolution reaction, where the concentrations of the ions are
raised to the power of their respective coefficients in the balanced equation.
Q6 QUESTION 6 OF 121
The solubility of AgBr will be greatest in
CORRECT ANSWER
0.1 M NaCl
RATIONALE
In the presence of NaCl, the common ion effect reduces the solubility of AgBr due to the increased concentration of Br⁻ ions, while the ionic strength of the
solution enhances AgBr dissociation. This dynamic allows for greater solubility of AgBr compared to pure water or other dilute solutions.
Q7 QUESTION 7 OF 121
The solubility of CaS will be greatest in
CORRECT ANSWER
0.1 M Mg(NO3)2
RATIONALE
Calcium sulfide (CaS) exhibits increased solubility in solutions containing Mg(NO3)2 due to the common ion effect, where the presence of magnesium ions
reduces the concentration of sulfide ions through selective precipitation. This dynamic shifts the dissolution equilibrium of CaS to favor greater solubility in the
presence of magnesium ions.
Q8 QUESTION 8 OF 121
What is the effect of adding NaOH to a saturated solution of Ca(OH)2?
CORRECT ANSWER
[Ca2+] will decrease, [OH-] will increase, there will be more Ca(OH)2(s)
RATIONALE
Adding NaOH to a saturated Ca(OH)2 solution increases the hydroxide ion concentration, driving the equilibrium to the left according to Le Chatelier's
principle, which reduces the solubility of Ca(OH)2 and precipitates more solid. this results in decreased calcium ion concentration while hydroxide ion
concentration rises.
Q9 QUESTION 9 OF 121
Page 3
, The Ksp of NiCO3 is 1.4 x 10-7 at 25 °C. What is the molar solubility of NiCO3 at 25 °C?
CORRECT ANSWER
Ksp = (S)(S)
Ni+ = S
CO3- = S
Ksp = (S)^2
1.47 * 10^-7 = (S)^2
square root (1.47 * 10^-7)
S = 3.7 * 10^-4
RATIONALE
The solubility product constant (Ksp) quantifies the equilibrium between solid NiCO3 and its ions in solution, allowing for the calculation of molar solubility
through the relationship Ksp = [Ni²⁺][CO₃²⁻]. By setting both ion concentrations equal to S and solving the equation Ksp = S², the molar solubility is determined
as 3.7 x 10⁻⁴ M.
Q10 QUESTION 10 OF 121
The Ksp of AgBr is 5.0 x 10-13 at 25 °C. What is the molar solubility of AgBr at 25 °C?
CORRECT ANSWER
Ksp = (S)(S)
Ag+ = S
Br- = S
Ksp = (S)^2
5.0 * 10^-13 = (S)^2
square root (5.0 * 10^-13)
S = 7.1 * 10^-7
RATIONALE
The molar solubility of AgBr can be calculated using the solubility product constant (Ksp), which defines the relationship between the ion concentrations at
equilibrium; since AgBr dissociates into Ag+ and Br- in a 1:1 ratio, the Ksp expression simplifies to S^2, allowing for the determination of S by taking the square
root of Ksp. Thus, substituting Ksp into the equation yields a molar solubility of approximately 7.1 x 10^-7 mol/L.
Q11 QUESTION 11 OF 121
The Ksp for CaF2 is 4.0 x 10-11. What is the molar solubility of CaF2?
CORRECT ANSWER
Ksp = (S)(2S)^2
Ca2+ = S
F2- = 2S
Ksp = (4S)^3
4.0 * 10^- = (4S)^
1.0 * 10^-11 = S^3
cube root (1.0 * 10 ^-11)
S = 2.2 * 10^-4
RATIONALE
The molar solubility of CaF2 is derived from its solubility product constant (Ksp), where the relationship between ion concentrations in a saturated solution
allows for the calculation of solubility using the formula Ksp = [Ca²⁺][F⁻]². By substituting the variable S for Ca²⁺ and 2S for F⁻, and solving the resulting equation,
the molar solubility is determined.
Page 4
