Ap chem 프린스턴 화학 summary part 7

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Use the following information to answer questions 1-5.

The following reaction is found to be at equilibrium at 25°C:

2 SO₃(g) ⇌ O₂(g) + 2 SO₂(g)     ΔH = -198 kJ/mol

Question 1:

What is the expression for the equilibrium constant, $K_c$?

Options:

• (A) $\dfrac{[SO_3]^2}{[O_2][SO_2]^2}$
• (B) $\dfrac{2[SO_3]}{[O_2]2[SO_2]}$
• (C) $\dfrac{[O_2][SO_2]^2}{[SO_3]^2}$
• (D) $\dfrac{[O_2]2[SO_2]}{2[SO_3]}$

View Answer

Correct Answer: (C)
        

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Question 2:

Which of the following would cause the reverse reaction to speed up?

Options:

• (A) Adding more $SO_3$
• (B) Raising the pressure
• (C) Lowering the temperature
• (D) Removing some $SO_2$

View Answer

Correct Answer: (B)
        

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Question 3:

The value for $K_c$ at 25°C is 8.1. What must happen in order for the reaction to reach equilibrium if the initial concentrations of all three species were 2.0 M?

Options:

• (A) The rate of the forward reaction would increase, and $[SO_3]$ would decrease.
• (B) The rate of the reverse reaction would increase, and $[SO_3]$ would decrease.
• (C) Both the rate of the forward and reverse reactions would increase, and the value for the equilibrium constant would also increase.
• (D) No change would occur in either the rate of reaction or the concentrations of any of the species.

View Answer

Correct Answer: (A)
        

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Question 4:

Which of the following would cause a reduction in the value for the equilibrium constant?

Options:

• (A) Increasing the amount of $SO_3$
• (B) Reducing the amount of $O_2$
• (C) Raising the temperature
• (D) Lowering the temperature

View Answer

Correct Answer: (C)
        

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Question 5:

The solubility product, $K_{sp}$, of AgCl is $1.8 \times 10^{-10}$. Which of the following expressions is equal to the solubility of AgCl?

Options:

• (A) $(1.8 \times 10^{-10})^2$
• (B) $1.8 \times 10^{-10}$
• (C) $\dfrac{1.8 \times 10^{-10}}{2}$
• (D) $\sqrt{1.8 \times 10^{-10}}$

View Answer

Correct Answer: (D)
        

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Question 6:

Which of the following expressions is equal to the $K_{sp}$ of $Ag_2CO_3$?

Options:

• (A) $K_{sp} = [Ag^+][CO_3^{2-}]$
• (B) $K_{sp} = 2[Ag^+][CO_3^{2-}]$
• (C) $K_{sp} = [Ag^+]^2[CO_3^{2-}]$
• (D) $K_{sp} = [Ag^+]^2[CO_3^{2-}]^2$

View Answer

Correct Answer: (C)
        

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Question 7:

If the solubility of $BaF_2$ is equal to $x$, which of the following expressions is equal to the solubility product, $K_{sp}$, for $BaF_2$?

Options:

• (A) $x^2$
• (B) $2x^2$
• (C) $2x^3$
• (D) $4x^3$

View Answer

Correct Answer: (D)
        

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Use the following information to answer questions 8-10:

150 mL of saturated SrF₂ solution is present in a 250 mL beaker at room temperature. The molar solubility of SrF₂ at 298 K is 1.0 × 10^-3 M.

Question 8:

What are the concentrations of Sr²⁺ and F⁻ in the beaker?

Options:

• (A) $[Sr^{2+}] = 1.0 \times 10^{-3} M, [F^-] = 1.0 \times 10^{-3} M$
• (B) $[Sr^{2+}] = 1.0 \times 10^{-3} M, [F^-] = 2.0 \times 10^{-3} M$
• (C) $[Sr^{2+}] = 2.0 \times 10^{-3} M, [F^-] = 1.0 \times 10^{-3} M$
• (D) $[Sr^{2+}] = 2.0 \times 10^{-3} M, [F^-] = 2.0 \times 10^{-3} M$

View Answer

Correct Answer: (B)
        

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Question 9:

If some of the solution evaporates overnight, which of the following will occur?

Options:

• (A) The mass of the solid and the concentration of the ions will stay the same.
• (B) The mass of the solid and the concentration of the ions will increase.
• (C) The mass of the solid will decrease, and the concentration of the ions will stay the same.
• (D) The mass of the solid will increase, and the concentration of the ions will stay the same.

View Answer

Correct Answer: (D)
        

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Question 10:

How could the concentration of Sr²⁺ ions in solution be decreased?

Options:

• (A) Adding some NaF to the beaker
• (B) Adding some Sr(NO₃)₂ to the beaker
• (C) Raising the temperature
• (D) Lowering the temperature

View Answer

Correct Answer: (A)
        

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Question 11:

For a reaction involving nitrogen monoxide inside a sealed flask, the value for the reaction quotient ($Q$) was found to be $1.1 \times 10^2$ at a given point. If, after this point, the amount of NO gas in the flask increased, which reaction is most likely taking place in the flask?

Options:

• (A) NOBr(g) ⇌ NO(g) + ½ Br₂(g), $K_c = 3.4 \times 10^{-2}$
• (B) 2 NOCl(g) ⇌ 2 NO(g) + Cl₂(g), $K_c = 1.6 \times 10^{-5}$
• (C) 2 NO(g) + 2 H₂(g) ⇌ N₂(g) + 2 H₂O(g), $K_c = 4.0 \times 10^6$
• (D) N₂(g) + O₂(g) ⇌ 2 NO(g), $K_c = 4.2 \times 10^2$

View Answer

Correct Answer: (D)
        

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Question 12:

2 HI(g) + Cl₂(g) ⇌ 2 HCl(g) + I₂(g) + energy

A gaseous reaction occurs and comes to equilibrium, as shown above. Which of the following changes to the system will serve to increase the number of moles of I₂ present at equilibrium?

Options:

• (A) Increasing the volume at constant temperature
• (B) Decreasing the volume at constant temperature
• (C) Increasing the temperature at constant volume
• (D) Decreasing the temperature at constant volume

View Answer

Correct Answer: (D)
        

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Question 13:

2 NOBr(g) ⇌ 2 NO(g) + Br₂(g)

The reaction above came to equilibrium at a temperature of 100°C. At equilibrium, the partial pressure due to NOBr was 4 atm, the partial pressure due to NO was 4 atm, and the partial pressure due to Br₂ was 2 atm. What is the equilibrium constant, $K_p$, for this reaction at 100°C?

Options:

• (A) $\frac{1}{4}$
• (B) $\frac{1}{2}$
• (C) 1
• (D) 2

View Answer

Correct Answer: (D)
        

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Question 14:

Br₂(g) + I₂(g) ⇌ 2 IBr(g)

At 150°C, the equilibrium constant, $K_p$, for the reaction shown above has a value of 300. This reaction was allowed to reach equilibrium in a sealed container and the partial pressure due to IBr(g) was found to be 3 atm. Which of the following could be the partial pressures due to Br₂(g) and I₂(g) in the container?

Options:

• (A) $P_{Br₂} = 0.1 \, atm, P_{I₂} = 0.3 \, atm$
• (B) $P_{Br₂} = 0.3 \, atm, P_{I₂} = 1 \, atm$
• (C) $P_{Br₂} = 1 \, atm, P_{I₂} = 1 \, atm$
• (D) $P_{Br₂} = 1 \, atm, P_{I₂} = 3 \, atm$

View Answer

Correct Answer: (A)
        

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Use the information below to answer questions 15-17:

Silver sulfate, Ag₂SO₄, has a solubility product constant of 1.0 × 10^-5. The below diagram shows the products of a precipitation reaction in which some silver sulfate was formed.

• K⁺ ions floating in the solution.
• NO₃^- ions distributed throughout the solution.
• Ag⁺ ions also present in the solution.
• A solid precipitate of Ag₂SO₄ settled at the bottom of the beaker.

Question 15:

What is the identity of the excess reactant?

Options:

• (A) AgNO₃
• (B) Ag₂SO₄
• (C) KNO₃
• (D) K₂SO₄

View Answer

Correct Answer: (A)
        

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Question 16:

If the beaker above were left uncovered for several hours

Options:

• (A) Some of the Ag₂SO₄ would dissolve
• (B) Some of the spectator ions would evaporate into the atmosphere
• (C) The solution would become electrically imbalanced
• (D) Additional Ag₂SO₄ would precipitate

View Answer

Correct Answer: (D)
        

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Question 17:

Which ion concentrations below would have led the precipitate to form?

Options:

• (A) $[Ag^+] = 0.01 M [SO_4^{2-}] = 0.01 M$
• (B) $[Ag^+] = 0.10 M [SO_4^{2-}] = 0.01 M$
• (C) $[Ag^+] = 0.01 M [SO_4^{2-}] = 0.10 M$
• (D) This is impossible to determine without knowing the total volume of the solution.

View Answer

Correct Answer: (B)
        

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Question 18:

PCl₃(g) + Cl₂(g) ⇌ PCl₅(g) ΔH = -92.5 kJ/mol

In which of the following ways could the reaction above be manipulated to create more product?

Options:

• (A) Decreasing the concentration of PCl₃
• (B) Increasing the pressure
• (C) Increasing the temperature
• (D) None of the above

View Answer

Correct Answer: (B)
        

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Question 19:

A sample of solid MgCl₂ would be most soluble in which of the following solutions?

Options:

• (A) LiOH (aq)
• (B) CBr₄ (aq)
• (C) Mg(NO₃)₂ (aq)
• (D) AlCl₃ (aq)

View Answer

Correct Answer: (A)
        

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Question 20:

For a reaction involving sulfur dioxide and oxygen inside a sealed flask, the value for the reaction quotient $Q$ was found to be $2.5 \times 10^1$ at a given point. If, after this point, the amount of SO₂ gas in the flask increased, which reaction is most likely taking place in the flask?

Options:

• (A) SO₂(g) + O₂(g) ⇌ 2 SO₃(g), $K_c = 1.2 \times 10^2$
• (B) 2 SO₃(g) ⇌ 2 SO₂(g) + O₂(g), $K_c = 3.4 \times 10^{-3}$
• (C) SO₃(g) ⇌ SO₂(g) + ½ O₂(g), $K_c = 5.6 \times 10^{-1}$
• (D) 2 SO₂(g) + O₂(g) ⇌ 2 SO₃(g), $K_c = 2.0 \times 10^1$

View Answer

Answer Explanation:
The amount of SO2 is increasing, so the reaction is shifting towards the reactants (left). The most likely reaction is the reverse reaction where SO3 decomposes into SO2 and O2.
Correct Answer: (B)
        

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Question 21:

In a reaction between nitrogen and hydrogen to produce ammonia, the equilibrium constant $K_c$ is $4.5 \times 10^2$ at 450°C. If the reaction quotient $Q$ is $5.0 \times 10^2$, which direction will the reaction proceed?

Options:

• (A) The reaction will proceed towards the products.
• (B) The reaction will proceed towards the reactants.
• (C) The reaction is already at equilibrium.
• (D) The reaction will stop immediately.

View Answer

Answer Explanation:
Since \(Q > K_c\), the reaction will shift towards the reactants to decrease the concentration of products and reach equilibrium.
Correct Answer: (B)
        

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Question 22:

At a certain temperature, the equilibrium constant $K_c$ for the following reaction is $3.2 \times 10^{-4}$:

2 H₂(g) + O₂(g) ⇌ 2 H₂O(g)

If the concentrations of H₂ and O₂ are both increased, how will the equilibrium shift?

Options:

• (A) The equilibrium will shift towards the products.
• (B) The equilibrium will shift towards the reactants.
• (C) The equilibrium will remain unchanged.
• (D) The reaction will stop.

View Answer

Answer Explanation:
According to Le Chatelier's principle, increasing the concentration of the reactants will shift the equilibrium towards the products to counteract the change.
Correct Answer: (A)
        

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Question 23:

A reaction has the following equilibrium expression:

$K_c = \frac{[CO_2][H_2O]}{[CH_4][O_2]^2}$

Which of the following reactions corresponds to this equilibrium expression?

Options:

• (A) CH₄(g) + 2 O₂(g) ⇌ CO₂(g) + 2 H₂O(g)
• (B) CH₄(g) + O₂(g) ⇌ CO(g) + 2 H₂O(g)
• (C) CO₂(g) + 2 H₂O(g) ⇌ CH₄(g) + O₂(g)
• (D) CO(g) + H₂ ⇌ CH₄(g) + O₂(g)

View Answer

Answer Explanation:
The equilibrium expression shows the ratio of products to reactants with the correct stoichiometric coefficients matching the reaction: CH4(g) + 2 O2(g) ⇌ CO2(g) + 2 H2O(g).
Correct Answer: (A)
        

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Question 24:

The solubility product $K_{sp}$ of silver chloride (AgCl) is $1.8 \times 10^{-10}$. What is the molar solubility of AgCl in pure water?

Options:

• (A) $1.8 \times 10^{-5}$ M
• (B) $1.3 \times 10^{-5}$ M
• (C) $1.8 \times 10^{-10}$ M
• (D) $1.3 \times 10^{-10}$ M

View Answer

Answer Explanation:
The solubility product for AgCl can be written as \(K_{sp} = [Ag^+][Cl^-]\). For pure water, the concentration of \(Ag^+\) and \(Cl^-\) are equal, so \(K_{sp} = x^2\), where \(x\) is the molar solubility. Solving for \(x\):
\[ x = \sqrt{1.8 \times 10^{-10}} = 1.3 \times 10^{-5} \, \text{M} \]
Correct Answer: (B)
        

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Question 25:

The solubility product $K_{sp}$ of lead(II) fluoride (PbF₂) is $4.0 \times 10^{-8}$. What is the molar solubility of PbF₂ in pure water?

Options:

• (A) $1.0 \times 10^{-4}$ M
• (B) $2.0 \times 10^{-4}$ M
• (C) $1.6 \times 10^{-3}$ M
• (D) $1.0 \times 10^{-3}$ M

View Answer

Answer Explanation:
For PbF2, the dissociation is \(PbF_2(s) \rightarrow Pb^{2+}(aq) + 2 F^{-}(aq)\). Thus, the \(K_{sp}\) expression is:
\[ K_{sp} = [Pb^{2+}][F^{-}]^2 = x(2x)^2 = 4x^3 \]
Solving for \(x\) (the molar solubility):
\[ x = \sqrt[3]{\frac{K_{sp}}{4}} = \sqrt[3]{\frac{4.0 \times 10^{-8}}{4}} = 1.6 \times 10^{-3} \, \text{M} \]
Correct Answer: (C)
        

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Question 26:

The solubility product $K_{sp}$ of calcium hydroxide (Ca(OH)₂) is $5.5 \times 10^{-6}$. What is the molar solubility of Ca(OH)₂ in pure water?

Options:

• (A) $1.5 \times 10^{-2}$ M
• (B) $2.0 \times 10^{-2}$ M
• (C) $1.2 \times 10^{-2}$ M
• (D) $1.0 \times 10^{-2}$ M

View Answer

Answer Explanation:
For calcium hydroxide, \(Ca(OH)_2 \rightarrow Ca^{2+}(aq) + 2 OH^{-}(aq)\). The solubility product expression is:
\[ K_{sp} = [Ca^{2+}][OH^{-}]^2 = x(2x)^2 = 4x^3 \]
Solving for \(x\) (the molar solubility):
\[ x = \sqrt[3]{\frac{K_{sp}}{4}} = \sqrt[3]{\frac{5.5 \times 10^{-6}}{4}} = 1.2 \times 10^{-2} \, \text{M} \]
Correct Answer: (C)
        

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Question 27:

What is the $K_{sp}$ expression for the compound barium sulfate (BaSO₄)?

Options:

• (A) $K_{sp} = [Ba^{2+}] [SO_4^{2-}]$
• (B) $K_{sp} = [Ba^{2+}]^2 [SO_4^{2-}]^2$
• (C) $K_{sp} = [Ba^{2+}]^2 [SO_4^{2-}]$
• (D) $K_{sp} = [Ba^{2+}] [SO_4^{2-}]^2$

View Answer

Answer Explanation:
Barium sulfate dissociates as: \(BaSO_4(s) \rightarrow Ba^{2+}(aq) + SO_4^{2-}(aq)\). Therefore, the \(K_{sp}\) expression is:
\[ K_{sp} = [Ba^{2+}] [SO_4^{2-}] \]
Correct Answer: (A)
        

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Question 28:

If the solubility product $K_{sp}$ of magnesium carbonate (MgCO₃) is $3.5 \times 10^{-8}$, what is the concentration of carbonate ions $[CO_3^{2-}]$ in a saturated solution?

Options:

• (A) $5.9 \times 10^{-4}$ M
• (B) $1.9 \times 10^{-4}$ M
• (C) $3.5 \times 10^{-8}$ M
• (D) $1.0 \times 10^{-5}$ M

View Answer

Answer Explanation:
For \(MgCO_3\), the dissociation is: \(MgCO_3(s) \rightarrow Mg^{2+}(aq) + CO_3^{2-}(aq)\). The \(K_{sp}\) expression is:
\[ K_{sp} = [Mg^{2+}][CO_3^{2-}] = x^2 \]
Solving for \(x\) (the concentration of \(CO_3^{2-}\)):
\[ x = \sqrt{3.5 \times 10^{-8}} = 1.9 \times 10^{-4} \, \text{M} \]
Correct Answer: (B)
        

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Question 29:

Which of the following will decrease the solubility of calcium phosphate $(Ca_3(PO_4)_2)$ in water?

Options:

• (A) Adding HCl
• (B) Adding Ca(NO₃)₂
• (C) Increasing temperature
• (D) Adding Na₃PO₄

View Answer

Answer Explanation:
Adding a common ion, such as \(Ca^{2+}\) or \(PO_4^{3-}\), will decrease the solubility of \(Ca_3(PO_4)_2\) by shifting the equilibrium towards the solid. Therefore, adding \(Ca(NO_3)_2\) or \(Na_3PO_4\) will decrease the solubility.
Correct Answer: (B)
        

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Question 30:

The $K_{sp}$ of copper(II) hydroxide $(Cu(OH)_2)$ is $2.2 \times 10^{-20}$. If the concentration of $OH^{-}$ ions in a solution is $1.0 \times 10^{-6}$ M, what is the concentration of $Cu^{2+}$ ions?

Options:

• (A) $2.2 \times 10^{-8}$ M
• (B) $2.2 \times 10^{-14}$ M
• (C) $1.1 \times 10^{-7}$ M
• (D) $4.4 \times 10^{-20}$ M

View Answer

Answer Explanation:
The solubility product expression for \(Cu(OH)_2\) is:
\[ K_{sp} = [Cu^{2+}][OH^{-}]^2 \]
Given that \([OH^{-}] = 1.0 \times 10^{-6} \, \text{M}\), we can solve for \([Cu^{2+}]\):
\[ [Cu^{2+}] = \frac{K_{sp}}{[OH^{-}]^2} = \frac{2.2 \times 10^{-20}}{(1.0 \times 10^{-6})^2} = 2.2 \times 10^{-8} \, \text{M} \]
Correct Answer: (A)
        

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Question 31:

The solubility product $K_{sp}$ of barium sulfate (BaSO₄) is $1.1 \times 10^{-10}$. If the ion product $Q$ for a given solution is $2.5 \times 10^{-10}$, what will happen when barium chloride (BaCl₂) is added to the solution?

Options:

• (A) BaSO₄ will precipitate out of the solution
• (B) The solution will remain unsaturated
• (C) BaSO₄ will dissolve completely
• (D) No change will occur

View Answer

Answer Explanation:
Since \(Q > K_{sp}\), the solution is supersaturated, and precipitation will occur to reduce the ion concentrations. Thus, BaSO4 will precipitate out of the solution.
Correct Answer: (A)
        

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Question 32:

The $K_{sp}$ of calcium carbonate (CaCO₃) is $4.5 \times 10^{-9}$. If the concentration of calcium ions is $3.0 \times 10^{-3}$ M and the concentration of carbonate ions is $2.0 \times 10^{-3}$ M, what is the value of the ion product $Q$, and will precipitation occur?

Options:

• (A) $Q = 6.0 \times 10^{-6}$, precipitation will occur
• (B) $Q = 6.0 \times 10^{-9}$, precipitation will not occur
• (C) $Q = 6.0 \times 10^{-6}$, precipitation will not occur
• (D) $Q = 6.0 \times 10^{-9}$, precipitation will occur

View Answer

Answer Explanation:
The ion product \(Q\) is calculated as:
\[ Q = [Ca^{2+}][CO_3^{2-}] = (3.0 \times 10^{-3})(2.0 \times 10^{-3}) = 6.0 \times 10^{-6} \]
Since \(Q > K_{sp}\), precipitation will occur.
Correct Answer: (A)
        

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Question 33:

The solubility product $K_{sp}$ of magnesium hydroxide (Mg(OH)₂) is $8.9 \times 10^{-12}$. What is the molar solubility of Mg(OH)₂ in a solution where the concentration of hydroxide ions is fixed at $1.0 \times 10^{-4}$ M?

Options:

• (A) $8.9 \times 10^{-12}$ M
• (B) $8.9 \times 10^{-8}$ M
• (C) $8.9 \times 10^{-4}$ M
• (D) $8.9 \times 10^{-6}$ M

View Answer

Answer Explanation:
For \(Mg(OH)_2\), the \(K_{sp}\) expression is:
\[ K_{sp} = [Mg^{2+}][OH^{-}]^2 \]
Given \([OH^{-}] = 1.0 \times 10^{-4}\), solving for \([Mg^{2+}]\):
\[ [Mg^{2+}] = \frac{K_{sp}}{[OH^{-}]^2} = \frac{8.9 \times 10^{-12}}{(1.0 \times 10^{-4})^2} = 8.9 \times 10^{-4} \, \text{M} \]
Correct Answer: (C)
        

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Question 34:

The ion product $Q$ for a solution of iron(III) hydroxide (Fe(OH)₃) is $4.0 \times 10^{-36}$. If the $K_{sp}$ of Fe(OH)₃ is $2.5 \times 10^{-39}$, what can be concluded about the solution?

Options:

• (A) The solution is supersaturated, and Fe(OH)₃ will precipitate
• (B) The solution is saturated, and no precipitation will occur
• (C) The solution is unsaturated, and Fe(OH)₃ will dissolve
• (D) The solution is supersaturated, but no precipitation will occur

View Answer

Answer Explanation:
Since \(Q > K_{sp}\), the solution is supersaturated, and precipitation will occur to reduce the ion concentrations.
Correct Answer: (A)
        

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Question 35:

If the solubility product $K_{sp}$ of silver sulfate (Ag₂SO₄) is $1.2 \times 10^{-5}$, what is the molar solubility of silver sulfate in water?

Options:

• (A) $1.0 \times 10^{-2}$ M
• (B) $1.2 \times 10^{-2}$ M
• (C) $1.6 \times 10^{-3}$ M
• (D) $4.0 \times 10^{-2}$ M

View Answer

Correct Answer: (B)
        

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Question 36:

For a solution of calcium fluoride (CaF₂), the ion product $Q$ is $1.0 \times 10^{-10}$ and the $K_{sp}$ is $3.9 \times 10^{-11}$. What will happen in the solution?

Options:

• (A) CaF₂ will precipitate
• (B) CaF₂ will dissolve
• (C) No change will occur
• (D) The solution will remain supersaturated

View Answer

Answer Explanation:
Since \(Q > K_{sp}\), the solution is supersaturated, and calcium fluoride (CaF2) will precipitate out of the solution.
Correct Answer: (A)
        

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Question 37:

The solubility product $K_{sp}$ of lead(II) chloride (PbCl₂) is $1.7 \times 10^{-5}$.  what will happen when more chloride ions are added to the solution?

Options:

• (A) PbCl₂ will precipitate
• (B) PbCl₂ will dissolve further
• (C) The solution will become unsaturated
• (D) No change will occur

View Answer

Answer Explanation:
Since \(Q < K_{sp}\), the solution is unsaturated, and no precipitation will occur initially. However, adding more chloride ions will increase \(Q\), and once \(Q\) exceeds \(K_{sp}\), PbCl2 will precipitate.
Correct Answer: (A)
        

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Question 38:

The solubility product $K_{sp}$ of silver bromide (AgBr) is $5.0 \times 10^{-13}$. If a solution has an ion product $Q$ of $3.0 \times 10^{-13}$, what can be concluded about the solution?

Options:

• (A) The solution is supersaturated, and AgBr will precipitate
• (B) The solution is saturated, and no precipitation will occur
• (C) The solution is unsaturated, and AgBr will dissolve
• (D) The solution is supersaturated, but no precipitation will occur

View Answer

Answer Explanation:
Since \(Q < K_{sp}\), the solution is unsaturated, meaning that more AgBr can dissolve until \(Q = K_{sp}\).
Correct Answer: (C)
        

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Question 39:

A solution contains $5.0 \times 10^{-3}$ M of Ba²⁺ and $1.0 \times 10^{-2}$ M of SO₄^2-. The $K_{sp}$ of barium sulfate (BaSO₄) is $1.1 \times 10^{-10}$. What is the value of the ion product $Q$, and will precipitation occur?

Options:

• (A) $Q = 5.0 \times 10^{-5}$, precipitation will not occur
• (B) $Q = 5.0 \times 10^{-5}$, precipitation will occur
• (C) $Q = 5.0 \times 10^{-3}$, precipitation will occur
• (D) $Q = 5.0 \times 10^{-3}$, precipitation will not occur

View Answer

Answer Explanation:
The ion product \(Q\) is calculated as:
\[ Q = [Ba^{2+}][SO_4^{2-}] = (5.0 \times 10^{-3})(1.0 \times 10^{-2}) = 5.0 \times 10^{-5} \]
Since \(Q > K_{sp}\), precipitation will occur.
Correct Answer: (B)
        

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Question 40:

The $K_{sp}$ of calcium hydroxide (Ca(OH)₂) is $6.5 \times 10^{-6}$. If the concentration of hydroxide ions in a solution is $2.0 \times 10^{-3}$ M, what is the molar solubility of Ca(OH)₂ in this solution?

Options:

• (A) $1.6 \times 10^{-4}$ M
• (B) $1.6 \times 10^{-2}$ M
• (C) $6.5 \times 10^{-2}$ M
• (D) $3.2 \times 10^{-4}$ M

View Answer

Answer Explanation:
For \(Ca(OH)_2\), the \(K_{sp}\) expression is:
\[ K_{sp} = [Ca^{2+}][OH^{-}]^2 \]
Given \([OH^{-}] = 2.0 \times 10^{-3}\), solving for \([Ca^{2+}]\):
\[ [Ca^{2+}] = \frac{K_{sp}}{[OH^{-}]^2} = \frac{6.5 \times 10^{-6}}{(2.0 \times 10^{-3})^2} = 1.6 \times 10^{-1} \, \text{M} \]
Correct Answer: (A)
        

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Question 41:

A solution is prepared by mixing $1.0 \times 10^{-3}$ M NaCl and $1.0 \times 10^{-3}$ M AgNO₃. If the solubility product $K_{sp}$ of silver chloride (AgCl) is $1.8 \times 10^{-10}$, what will happen in this solution?

Options:

• (A) No reaction will occur
• (B) AgCl will dissolve
• (C) AgCl will precipitate
• (D) The solution will become supersaturated

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Answer Explanation:
The ion product \(Q\) is calculated as:
\[ Q = [Ag^{+}][Cl^{-}] = (1.0 \times 10^{-3})(1.0 \times 10^{-3}) = 1.0 \times 10^{-6} \]
Since \(Q > K_{sp}\), AgCl will precipitate out of the solution.
Correct Answer: (C)
        

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