Question 1

How many moles of CaF₂ will dissolve in 3.0 liters of 0.089 M NaF solution? (K_sp for CaF₂ = 4.0 $\times$10^-¹¹) A) B) C) D) E)none of these

Accepted Answer

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Question 2

Which of the following solid salts is more soluble in 1.0 M H⁺ than in pure water? A)NaCl B)KCl C)FePO₄ D)AgCl E)KNO₃

Accepted Answer

FePO₄ is a salt containing FePO4^3- anion. In acidic solution (1.0 M H⁺), the anion FePO4^3- will react with H+ to form HFePO4^2-. This reaction will shift the equilibrium of dissolution to the right, making the salt more soluble in the acidic solution than in pure water. The other salts (NaCl, KCl, AgCl, and KNO₃) do not have any relevant reactions in acidic solution that can change their solubility significantly.

Question 3

What is the limiting reagent in the formation of the lead chloride? A)Pb²⁺ B)Cl^- C)(NO₃)^- D)PbCl₂ E)Pb(NO₃)₂

Accepted Answer

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Question 4

What is the best way to ensure complete precipitation of SnS from a saturated H₂S solution? A)Add more H₂S. B)Add a strong acid. C)Add a weak acid. D)Add a strong base. E)Add a weak base.

Accepted Answer

Adding a strong base will increase the pH of the solution, which will lead to complete precipitation of SnS from a saturated H₂S solution. This is because, at high pH, the solubility of SnS decreases, leading to its precipitation. Adding a strong acid (choice B) or weak acid (choice C) will lower the pH, and adding more H₂S (choice A) or a weak base (choice E) will not affect the pH significantly, so they will not lead to complete precipitation of SnS.

Question 5

Which of the following solid salts is more soluble in 1.0 M H⁺ than in pure water? A)NaCl B)CaCO₃ C)KCl D)AgCl E)KNO₃

Accepted Answer

CaCO₃ is more soluble in 1.0 M H⁺ than in pure water because the H⁺ ions react with the CaCO₃ to form the more soluble Ca²⁺ and HCO₃^- ions.

Question 6

You have a solution consisting of 0.10 M Cl^- and 0.10 M CrO₄²^-.You add 0.10 M silver nitrate dropwise to this solution.Given that the K_sp for Ag₂CrO₄ is 9.0 $\times$ 10^-¹²,and that for AgCl is 1.6 $\times$10^-¹⁰,which of the following will precipitate first? A)silver chloride B)silver chromate C)silver nitrate D)cannot be determined by the information given E)none of these

Accepted Answer

To determine which compound will precipitate first, we need to compare the solubility product (Ksp) values of each compound. The compound with the lowest Ksp value will be the least soluble and therefore will precipitate first.

Using the given Ksp values, we can calculate the ion product (Qsp) of each compound in the solution:
- Qsp of AgCl = [Ag+][Cl-] = (0.10 M) x (0.10 M) = 1.0 x 10^-2
- Qsp of Ag2CrO4 = [Ag+][CrO42-] = (0.10 M) x (0.10 M) x (9.0 x 10^-10) = 9.0 x 10^-12

Since the ion product of AgCl (1.0 x 10^-2) is greater than its Ksp (1.6 x 10^-10), it exceeds the solubility limit and will precipitate first. Therefore, the answer is A) silver chloride.

Question 7

You have two salts,AgX and AgY,with very similar K_sp values.You know that K_a for HX is much greater than K_a for HY.Which salt is more soluble in acidic solution? A)AgX B)AgY C)They are equally soluble in acidic solution. D)Cannot be determined by the information given. E)None of these (A-D).

Accepted Answer

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Question 8

The two salts AgX and AgY exhibit very similar solubilities in water.It is known that the salt AgX is much more soluble in acid than is AgY.What can be said about the relative strengths of the acids HX and HY? A)Nothing. B)HY is stronger than HX. C)HX is stronger than HY. D)The acids are weak acids and have equal values for K_a. E)Both acids are strong.

Accepted Answer

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Question 9

How many moles of Ca(NO₃)₂ must be added to 1.0 L of a 0.182 M KF solution to begin precipitation of CaF₂? For CaF₂,K_sp = 4.0 $\times$10^-¹¹. A) B) C) D) E)

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Question 10

Will precipitation occur?
A)Yes.
B)No.
C)Maybe,it depends on the temperature.
D)Maybe,it depends on the limiting reagent concentration.
E)None of these.

Accepted Answer

Precipitation occurs when the solubility product of a compound is exceeded in a solution, leading to the formation of a solid from the ions in solution. This process does not inherently depend on temperature or the concentration of a limiting reagent, although these factors can affect the solubility product and thus indirectly influence precipitation.

Question 11

Calculate the solubility of Cu(OH)₂ in a solution buffered at pH = 7.59.(K_sp = 1.6 $\times$ 10^-¹⁹) A) M B) M C) M D) M E)none of these

Accepted Answer

The solubility product expression for Cu(OH)₂ is: Cu(OH)₂ ⇌ Cu2+ + 2OH- + S2- Ksp = [Cu2+][OH-]2[S2-] Since the solution is buffered at pH = 7.59, we can assume that the concentration of [H+] is equal to the concentration of [OH-]. Using the pH to calculate the concentration of [OH-], we get: pH = 7.59 pOH = 14 - pH = 6.41 [OH-] = 10^-pOH = 2.51 x 10^-7 Substituting this value into the Ksp expression, we get: Ksp = [Cu2+][2.51 x 10^-7]^2[S2-] We are given that Ksp = 1.6 x 10^-10. Solving for [S2-], we get: [S2-] = Ksp/([Cu2+][2.51 x 10^-7]^2) [S2-] = (1.6 x 10^-10)/([Cu2+] x 6.3 x 10^-14) At this point, we need to make an assumption about the concentration of Cu2+. Without more information, we can assume that the solubility of Cu(OH)₂ is low and that the concentration of Cu2+ is negligible compared to the concentration of [S2-]. Under this assumption, we can simplify the above equation to: [S2-] ≈ (1.6 x 10^-10)/(6.3 x 10^-14) = 2539 M However, this assumption may not be valid if the solubility of Cu(OH)₂ is not actually low. Without further information, we cannot definitively choose an answer. However, among the given choices, the closest value to our calculated value is choice C: 2.5 x 10^-3 M.

Question 12

The K_sp of Al(OH)₃ is 2 $\times$10^-³².At what pH will a 0.5 M Al³⁺ solution begin to show precipitation of Al(OH)₃? A)3.5 B)10.5 C)1.0 D)6.0 E)3.1

Accepted Answer

The Ksp expression for Al(OH)3 is:Ksp = [Al3+][OH-]3At the point of precipitation, the ion product [Al3+][OH-]3 will be equal to Ksp.Substituting the given values into the Ksp expression, we get:2 x 10^-32 = [0.5 M][OH-]3Solving for [OH-], we get:[OH-] = 2 x 10^-11 MTo find the pH at which precipitation will begin, we use the pOH equation:pOH = -log[OH-]pOH = -log(2 x 10^-11) = 10.7pH = 14 - pOH = 14 - 10.7 = 3.3Therefore, the 0.5 M Al3+ solution will begin to show precipitation of Al(OH)3 at a pH of 3.3, which is closest to choice A) 3.5.

Question 13

The solubility of AgCl in water is _____ the solubility of AgCl in strong acid at the same temperature.
A)greater than
B)less than
C)about the same as
D)cannot be determined
E)much different from

Accepted Answer

The solubility of AgCl in water is about the same as in a strong acid because AgCl is a neutral salt and its solubility is not significantly affected by the presence of strong acids.

Question 14

Given the following K_sp values,which statement about solubility in mol/L in water is correct? A)PbCrO₄,Zn(OH)₂,and Pb(OH)₂ have equal solubilities in water. B)PbCrO₄ has the lowest solubility in water. C)The solubility of MnS in water will not be pH dependent. D)MnS has the highest molar solubility in water. E)A saturated PbCrO₄ solution will have a higher [Pb²⁺] than a saturated Pb(OH)₂ solution.

Accepted Answer

The solubility values indicate that PbCrO4 has the lowest solubility in water (2.0 x 10^-14 mol/L) compared to the other compounds listed.

Note: The other statements are incorrect. The solubility of MnS could be pH dependent, and the molar solubility of MnS is not the highest among the listed compounds. Additionally, it cannot be determined from the given data whether one saturated solution would have a higher concentration of a particular ion than another saturated solution.

Question 15

The K_sp for BaF₂ is 2.4 $\times$10^-⁵.When 10 mL of 0.01 M NaF is mixed with 10 mL of 0.01 M BaNO₃,will a precipitate form? A)No,because Q is 1 $\times$ 10^-¹² and since it is less than K_sp no precipitate will form. B)Yes,because Q is 1 $\times$ 10^-¹² and since it is less than K_sp a precipitate will form. C)No,because Q is 1.25 $\times$10^-⁷ and since it is less than K_sp no precipitate will form. D)Yes,because Q is 1.25 $\times$ 10^-⁷ and since it is less than K_sp a precipitate will form. E)None of the above.

Accepted Answer

The reaction quotient (Q) is calculated based on the concentrations of the ions in solution. After mixing, the concentration of each ion is halved due to dilution, making the concentration of F- and Ba2+ each 0.005 M. The Q for the reaction Ba2+ + 2F- -> BaF2 is calculated as [Ba2+][F-]^2 = (0.005)(0.005)^2 = 1.25 x 10^-7. Since this value is less than the Ksp of BaF2 (2.4 x 10^-5), no precipitate will form, indicating that the solution is unsaturated with respect to BaF2.

Question 16

A 50.0-mL sample of 0.100 M Ca(NO₃)₂ is mixed with 50.00 mL of 0.200 M NaF.When the system has come to equilibrium,which of the following sets of conditions will hold? The K_sp for CaF₂ is 4.0 $\times$ 10^-¹¹. Moles Solid CaF₂ A) B) C) D) E)

Accepted Answer

We can use the equilibrium constant expression for the reaction:

Ca(NOS1U1B13S1U1B0)S1U1B12S1U1B0 + 2 NaFS1U1B13S1U1B0 ⇌ CaFS1U1B12S1U1B0 + 2 NaNO3S1U1B10

K = [CaFS1U1B12S1U1B0][NaNO3S1U1B10]^2 / [Ca(NOS1U1B13S1U1B0)S1U1B12S1U1B0][NaFS1U1B13S1U1B0]^2

Since the initial concentrations of Ca(NOS1U1B13S1U1B0)S1U1B12S1U1B0 and NaFS1U1B13S1U1B0 are equal and the reaction stoichiometry is 1:2, we can assume that x moles of CaFS1U1B12S1U1B0 will be formed and 2x moles of NaNO3S1U1B10 will be formed at equilibrium. The initial moles of Ca(NOS1U1B13S1U1B0)S1U1B12S1U1B0 and NaFS1U1B13S1U1B0 in the mixture are:

moles Ca(NOS1U1B13S1U1B0)S1U1B12S1U1B0 = 0.100 mol/L x 0.050 L = 0.0050 mol
moles NaFS1U1B13S1U1B0 = 0.200 mol/L x 0.050 L = 0.0100 mol

After the reaction reaches equilibrium, the concentrations of CaFS1U1B12S1U1B0 and NaNO3S1U1B10 are:

[CaFS1U1B12S1U1B0] = x / (0.050 L + 0.050 L) = x / 0.100 L
[NaNO3S1U1B10] = 2x / (0.050 L + 0.050 L) = 2x / 0.100 L

[Ca(NOS1U1B13S1U1B0)S1U1B12S1U1B0] and [NaFS1U1B13S1U1B0] will be decreased by x and 2x moles, respectively, so their concentrations will be:

[Ca(NOS1U1B13S1U1B0)S1U1B12S1U1B0] = (0.0050 mol - x) / 0.100 L
[NaFS1U1B13S1U1B0] = (0.0100 mol - 2x) / 0.100 L

Substituting these expressions into the equilibrium constant expression and solving for x, we get:

4.0 F1F1F1S1F1F1F10 10S1U1P1-S1S1P0S1U1P111S1S1P0 = (x / 0.100)^1 (2x / 0.100)^2 / [(0.0050 - x) / 0.100]^1 [(0.0100 - 2x) / 0.100]^2

Simplifying and solving for x, we get:

x = 0.0557 mol

Therefore, the moles of CaFS1U1B12S1U1B0 formed at equilibrium is 0.0557 mol, which corresponds to choice C.

Question 17

Determine the equilibrium concentration of the chloride ion. A)3.9 $\times$ 10^-⁴ B)8.0 $\times$ 10^-⁶ C)2.8 $\times$ 10^-³ D)6.1 $\times$ 10^-² E)none of these

Accepted Answer

The equilibrium constant expression for the reaction is Kc = [Ag+(aq)][Cl-(aq)]/[AgCl(s)]. The initial concentration of AgNO3 is 3.9×10^-3 M, and assuming that all of it dissociates, the initial concentration of Ag+ and Cl- is also 3.9×10^-3 M. Letting x be the equilibrium concentration of Cl-, the equilibrium concentrations of Ag+ and AgCl(s) will also be x. Substituting the values into the equilibrium constant expression and solving for x gives x = 2.8×10^-3 M, which is choice C.

Question 18

The following questions refer to the following system: 3.5 $\times$ 10S1U1P12S1S1P0 mL of 3.2 M Pb(NOS1U1B13S1U1B0)S1U1B12S1U1B0 and 2.0 $\times$ 10S1U1P12S1S1P0 mL of 0.020 M NaCl are added together.KS1U1B1spS1U1B0 for the lead chloride is 1.6 $\times$ 10S1U1P1-S1S1P0S1U1P15S1S1P0. -Determine the ion product. A)1.1 $\times$ 10^-⁴ B)1.5 $\times$ 10^-² C)7.8 $\times$ 10^-³ D)8.1 $\times$ 10^-⁴ E)none of these

Accepted Answer

The ion product is the product of the concentrations of the lead and chloride ions in solution. The concentration of the lead ions can be calculated from the concentration of the lead nitrate solution and the stoichiometry of the reaction. The concentration of the chloride ions can be calculated from the concentration of the sodium chloride solution and the stoichiometry of the reaction. The ion product is then calculated by multiplying the concentrations of the lead and chloride ions.[Pb2+][Cl-] = (3.2 M)(2.0 mL)(1/50 mL)(1.6 x 10^-5) = 1.1 x 10^-4

Question 19

The solubility in mol/L of M(OH)₂ in 0.053 M KOH is 1.0 $\times$ 10^-⁵ mol/L.What is the K_sp for M(OH)₂? A) B) C) D) E)

Accepted Answer

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Question 20

The best explanation for the dissolution of ZnS in dilute HCl is that: A)The zinc ion is amphoteric. B)The sulfide-ion concentration is decreased by the formation of H₂S. C)the sulfide-ion concentration is decreased by oxidation to sulfur. D)the zinc-ion concentration is decreased by the formation of a chloro complex. E)The solubility product of ZnCl₂ is less than that of ZnS.

Accepted Answer

Dilute HCl reacts with ZnS to form H₂S, which decreases the concentration of sulfide ions and helps dissolve the ZnS. This explains the dissolution of ZnS in dilute HCl.

Quiz 16: Solubility and Complex Ion Equilibria

How many moles of CaF₂ will dissolve in 3.0 liters of 0.089 M NaF solution? (K_sp for CaF₂ = 4.0 $\times$ 10^-¹¹)

Which of the following solid salts is more soluble in 1.0 M H⁺ than in pure water?

What is the limiting reagent in the formation of the lead chloride?

What is the best way to ensure complete precipitation of SnS from a saturated H₂S solution?

Which of the following solid salts is more soluble in 1.0 M H⁺ than in pure water?

You have a solution consisting of 0.10 M Cl^- and 0.10 M CrO₄²^-.You add 0.10 M silver nitrate dropwise to this solution.Given that the K_sp for Ag₂CrO₄ is 9.0 $\times$ 10^-¹²,and that for AgCl is 1.6 $\times$ 10^-¹⁰,which of the following will precipitate first?

You have two salts,AgX and AgY,with very similar K_sp values.You know that K_a for HX is much greater than K_a for HY.Which salt is more soluble in acidic solution?

The two salts AgX and AgY exhibit very similar solubilities in water.It is known that the salt AgX is much more soluble in acid than is AgY.What can be said about the relative strengths of the acids HX and HY?

How many moles of Ca(NO₃) ₂ must be added to 1.0 L of a 0.182 M KF solution to begin precipitation of CaF₂? For CaF₂,K_sp = 4.0 $\times$ 10^-¹¹.

Will precipitation occur?

Calculate the solubility of Cu(OH) ₂ in a solution buffered at pH = 7.59.(K_sp = 1.6 $\times$ 10^-¹⁹)

The K_sp of Al(OH) ₃ is 2 $\times$ 10^-³².At what pH will a 0.5 M Al³⁺ solution begin to show precipitation of Al(OH) ₃?

The solubility of AgCl in water is _____ the solubility of AgCl in strong acid at the same temperature.

Given the following K_sp values,which statement about solubility in mol/L in water is correct?

The K_sp for BaF₂ is 2.4 $\times$ 10^-⁵.When 10 mL of 0.01 M NaF is mixed with 10 mL of 0.01 M BaNO₃,will a precipitate form?

Determine the equilibrium concentration of the chloride ion.

The following questions refer to the following system: 3.5 $\times$ 10² mL of 3.2 M Pb(NO₃) ₂ and 2.0 $\times$ 10² mL of 0.020 M NaCl are added together.K_sp for the lead chloride is 1.6 $\times$ 10^-⁵. -Determine the ion product.

The solubility in mol/L of M(OH) ₂ in 0.053 M KOH is 1.0 $\times$ 10^-⁵ mol/L.What is the K_sp for M(OH) ₂?

The best explanation for the dissolution of ZnS in dilute HCl is that:

Quiz 16: Solubility and Complex Ion Equilibria

How many moles of CaF2 will dissolve in 3.0 liters of 0.089 M NaF solution? (Ksp for CaF2 = 4.0 ×\times× 10-11)

Which of the following solid salts is more soluble in 1.0 M H+ than in pure water?

What is the limiting reagent in the formation of the lead chloride?

What is the best way to ensure complete precipitation of SnS from a saturated H2S solution?

Which of the following solid salts is more soluble in 1.0 M H+ than in pure water?

You have a solution consisting of 0.10 M Cl- and 0.10 M CrO42-.You add 0.10 M silver nitrate dropwise to this solution.Given that the Ksp for Ag2CrO4 is 9.0 ×\times× 10-12,and that for AgCl is 1.6 ×\times× 10-10,which of the following will precipitate first?

You have two salts,AgX and AgY,with very similar Ksp values.You know that Ka for HX is much greater than Ka for HY.Which salt is more soluble in acidic solution?

The two salts AgX and AgY exhibit very similar solubilities in water.It is known that the salt AgX is much more soluble in acid than is AgY.What can be said about the relative strengths of the acids HX and HY?

How many moles of Ca(NO3) 2 must be added to 1.0 L of a 0.182 M KF solution to begin precipitation of CaF2? For CaF2,Ksp = 4.0 ×\times× 10-11.

Will precipitation occur?

Calculate the solubility of Cu(OH) 2 in a solution buffered at pH = 7.59.(Ksp = 1.6 ×\times× 10-19)

The Ksp of Al(OH) 3 is 2 ×\times× 10-32.At what pH will a 0.5 M Al3+ solution begin to show precipitation of Al(OH) 3?

The solubility of AgCl in water is _____ the solubility of AgCl in strong acid at the same temperature.

Given the following Ksp values,which statement about solubility in mol/L in water is correct?

The Ksp for BaF2 is 2.4 ×\times× 10-5.When 10 mL of 0.01 M NaF is mixed with 10 mL of 0.01 M BaNO3,will a precipitate form?

A 50.0-mL sample of 0.100 M Ca(NO3) 2 is mixed with 50.00 mL of 0.200 M NaF.When the system has come to equilibrium,which of the following sets of conditions will hold? The Ksp for CaF2 is 4.0 ×\times× 10-11. Moles Solid CaF2

Determine the equilibrium concentration of the chloride ion.

The following questions refer to the following system: 3.5 ×\times× 102 mL of 3.2 M Pb(NO3) 2 and 2.0 ×\times× 102 mL of 0.020 M NaCl are added together.Ksp for the lead chloride is 1.6 ×\times× 10-5. -Determine the ion product.

The solubility in mol/L of M(OH) 2 in 0.053 M KOH is 1.0 ×\times× 10-5 mol/L.What is the Ksp for M(OH) 2?

The best explanation for the dissolution of ZnS in dilute HCl is that:

How many moles of CaF₂ will dissolve in 3.0 liters of 0.089 M NaF solution? (K_sp for CaF₂ = 4.0 $\times$ 10^-¹¹)

Which of the following solid salts is more soluble in 1.0 M H⁺ than in pure water?

What is the best way to ensure complete precipitation of SnS from a saturated H₂S solution?

Which of the following solid salts is more soluble in 1.0 M H⁺ than in pure water?

You have a solution consisting of 0.10 M Cl^- and 0.10 M CrO₄²^-.You add 0.10 M silver nitrate dropwise to this solution.Given that the K_sp for Ag₂CrO₄ is 9.0 $\times$ 10^-¹²,and that for AgCl is 1.6 $\times$ 10^-¹⁰,which of the following will precipitate first?

You have two salts,AgX and AgY,with very similar K_sp values.You know that K_a for HX is much greater than K_a for HY.Which salt is more soluble in acidic solution?

How many moles of Ca(NO₃) ₂ must be added to 1.0 L of a 0.182 M KF solution to begin precipitation of CaF₂? For CaF₂,K_sp = 4.0 $\times$ 10^-¹¹.

Calculate the solubility of Cu(OH) ₂ in a solution buffered at pH = 7.59.(K_sp = 1.6 $\times$ 10^-¹⁹)

The K_sp of Al(OH) ₃ is 2 $\times$ 10^-³².At what pH will a 0.5 M Al³⁺ solution begin to show precipitation of Al(OH) ₃?

Given the following K_sp values,which statement about solubility in mol/L in water is correct?

The K_sp for BaF₂ is 2.4 $\times$ 10^-⁵.When 10 mL of 0.01 M NaF is mixed with 10 mL of 0.01 M BaNO₃,will a precipitate form?

The following questions refer to the following system: 3.5 $\times$ 10² mL of 3.2 M Pb(NO₃) ₂ and 2.0 $\times$ 10² mL of 0.020 M NaCl are added together.K_sp for the lead chloride is 1.6 $\times$ 10^-⁵. -Determine the ion product.

The solubility in mol/L of M(OH) ₂ in 0.053 M KOH is 1.0 $\times$ 10^-⁵ mol/L.What is the K_sp for M(OH) ₂?