Question 1

Calculate the solubility of zinc hydroxide, Zn(OH)₂, in 1.00 M NaOH. K_sp = 3.0 × 10^-16 for Zn(OH)₂, K_f = 3.0 × 10¹⁵ for Zn(OH)₄^2- A) 0.60 M B) 0.52 M C) 0.37 M D) 0.32 M E) 0.24 M

Accepted Answer

The answer of Calculate the solubility of zinc hydroxide, Zn(OH)₂,...

Question 2

Consider the dissolution of MnS in water How is the solubility of manganese(II) sulfide affected by the addition of aqueous potassium hydroxide to the system? A) The solubility will be unchanged. B) The solubility will decrease. C) The solubility will increase. D) The amount of KOH added must be known before its effect can be predicted. E) The pK_a of H₂S is needed before a reliable prediction can be made.

Accepted Answer

The addition of aqueous potassium hydroxide to the system will result in the formation of insoluble manganese(II) hydroxide, according to the following reaction:
MnS + 2KOH → Mn(OH)2 + K2S
As a result, the concentration of manganese(II) ions in solution will decrease, causing a decrease in the solubility of MnS.

Question 3

A solution is prepared by adding 4.50 mol of sodium hydroxide to 1.00 L of 1.00 M Co(NO₃)₂. What is the equilibrium concentration of cobalt ions? K_f= 5.0 × 10⁹ for Co(OH)₄^2- A) 1.1 × 10^-2 M B) 1.4 × 10^-5 M C) 3.2 × 10^-9 M D) 2.0 × 10^-10 M E) 4.9 × 10^-13 M

Accepted Answer

First, we can write the balanced chemical equation for the reaction:

Co(NOS1U1B13S1U1B0)S1U1B12S1U1B0 + 2 NaOH → Co(OH)S1U1B14S1U1B0S1U1P12-S1S1P0 + Na2(NOS1U1B13S1U1B0)S1U1B12S1U1B0

Since we added 4.50 mol of NaOH to a 1.00 L solution of 1.00 M Co(NOS1U1B13S1U1B0)S1U1B12S1U1B0, we now have a total volume of 1.00 L + 4.50 L = 5.50 L of solution. The initial concentration of Co(NOS1U1B13S1U1B0)S1U1B12S1U1B0 was 1.00 M, so there are 1.00 mol/L × 1.00 L = 1.00 mol of Co(NOS1U1B13S1U1B0)S1U1B12S1U1B0 in the solution.

We can then use the equilibrium constant (KS1U1B1f S1U1B0= 5.0 × 10S1U1P19S1S1P0) to calculate the concentration of Co(OH)S1U1B14S1U1B0S1U1P12-S1S1P0 at equilibrium. We can assume that the amount of Co(NOS1U1B13S1U1B0)S1U1B12S1U1B0 that reacts with NaOH is small compared to the initial amount, so the change in concentration of Co(NOS1U1B13S1U1B0)S1U1B12S1U1B0 is negligible.

Let x be the equilibrium concentration of Co(OH)S1U1B14S1U1B0S1U1P12-S1S1P0, in mol/L. Then the equilibrium concentrations of NaOH and Na2(NOS1U1B13S1U1B0)S1U1B12S1U1B0 are both 4.50 mol/5.50 L = 0.818 M. The equilibrium concentration of Co(NOS1U1B13S1U1B0)S1U1B12S1U1B0 is approximately 1.00 M, as mentioned above. Using the equilibrium constant expression:

KS1U1B1f S1U1B0= [Co(OH)S1U1B14S1U1B0S1U1P12-S1S1P0][Na2(NOS1U1B13S1U1B0)S1U1B12S1U1B0]/[Co(NOS1U1B13S1U1B0)S1U1B12S1U1B0][NaOH]S1U1P2S1S1P0

We can plug in the concentrations and the value of the equilibrium constant:

5.0 × 10S1U1P19S1S1P0 = x × (0.818 M)S1U1P2S1S1P0 / (1.00 M)S1U1P1S1S1P0

Solving for x, we get:

x = 3.2 × 10S1U1P1-9S1S1P0 M

Therefore, the equilibrium concentration of cobalt ions is 3.2 × 10S1U1P1-9S1S1P0 M, which corresponds to choice (C).

Question 4

Use the following information to calculate the solubility product constant, K_sp, for PbCl₂. A saturated solution of PbCl₂ in water was prepared and filtered. From the filtrate, 1.0 L was measured out into a beaker and evaporated to dryness. The solid PbCl₂residue recovered in the beaker amounted to 0.0162 moles. A) K_sp = 6.9 × 10^-8 B) K_sp = 4.3 × 10^-6 C) K_sp = 1.7 × 10^-5 D) K_sp = 2.6 × 10^-4 E) K_sp = 3.2 × 10^-2

Accepted Answer

The answer of Use the following information to calculate the...

Question 5

Calculate the solubility of magnesium sulfate, MgSO₄, when placed into a 0.10 M MgCl₂solution. K_sp = 5.9 × 10^-3 A) 4.2 × 10^-2M B) 5.9 × 10^-2M C) 7.7 × 10^-2M D) 3.5 × 10^-5M E) 3.5 × 10^-6M

Accepted Answer

The answer of Calculate the solubility of magnesium sulfate, MgSO₄,...

Question 6

The salts X(NO₃)₂and Y(NO₃)₂ (where X⁺ and Y⁺ are metal ions) are dissolved in water to give a solution which is 0.1 M in each of them. Using the K_sp values listed below, decide which aqueous reagent, if any, will definitely precipitate X⁺ before precipitating Y⁺ from solution. Given K_spvalues: XCl₂, 1 × 10^-5YCl₂, 1 × 10^-10X(OH)₂, 1 × 10^-10Y(OH)₂, 1 × 10^-5 A) 1 M NaNO₃ B) 1 M HCl C) 1 M HNO₃ D) 1 M NaCl E) None of the above reagents will accomplish the precipitation.

Accepted Answer

The answer of The salts X(NO₃)₂and Y(NO₃)₂ (where X⁺...

Question 7

A lab technician adds 0.015 mol of KOH to 1.00 L of 0.0010 M Ca(NO₃)₂. K_sp = 6.5 × 10^-6 for Ca(OH)₂). Which of the following statements is correct? A) Calcium hydroxide precipitates until the solution is saturated. B) The solution is unsaturated and no precipitate forms. C) The concentration of calcium ions is reduced by the addition of the hydroxide ions. D) One must know K_sp for calcium nitrate to make meaningful predictions on this system. E) The presence of KOH will raise the solubility of Ca(NO₃)₂.

Accepted Answer

The concentration of Ca(NO₃)₂ is only 0.0010 M, which is less than the solubility product constant for Ca(OH)₂ (6.5 × 10^-6). Therefore, the addition of KOH will not result in the precipitation of Ca(OH)₂ and the solution remains unsaturated. Choices A, C, D, and E are incorrect.

Question 8

Barium sulfate (BaSO₄) is a slightly soluble salt, with K_sp = 1.1 × 10^-10. What mass of Ba²⁺ ions will be present in 1.0 L of a saturated solution of barium sulfate? A) < 10^-7 g B) 1.0 × 10^-5 g C) 0.0014 g D) 0.0024 g E) > 0.05 g

Accepted Answer

Using the solubility product expression, we can find the concentration of Ba2+ ions in a saturated solution of BaSO4: K_sp = [Ba2+][SO42-] 1.1 × 10^-10 = x(2x) where x is the concentration of Ba2+ ions and 2x is the concentration of SO42- ions (since the stoichiometric ratio between BaSO4 and Ba2+ and SO42- is 1:1). Solving for x, we get: x = 1.04 × 10^-5 M Therefore, the molar mass of BaSO4 is 233.4 g/mol. In 1.0 L of solution, there will be 1.04 × 10^-5 mol of Ba2+ ions, which corresponds to a mass of: 1.04 × 10^-5 mol × 137.3 g/mol = 0.014 g Note that this assumes that only Ba2+ ions are present in solution (ignoring the small amount of SO42- ions present due to the slight solubility of BaSO4). The actual mass of Ba²⁺ ions present in solution may be slightly higher. Therefore, the best answer choice is C, 0.0014 g.

Question 9

Calculate the solubility of silver chromate, Ag₂CrO₄, in 0.005 M Na₂CrO₄. K_sp = 2.6 × 10^-12 A) 1.4 × 10^-4M B) 3.4 × 10^-5M C) 1.1 × 10^-5M D) 1.6 × 10^-6M E) < 1.0 × 10^-6M

Accepted Answer

The solubility product expression for silver chromate is:

Ksp = [Ag2CrO4] = [Ag+]^2[CrO4 2-]

The balanced equation for the dissolution of Ag2CrO4 in water is:

Ag2CrO4(s) ⇌ 2Ag+(aq) + CrO4 2-(aq)

Since we are given an initial concentration of Na2CrO4, we can assume that there is excess CrO4 2- in solution, and that the dissolution reaction will proceed until equilibrium is reached. At equilibrium, we can assume that:

[Ag+] = 2x
[CrO4 2-] = (0.005 M) - 2x

Substituting these expressions into the Ksp expression, we get:

Ksp = (2x)^2[(0.005 M) - 2x] = 4x^2(0.005 - 2x)

Solving for x, we get:

4x^2(0.005 - 2x) = 2.6 × 10^-12
8x^3 - 0.02x^2 + 2.6 × 10^-12 = 0

Using a calculator or solver tool, we find that x = 1.05 × 10^-5.

Therefore, [Ag+] = 2x = 2.1 × 10^-5 M, and the solubility of Ag2CrO4 is:

[Ag2CrO4] = [Ag+]^2[CrO4 2-] = (2.1 × 10^-5)^2(0.005 - 2x) = 1.1 × 10^-5 M

So the correct answer is C.

Question 10

Which of the following substances has the greatest solubility in water? A) Ba(IO₃)₂, K_sp = 1.5 × 10^-9 B) PbF₂, K_sp = 3.6 × 10^-8 C) SrSO₄, K_s_p = 3.2 × 10^-7 D) CuCl, K_sp = 1.9 × 10^-7 E) CdS, K_sp = 1.0 × 10^-24

Accepted Answer

The answer of Which of the following substances has the...

Question 11

Use the following information to calculate the solubility product constant, K_sp, for CuCl. A saturated solution of CuCl in water was prepared and filtered. From the filtrate, 1.0 L was measured out into a beaker and evaporated to dryness. The solid CuClresidue recovered in the beaker was found to weigh 0.041g. A) K_sp =1.7 × 10^-9 B) K_sp = 1.7 × 10^-7 C) K_sp = 1.7 × 10^-5 D) K_sp = 4.3 × 10^-4 E) K_sp = 2.1 × 10^-2

Accepted Answer

The answer of Use the following information to calculate the...

Question 12

A lab technician adds 0.20 mol of NaF to 1.00 L of 0.35 M cadmium nitrate, Cd(NO₃)₂. Which of the following statements is correct? K_sp = 6.44 × 10^-3 for CdF₂. A) Cadmium fluoride precipitates until the solution is saturated. B) The solution is unsaturated and no precipitate forms. C) The solubility of cadmium fluoride is increased by the presence of additional fluoride ions. D) One must know K_sp for cadmium nitrate to make meaningful predictions on this system. E) The presence of NaF will raise the solubility of Cd(NO₃)₂.

Accepted Answer

The addition of NaF will increase the concentration of fluoride ions in the solution, which will increase the solubility product of CdF2. Since the solubility product of CdF2 is less than the product of the ion concentrations in the solution (Q), CdF2 will precipitate until the solution is saturated. Hence, choice A is correct.

Question 13

Assuming that the total volume does not change after 0.200 g of KCl is added to 1.0 L of a saturated aqueous solution of AgCl, calculate the number of moles of Ag⁺ ion in the solution after equilibrium has been reestablished. For AgCl, K_sp = 1.8 × 10^{- 10}. A) 1.8 × 10^-10 mol Ag⁺ B) 9.0 × 10^-10mol Ag⁺ C) 9.0 × 10^-9mol Ag⁺ D) 6.7 × 10^-8mol Ag⁺ E) 1.3 × 10^-5mol Ag⁺

Accepted Answer

First, write the chemical equation for the dissolution of AgCl in water:

AgCl (s) ⇌ Ag+ (aq) + Cl- (aq)

The dissociation equilibrium constant for this reaction is given:

Ksp = [Ag+][Cl-]

The addition of KCl (which fully dissociates into K+ and Cl-) will cause an increase in the concentration of Cl- ions in the solution. This can lead to more AgCl dissolving according to the above equation. However, since the solution was initially saturated with AgCl, any additional AgCl that dissolves will immediately precipitate out again. This means that the concentration of Ag+ ions will remain the same, but the concentration of Cl- ions will increase until the solubility product for AgCl is reached again:

Ksp = [Ag+][Cl-] = 1.8 × 10^-10

Let x be the increase in the concentration of Cl- ions due to the addition of KCl. Then, the equilibrium concentrations of Ag+ and Cl- ions are:

[Ag+] = [Cl-] = x

Since the total volume of the solution does not change, the total number of moles of Ag+ and Cl- ions before and after the addition of KCl must be the same. Initially, there is 1.8 × 10^-10 mol of Ag+ ion in the solution (from the dissolution of AgCl), and 1.8 × 10^-10 mol of Cl- ion (from the dissociation of AgCl). After the addition of KCl, there will be an additional x mol of Cl- ion, giving a total of:

1.8 × 10^-10 mol Ag+ ion
(1.8 × 10^-10 + x) mol Cl- ion

Since KCl fully dissociates in water, the additional x mol of Cl- ion comes from the dissociation of 0.200 g of KCl. First, calculate the molar mass of KCl:

K = 39.10 g/mol
Cl = 35.45 g/mol
Molar mass of KCl = 39.10 + 35.45 = 74.55 g/mol

Convert the mass of KCl to moles:

0.200 g KCl × 1 mol KCl/74.55 g KCl = 0.00268 mol KCl

Since KCl dissociates completely into K+ and Cl- ions, there will be 0.00268 mol of Cl- ion added to the solution. This gives:

(1.8 × 10^-10 + x) mol Cl- ion = 1.8 × 10^-10 mol Cl- ion + 0.00268 mol Cl- ion

Solving for x:

x = 0.00268 mol Cl- ion

Since [Ag+] = [Cl-] = x, the concentration of Ag+ ion after equilibrium has been reestablished is also 0.00268 mol/L. The total number of moles of Ag+ and Cl- ions in the solution is:

2 × 0.00268 mol = 0.00536 mol

However, the question asks for the number of moles of Ag+ and Cl- ions COMBINED in the form of AgCl. Since AgCl dissociates into one Ag+ ion and one Cl- ion, the number of moles of AgCl present at equilibrium is:

0.00536 mol Ag+ ion × (1 mol AgCl/1 mol Ag+ ion) = 0.00536 mol AgCl

Therefore, the number of moles of Ag+ and Cl- ions in the solution after equilibrium has been reestablished is 0.00536 mol AgCl, or approximately 6.7 × 10^-8 mol AgCl. The answer is D.

Question 14

The salts X(NO₃)₂and Y(NO₃)₂ (where X⁺ and Y⁺ are metal ions) are dissolved in water to give a solution which is 0.1 M in each of them. Which of the answers gives the concentration of chloride ions will precipitate the most YCl₂ without precipitating any XCl₂? Given K_spvalues: XCl₂, 2 × 10^-5YCl₂, 1 × 10^-¹⁰ A) 1 M Cl^- B) 0.1 M Cl^- C) 0.01 M Cl^- D) 0.001 M Cl^- E) 0.0001 M Cl^-

Accepted Answer

The answer of The salts X(NO₃)₂and Y(NO₃)₂ (where X⁺...

Question 15

Calculate the solubility of lead(II) iodide, PbI₂, in 0.025 M KI. K_sp = 7.9 × 10^-9 A) 4.5 × 10^-2M B) 2.8 × 10^-2M C) 8.9 × 10^-5M D) 5.0 × 10^-5M E) 1.3 × 10^-5M

Accepted Answer

The answer of Calculate the solubility of lead(II) iodide, PbI₂,...

Question 16

The salts X(NO₃)₂and Y(NO₃)₂ (where X⁺ and Y⁺ are metal ions) are dissolved in water to give a solution which is 0.1 M in each of them. Using the K_sp values listed below, decide which aqueous reagent, if any, will definitely precipitate X⁺ before precipitating Y⁺ from solution. Given K_spvalues: XCl₂, 1 × 10^-5YCl₂, 1 × 10^-10X(OH)₂, 1 × 10^-10Y(OH)₂, 1 × 10^-5 A) 1 M NH₃ B) 1 M HCl C) 1 M HNO₃ D) 1 M NaCl E) None of the above reagents will accomplish the precipitation.

Accepted Answer

The answer of The salts X(NO₃)₂and Y(NO₃)₂ (where X⁺...

Question 17

The salts X(NO₃)₂and Y(NO₃)₂ (where X⁺ and Y⁺ are metal ions) are dissolved in water to give a solution which is 0.1 M in each of them. Using the K_sp values listed below, decide which aqueous reagent, if any, will definitely precipitate X⁺ before precipitating Y⁺ from solution. Given K_spvalues: XCl₂, 1 × 10^-5YCl₂, 1 × 10^-10X(OH)₂, 1 × 10^-10Y(OH)₂, 1 × 10^-5 A) 1 M NaCl B) 1 M HCl C) 1 M HNO₃ D) 1 M NaOH E) None of the above reagents will accomplish the precipitation.

Accepted Answer

To determine which reagent will precipitate X before Y, we need to compare the Ksp values of the possible precipitates. The smaller the Ksp value, the less soluble the compound and the more likely it will precipitate first. 
Using the given Ksp values: 
- Ksp for XCl2 is 1 × 10^−5 
- Ksp for YCl2 is 1 × 10^−10 
- Ksp for X(OH)2 is 1 × 10^−10 
- Ksp for Y(OH)2 is 1 × 10^−5

Comparing the Ksp values, we can see that X will precipitate before Y if we add a strong base that will deprotonate X(OH)2 and drive its Ksp value down. The only option in the given reagents that fits this criteria is 1 M NaOH, which will deprotonate X(OH)2 to form the insoluble X(OH)2 precipitate before Y(OH)2 precipitates. Therefore, the answer is D.

Question 18

A solution is prepared by mixing 50.0 mL of 0.50 M Cu(NO₃)₂with 50.0 mL of 0.50 M Co(NO₃)₂. Sodium hydroxide is added to the mixture. Which hydroxide precipitates first and what concentration of hydroxide ions present in solution will accomplish the separation? K_sp = 2.2 × 10^-20 for Cu(OH)₂, K_sp = 1.3 × 10^-15for Co(OH)₂ A) Co(OH)₂; [OH^-] = 6.9 × 10^-6 M B) Co(OH)₂; [OH^-] = 2.6 × 10^-7 M C) Cu(OH)₂; [OH^-] = 1.8 × 10^-7 M D) Cu(OH)₂; [OH^-] = 1.1 × 10^-9 M E) Cu(OH)₂; [OH^-] = 1.0 × 10^-17 M

Accepted Answer

The answer of A solution is prepared by mixing 50.0...

Question 19

What is the maximum amount of sodium sulfate that can be added to 1.00 L of 0.0020 M Ca(NO₃)₂before precipitation of calcium sulfate begins? K_sp = 2.4 × 10^-5 for calcium sulfate. A) 1.2 × 10^-2 mol B) 4.9 × 10^-3 mol C) 3.5 × 10^-3 mol D) 1.2 × 10^-5 mol E) 4.8 × 10^-8 mol

Accepted Answer

The answer of What is the maximum amount of sodium...

Question 20

What is the maximum mass of KCl that can be added to1.0 L of a 0.010 M lead(II) chloride solution without causing any precipitation of lead(II) chloride? Assume that addition of KCl does not affect the solution volume. For lead(II) chloride, K_sp = 1.6 × 10^-5. A) 3.0 g B) 1.5 g C) 0.8 g D) 1.0 g E) 0.2 g

Accepted Answer

The answer of What is the maximum mass of KCl...

Quiz 19: Ionic Equilibria in Aqueous Systems

Calculate the solubility of zinc hydroxide, Zn(OH) ₂, in 1.00 M NaOH. K_sp = 3.0 × 10^-16 for Zn(OH) ₂, K_f = 3.0 × 10¹⁵ for Zn(OH) ₄^2-

Consider the dissolution of MnS in water How is the solubility of manganese(II) sulfide affected by the addition of aqueous potassium hydroxide to the system?

A solution is prepared by adding 4.50 mol of sodium hydroxide to 1.00 L of 1.00 M Co(NO₃) ₂. What is the equilibrium concentration of cobalt ions? K_f= 5.0 × 10⁹ for Co(OH) ₄^2-

Calculate the solubility of magnesium sulfate, MgSO₄, when placed into a 0.10 M MgCl₂solution. K_sp = 5.9 × 10^-3

A lab technician adds 0.015 mol of KOH to 1.00 L of 0.0010 M Ca(NO₃) ₂. K_sp = 6.5 × 10^-6 for Ca(OH) ₂) . Which of the following statements is correct?

Barium sulfate (BaSO₄) is a slightly soluble salt, with K_sp = 1.1 × 10^-10. What mass of Ba²⁺ ions will be present in 1.0 L of a saturated solution of barium sulfate?

Calculate the solubility of silver chromate, Ag₂CrO₄, in 0.005 M Na₂CrO₄. K_sp = 2.6 × 10^-12

Which of the following substances has the greatest solubility in water?

A lab technician adds 0.20 mol of NaF to 1.00 L of 0.35 M cadmium nitrate, Cd(NO₃) ₂. Which of the following statements is correct? K_sp = 6.44 × 10^-3 for CdF₂.

Assuming that the total volume does not change after 0.200 g of KCl is added to 1.0 L of a saturated aqueous solution of AgCl, calculate the number of moles of Ag⁺ ion in the solution after equilibrium has been reestablished. For AgCl, K_sp = 1.8 × 10^{- 10}.

Calculate the solubility of lead(II) iodide, PbI₂, in 0.025 M KI. K_sp = 7.9 × 10^-9

What is the maximum amount of sodium sulfate that can be added to 1.00 L of 0.0020 M Ca(NO₃) ₂before precipitation of calcium sulfate begins? K_sp = 2.4 × 10^-5 for calcium sulfate.

What is the maximum mass of KCl that can be added to1.0 L of a 0.010 M lead(II) chloride solution without causing any precipitation of lead(II) chloride? Assume that addition of KCl does not affect the solution volume. For lead(II) chloride, K_sp = 1.6 × 10^-5.

Quiz 19: Ionic Equilibria in Aqueous Systems

Calculate the solubility of zinc hydroxide, Zn(OH) 2, in 1.00 M NaOH. Ksp = 3.0 × 10-16 for Zn(OH) 2, Kf = 3.0 × 1015 for Zn(OH) 42-

Consider the dissolution of MnS in water How is the solubility of manganese(II) sulfide affected by the addition of aqueous potassium hydroxide to the system?

A solution is prepared by adding 4.50 mol of sodium hydroxide to 1.00 L of 1.00 M Co(NO3) 2. What is the equilibrium concentration of cobalt ions? Kf = 5.0 × 109 for Co(OH) 42-

Calculate the solubility of magnesium sulfate, MgSO4, when placed into a 0.10 M MgCl2 solution. Ksp = 5.9 × 10-3

A lab technician adds 0.015 mol of KOH to 1.00 L of 0.0010 M Ca(NO3) 2. Ksp = 6.5 × 10-6 for Ca(OH) 2) . Which of the following statements is correct?

Barium sulfate (BaSO4) is a slightly soluble salt, with Ksp = 1.1 × 10-10. What mass of Ba2+ ions will be present in 1.0 L of a saturated solution of barium sulfate?

Calculate the solubility of silver chromate, Ag2CrO4, in 0.005 M Na2CrO4. Ksp = 2.6 × 10-12

Which of the following substances has the greatest solubility in water?

A lab technician adds 0.20 mol of NaF to 1.00 L of 0.35 M cadmium nitrate, Cd(NO3) 2. Which of the following statements is correct? Ksp = 6.44 × 10-3 for CdF2.

Assuming that the total volume does not change after 0.200 g of KCl is added to 1.0 L of a saturated aqueous solution of AgCl, calculate the number of moles of Ag+ ion in the solution after equilibrium has been reestablished. For AgCl, Ksp = 1.8 × 10- 10.

Calculate the solubility of lead(II) iodide, PbI2, in 0.025 M KI. Ksp = 7.9 × 10-9

What is the maximum amount of sodium sulfate that can be added to 1.00 L of 0.0020 M Ca(NO3) 2 before precipitation of calcium sulfate begins? Ksp = 2.4 × 10-5 for calcium sulfate.

What is the maximum mass of KCl that can be added to1.0 L of a 0.010 M lead(II) chloride solution without causing any precipitation of lead(II) chloride? Assume that addition of KCl does not affect the solution volume. For lead(II) chloride, Ksp = 1.6 × 10-5 .

Calculate the solubility of zinc hydroxide, Zn(OH) ₂, in 1.00 M NaOH. K_sp = 3.0 × 10^-16 for Zn(OH) ₂, K_f = 3.0 × 10¹⁵ for Zn(OH) ₄^2-

A solution is prepared by adding 4.50 mol of sodium hydroxide to 1.00 L of 1.00 M Co(NO₃) ₂. What is the equilibrium concentration of cobalt ions? K_f= 5.0 × 10⁹ for Co(OH) ₄^2-

Calculate the solubility of magnesium sulfate, MgSO₄, when placed into a 0.10 M MgCl₂solution. K_sp = 5.9 × 10^-3

A lab technician adds 0.015 mol of KOH to 1.00 L of 0.0010 M Ca(NO₃) ₂. K_sp = 6.5 × 10^-6 for Ca(OH) ₂) . Which of the following statements is correct?

Barium sulfate (BaSO₄) is a slightly soluble salt, with K_sp = 1.1 × 10^-10. What mass of Ba²⁺ ions will be present in 1.0 L of a saturated solution of barium sulfate?

Calculate the solubility of silver chromate, Ag₂CrO₄, in 0.005 M Na₂CrO₄. K_sp = 2.6 × 10^-12

A lab technician adds 0.20 mol of NaF to 1.00 L of 0.35 M cadmium nitrate, Cd(NO₃) ₂. Which of the following statements is correct? K_sp = 6.44 × 10^-3 for CdF₂.

Assuming that the total volume does not change after 0.200 g of KCl is added to 1.0 L of a saturated aqueous solution of AgCl, calculate the number of moles of Ag⁺ ion in the solution after equilibrium has been reestablished. For AgCl, K_sp = 1.8 × 10^{- 10}.

Calculate the solubility of lead(II) iodide, PbI₂, in 0.025 M KI. K_sp = 7.9 × 10^-9

What is the maximum amount of sodium sulfate that can be added to 1.00 L of 0.0020 M Ca(NO₃) ₂before precipitation of calcium sulfate begins? K_sp = 2.4 × 10^-5 for calcium sulfate.

What is the maximum mass of KCl that can be added to1.0 L of a 0.010 M lead(II) chloride solution without causing any precipitation of lead(II) chloride? Assume that addition of KCl does not affect the solution volume. For lead(II) chloride, K_sp = 1.6 × 10^-5.