Question 1

The concentration of the complex ion in each of following solutions is 1.00 M. In which of the solutions will the concentration of the uncomplexed metal ion be the greatest? Hg(CN)₄^2- K_f = 9.3 × 10³⁸ Be(OH)₄^2- K_f = 4.0 × 10¹⁸ Zn(OH)₄^2- K_f = 3.0 × 10¹⁵ Cu(NH₃)₄²⁺ K_f = 5.6 × 10¹¹ CdI₄^2- K_f = 1.0 × 10⁶

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Question 2

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 CuCl residue recovered in the beaker was found to weigh 0.041g.

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Question 3

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? K_sp = 2.4 × 10^-5 for calcium sulfate.

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Question 4

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.

Accepted Answer

The balanced equation for the dissolution of AgCl is:

AgCl(s) ↔ Ag+(aq) + Cl-(aq)

The dissolution reaction is in equilibrium when the rate of the forward reaction, dissolution of AgCl(s), is equal to the rate of the reverse reaction, precipitation of AgCl(s). When the solution is saturated, the rate of dissolution is equal to the rate of precipitation, and the concentration of Ag+ and Cl- in the solution is constant.

When Ksp is known, the equilibrium concentrations of the ions in the solution can be calculated. The equilibrium constant expression for AgCl is:

Ksp = [Ag+][Cl-] = 1.8 × 10^-10

Assuming that no other sources of Ag+ or Cl- are present, the ion concentrations in the saturated AgCl solution are:

[Ag+] = [Cl-] = sqrt(Ksp) = sqrt(1.8 × 10^-10) = 1.34 × 10^-5 M

Now that we have established the equilibrium concentrations of Ag+ and Cl- in the solution, we can consider the effect of adding KCl to the solution. When KCl dissolves in water, it dissociates into K+ and Cl- ions:

KCl(s) → K+(aq) + Cl-(aq)

The addition of KCl does not affect the concentration of Cl- ions in the solution, because the concentration of Cl- is already at its maximum (saturated). The addition of more Cl- ions would cause more AgCl to precipitate until the concentration of Cl- ions was brought back to its original value.

The addition of K+ ions, however, does affect the equilibrium. K+ ions can combine with the Cl- ions already present to form KCl(s), reducing the concentration of Cl- ions in the solution. This causes more AgCl to dissolve until the concentration of Cl- ions is restored to its original value.

The amount of K+ ions added to the solution is:

nK+ = mKCl / MK+

where mKCl is the mass of KCl added and MK+ is the molar mass of K+.

mKCl = 0.200 g
MK+ = 39.1 g/mol

nK+ = (0.200 g / 39.1 g/mol) = 0.00511 mol

Since KCl is a 1:1 electrolyte, the number of moles of Cl- ions that are combined with K+ ions and removed from the solution is also 0.00511 mol. Thus, the new concentration of Cl- ions in the solution is:

[Cl-] = [Cl-]initial - nCl-

where [Cl-]initial is the initial concentration of Cl- in the solution, and nCl- is the number of moles of Cl- ions removed from the solution.

[Cl-]initial = sqrt(Ksp) = 1.34 × 10^-5 M
nCl- = 0.00511 mol

[Cl-] = (1.34 × 10^-5 M) - (0.00511 mol / 1 L) = 1.2889 × 10^-5 M

Since [Ag+] and [Cl-] are now different from their initial equilibrium values, the dissolution of AgCl must occur until a new equilibrium is established. The concentration of Ag+ ions in the solution will increase until the product of [Ag+] and [Cl-] is equal to Ksp.

[Ag+] = Ksp / [Cl-] = (1.8 × 10^-10) / (1.2889 × 10^-5 M) = 1.3967 × 10^-6 M

Therefore, the number of moles of Ag+ ions in the solution is:

nAg+ = [Ag+] × V = (1.3967 × 10^-6 mol / 1 L) × (1 L) = 1.3967 × 10^-6 mol

The number of moles of Cl- ions is:

nCl- = [Cl-] × V = (1.2889 × 10^-5 mol / 1 L) × (1 L) = 1.2889 × 10^-5 mol

The total number of moles of Ag+ and Cl- ions in the solution is:

nAg+ + nCl- = (1.3967 × 10^-6 mol) + (1.2889 × 10^-5 mol) = 1.4286 × 10^-5 mol

However, we were asked for the number of moles of AgCl ion (Ag+ + Cl-), not just the separate ions. So we need to add nAg+ and nCl-:

nAgCl = nAg+ + nCl- = 1.4286 × 10^-5 mol

Thus, the answer is D) 6.7 × 10^-8 mol AgCl.

Question 5

Consider the dissolution of MnS in water (K_sp = 3.0 × 10^-14). MnS(s) + H₂O(l) Mn²⁺(aq) + HS^-(aq) + OH^-(aq) How is the solubility of manganese(II) sulfide affected by the addition of aqueous potassium hydroxide to the system?

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Question 6

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

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Question 7

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?

Accepted Answer

Using the solubility product expression:
KS1U1B1spS1U1B0 = [Ba2+S1U1P][(SO4)2-S1U1P]
1.1 × 10S1U1P1-10S1S1P0 = x2(1 - 2x)
where x is the molar solubility of BaSO4.
Solving for x gives:
x = 1.4 × 10S1U1P4S1S1P0 M
So the concentration of Ba2+ ions is also 1.4 × 10S1U1P4S1S1P0 M.
Therefore, the mass of BaS1U1P12+S1S1P0 ions present in 1.0 L of solution is:
(1.4 × 10S1U1P4S1S1P0 M)(1.0 L)(2 mol BaS1U1P12+S1S1P0/mol) = 2.8 × 10S1U1P4S1S1P0 g
However, we need to take into account the slight solubility of BaSO4, which means that the actual mass of BaS1U1P12+S1S1P0 ions present will be less than this. The correct answer is therefore A) < 10S1U1P1-7S1S1P0 g, which is the only choice less than 2.8 × 10S1U1P4S1S1P0 g. 
The closest choice to this is C) 0.0014 g, which is the correct answer considering significant figures.

Question 8

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?

Accepted Answer

The given concentration of Ca(NO₃)₂ (0.0010 M) is less than the solubility product constant (K_sp) of calcium hydroxide (6.5 × 10^-6), which indicates that the solution is unsaturated and no precipitate of calcium hydroxide will form. The addition of KOH will increase the concentration of hydroxide ions, but this will not affect the solubility of calcium hydroxide, as it is a sparingly soluble salt. The given information is sufficient to make this prediction without the need for the solubility product constant of calcium nitrate. The option stating that calcium hydroxide will precipitate until the solution is saturated is incorrect, as the solution at the given concentration of calcium nitrate is unsaturated. The option that the concentration of calcium ions will be reduced by the addition of KOH is also incorrect, as both calcium and hydroxide ions will be present in the solution.

Question 9

The lab technician Anna Lytic adds 2.20 mol KOH to 1.00 L of 0.5 M Al(NO₃)₃. What is the concentration of aluminum ions after the aluminum nitrate has reacted with the potassium hydroxide? K_f = 3.0 × 10³³ for Al(OH)₄^-

Accepted Answer

The reaction between Al(NO3)3 and KOH forms Al(OH)3, which then reacts with excess OH- to form Al(OH)4-. Given the high Kf value, almost all Al3+ will convert to Al(OH)4-, leaving a very low concentration of free Al3+ ions.

Question 10

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

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 11

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 PbCl₂ residue recovered in the beaker amounted to 0.0162 moles.

Accepted Answer

The solubility of PbCl2 is 0.0162 M. The dissolution equation is PbCl2 ⇌ Pb^2+ + 2Cl^-. Therefore, [Pb^2+] = 0.0162 M and [Cl^-] = 2 × 0.0162 M = 0.0324 M. Ksp = [Pb^2+][Cl^-]^2 = (0.0162)(0.0324)^2 = 1.7 × 10^-5.

Question 12

Calculate the solubility of silver phosphate, Ag3PO4, in pure water. K_sp = 2.6 × 10^-18

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Question 13

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

Accepted Answer

Ag₂SO₄ has the highest K_sp value, indicating it has the greatest solubility in water among the given options.

Question 14

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-

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Question 15

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

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Question 16

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

Accepted Answer

NiCO₃ has the highest K_sp value among the given substances, indicating it has the greatest solubility in water.

Question 17

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

Accepted Answer

The solubility of PbI2 in a solution containing a common ion (I-) is calculated using the Ksp expression: Ksp = [Pb2+][I-]^2. In 0.025 M KI, [I-] is 0.025 M. Solving for [Pb2+], we get [Pb2+] = Ksp / [I-]^2 = 7.9 × 10^-9 / (0.025)^2 = 1.3 × 10^-5 M.

Question 18

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-

Accepted Answer

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Question 19

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.

Accepted Answer

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Question 20

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.

Accepted Answer

Cadmium fluoride precipitates until the solution is saturated.

Quiz 19: Ionic Equilibria in Aqueous Systems

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? K_sp = 2.4 × 10^-5 for calcium sulfate.

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.

Consider the dissolution of MnS in water (K_sp = 3.0 × 10^-14) . MnS(s) + H₂O(l) Mn²⁺(aq) + HS^-(aq) + OH^-(aq) How is the solubility of manganese(II) sulfide affected by the addition of aqueous potassium hydroxide to the system?

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

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 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?

The lab technician Anna Lytic adds 2.20 mol KOH to 1.00 L of 0.5 M Al(NO₃) ₃. What is the concentration of aluminum ions after the aluminum nitrate has reacted with the potassium hydroxide? K_f = 3.0 × 10³³ for Al(OH) ₄^-

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

Calculate the solubility of silver phosphate, Ag3PO4, in pure water. K_sp = 2.6 × 10^-18

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

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-

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

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

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

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-

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 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.

Quiz 19: Ionic Equilibria in Aqueous Systems

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.

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.

Consider the dissolution of MnS in water (Ksp = 3.0 × 10-14) . MnS(s) + H2O(l) Mn2+(aq) + HS-(aq) + OH-(aq) How is the solubility of manganese(II) sulfide affected by the addition of aqueous potassium hydroxide to the system?

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

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?

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?

The lab technician Anna Lytic adds 2.20 mol KOH to 1.00 L of 0.5 M Al(NO3) 3. What is the concentration of aluminum ions after the aluminum nitrate has reacted with the potassium hydroxide? Kf = 3.0 × 1033 for Al(OH) 4-

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

Calculate the solubility of silver phosphate, Ag3PO4, in pure water. Ksp = 2.6 × 10-18

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

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-

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

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

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

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-

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.

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.

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? K_sp = 2.4 × 10^-5 for calcium sulfate.

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.

Consider the dissolution of MnS in water (K_sp = 3.0 × 10^-14) . MnS(s) + H₂O(l) Mn²⁺(aq) + HS^-(aq) + OH^-(aq) How is the solubility of manganese(II) sulfide affected by the addition of aqueous potassium hydroxide to the system?

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

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 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?

The lab technician Anna Lytic adds 2.20 mol KOH to 1.00 L of 0.5 M Al(NO₃) ₃. What is the concentration of aluminum ions after the aluminum nitrate has reacted with the potassium hydroxide? K_f = 3.0 × 10³³ for Al(OH) ₄^-

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

Calculate the solubility of silver phosphate, Ag3PO4, in pure water. K_sp = 2.6 × 10^-18

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-

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

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-

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 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.