Question 1

Consider mixing 5.1×10^{- 6} moles of Na₂SO₄ and 1.2×10^{- 5} moles of BaCl₂ in enough water to form 1.0 L of an aqueous solution. BaSO₄ has limited solubility with a K _sp= 1.1×10^{- 10}. Na₂SO₄ (aq) + BaCl₂ (aq)→BaSO₄ (s) + 2 NaCl (s) Which statement below is true regarding the solubility reaction quotient, Q _sp, and whether or not a precipitate of BaSO₄ will form?

Accepted Answer

The reaction quotient, Q _sp, can be calculated using the concentrations of the products and reactants at any given moment during the mixing process. However, the solubility product constant, K _sp, is a constant value that represents the maximum value of Q _sp at which the solution is still in equilibrium and no precipitate of BaSO₄ will form. If Q _sp is less than K _sp, then the solution is not saturated with BaSO₄ and no precipitate will form. Therefore, the correct choice is D, Q _sp < K _sp and a precipitate of BaSO₄ will not form.

Question 2

A solution containing AgNO₃ is prepared. Which of the following solutions would add common ions if added to this solution?

Accepted Answer

All options introduce either Ag+ or NO3- ions, which are common ions to the original solution AgNO3, thus affecting its equilibrium.

Question 3

What is the solubility of PbI₂ ( K _sp = 7.1×10^{- 9}) in a 0.1 M solution of NaI?

Accepted Answer

The answer of What is the solubility of PbI₂ (...

Question 4

What is the solubility of AgCl (s) ( K _sp = 1.7×10^{- 10}) in a 0.30 M NaCl solution?

Accepted Answer

The solubility of AgCl in pure water is given by the Ksp value: 
Ksp = [Ag+][Cl-] = 1.7×10^-10
In a solution containing NaCl, the Na+ and Cl- ions will both be present and can affect the solubility of AgCl. To calculate the new solubility, we need to consider the common ion effect. The addition of NaCl will increase the concentration of Cl- ions in the solution, which will shift the equilibrium towards the solid AgCl. 
Using the common ion effect equation, we can find the new equilibrium concentration of Cl- ions:
[Cl-] = [NaCl] = 0.30 M
Now we can calculate the new solubility of AgCl:
Ksp = [Ag+][Cl-] = (x)(0.30)
1.7×10^-10 = x(0.30)
x = 5.7×10^-11 M
Therefore, the solubility of AgCl in a 0.30 M NaCl solution is 5.7×10^-11 M, which corresponds to answer choice D.

Question 5

What is the molar solubility of Ag₂SO₄ in the presence of 0.22 M aqueous solution of AgNO₃? K _sp(Ag₂SO₄) = 1.5×10^{- 5}

Accepted Answer

Using the solubility product constant (Ksp) equation and the ion coefficients for Ag₂SO₄, we can set up the equation: Ksp = [Ag+]^2 [₂SO₄^-]^3 1.5×10^{- 5} = (2x)^2 (x)^3 where x is the molar solubility of Ag₂SO₄. Solving for x, we get: x = 3.1×10^{- 4} M Therefore, the correct answer is option C.

Question 6

What is the concentration of F^- ions in a solution prepared by saturating a 0.0045 M Ba(NO₃)₂ solution with BaF₂? ( K _sp = 1.0×10^{- 6})

Accepted Answer

The solubility product expression for BaF2 is Ksp = [Ba2+][F-]^2. In a saturated solution, [Ba2+] = 0.0045 M due to the common ion effect from Ba(NO3)2. Solving for [F-] using Ksp = 1.0×10^-6, we find [F-] = 1.5×10^-2 M.

Question 7

The pH of a 0.050 M solution of CaCl₂ is adjusted such that it maintains a value of 8.0. Which statement below is true regarding whether a precipitate of Ca(OH)₂ will form? K _sp[Ca(OH)₂] = 6.5×10^{- 6}

Accepted Answer

Since the pH of the solution is adjusted to 8.0, there will be OH- ions present in the solution. The reaction that occurs is:
Ca2+ + 2OH- → Ca(OH)2
The Ksp expression for Ca(OH)2 is:
Ksp[Ca(OH)2] = [Ca2+][OH-]2 = 6.5×10-6
Since the pH is adjusted to 8.0, the concentration of OH- can be calculated using the Kw expression:
Kw = [H+][OH-] = 1.0×10-14
[OH-] = 1.0×10-14/[H+] = 1.0×10-14/10-8 = 1.0×10-6 M
Substituting this value in the Ksp expression gives:
Ksp[Ca(OH)2] = [Ca2+][OH-]2 = (1.0×10-6)2[Ca2+] = 1.0×10-12[Ca2+]
Since the concentration of Ca2+ is 0.050 M, the Qsp can be calculated as follows:
Qsp = [Ca2+][OH-]2 = (0.050)(1.0×10-6)2 = 5.0×10-14
Comparing Qsp and Ksp, we can see that Qsp < Ksp. This means that the solution is not saturated and a precipitate of Ca(OH)2 will not form. Therefore, the correct answer is B.

Question 8

Calculate the solubility of AgI in 0.10 M NaI solution at 25 ° C. K _sp for AgI is 8.5×10^{- 17} at 25 ° C.

Accepted Answer

First, we need to write out the equilibrium expression for the dissolution of AgI in water:

AgI(s) ⇌ Ag⁺(aq) + I⁻(aq)

Next, we need to add in the effect of the NaI solution:

AgI(s) ⇌ Ag⁺(aq) + I⁻(aq)

I⁻(aq) + Na⁺(aq) ⇌ NaI(aq)

The Ksp expression now becomes:

Ksp = [Ag⁺][I⁻]

However, we also need to consider the effect of the presence of Na⁺ ions on the solubility of AgI. This is done using the common ion effect, where the solubility of a salt decreases when a common ion is added to the solution.

Using an ICE table, we can calculate the concentration of I⁻ ions:

I⁻(aq) + Na⁺(aq) ⇌ NaI(aq)

I⁻(aq)      0.10  + x          x

Na⁺(aq)     0.10           x

The equilibrium concentration of I⁻ ions is equal to the initial concentration plus the concentration added from the NaI solution (x). The equilibrium concentration of Na⁺ ions is also equal to x.

Substituting these values into the Ksp expression:

Ksp = [Ag⁺][I⁻]

Ksp = (x)(8.5×10⁻¹⁷)

Substituting x = [I⁻] into the above expression, we get:

Ksp = (0.10 + [I⁻])(8.5×10⁻¹⁷)

Since the solubility of AgI is equal to the concentration of I⁻ ions at equilibrium, we can rearrange the equation to solve for the solubility:

Solubility = [I⁻] = (Ksp / (0.10 + [I⁻]))

There is no simple algebraic way to solve for the concentration of I⁻ ions, so we must use an iterative method or a calculator. Using an iterative method, we can make an initial guess that [I⁻] is equal to the initial concentration added by the NaI solution, which is 0.10 M. This gives:

Solubility = [I⁻] = (8.5×10⁻¹⁷ / (0.10 + 0.10)) = 4.25×10⁻⁹ M

This value is not exact, but it is close enough to the actual value that we can assume it is correct to the appropriate number of significant figures. The correct answer is option D: 8.5×10⁻¹⁶ M.

Question 9

What is the minimum concentration of Br^- ions necessary to precipitate PbBr₂ from a 0.080 M Pb(NO₃)₂ solution, if K _sp = 4.0×10^{- 5} for PbBr₂?

Accepted Answer

The answer of What is the minimum concentration of Br<sup>...

Question 10

Mg(OH)₂ ( K _sp = 8.9×10^{- 12}) is precipitated by the addition of strong base to a 0.10 M solution of MgCl₂. At what [OH^-] will Mg(OH)₂ begin to precipitate?

Accepted Answer

The solubility product constant expression for Mg(OH)2 is:

Ksp = [Mg2+][OH-]^2

Adding a strong base (in this case, OH-) to a solution containing Mg2+ ions will increase the concentration of OH- ions until they reach a concentration that is equal to or greater than the solubility product constant. At this point, Mg(OH)2 will begin to precipitate.

Starting with a 0.10 M solution of MgCl2, the initial concentration of Mg2+ ions will also be 0.10 M. We can use this value to calculate the minimum concentration of OH- ions needed to reach the solubility product constant and initiate precipitation:

Ksp = [Mg2+][OH-]^2
8.9×10^-12 = (0.10 M)([OH-]^2)
[OH-]^2 = (8.9×10^-12) / (0.10 M)
[OH-] = √(8.9×10^-12 / 0.10) = 9.4×10^-7 M

Therefore, the minimum concentration of OH- ions needed to initiate precipitation is 9.4×10^-7 M, which corresponds to option C.

Question 11

What is the molar solubility of PbBr₂ ( K _sp = 4.6×10^{- 6}) in 0.200 M NaBr?

Accepted Answer

The solubility product (Ksp) for PbBr2 is given as 4.6×10^-6. When PbBr2 dissolves, it dissociates into Pb^2+ and 2Br^-. In a solution that already contains 0.200 M NaBr, the concentration of Br^- ions is significantly increased due to the common ion effect, which suppresses the dissociation of PbBr2. The equilibrium expression for the dissolution of PbBr2 is Ksp = [Pb^2+][Br^-]^2. Since [Br^-] is large due to the presence of NaBr, the solubility of PbBr2 (which is the concentration of Pb^2+ at equilibrium) will be much lower than it would be in pure water. To find the molar solubility of PbBr2 in the presence of 0.200 M NaBr, we set up the equation 4.6×10^-6 = [Pb^2+](0.200)^2, solving for [Pb^2+] gives a value close to 1.2×10^-4 M, which corresponds to choice B.

Question 12

What is the molar solubility of Ag₂SO₄ in 0.11 M Na₂SO₄? K _sp(Ag₂SO₄) = 1.5×10^{- 5}

Accepted Answer

The answer of What is the molar solubility of Ag₂SO₄...

Question 13

Exhibit 14-7 Consider dissolving LaF 3 in water as shown below to answer the following question(s): LaFS1U1B13S1U1B0 (s) La S1U1P13+S1S1P0 (aq) + 3 F - (aq) K sp = 2.0*10S1U1P1- 19 S1S1P0 -Refer to Exhibit 14-7. The solubility product constant for LaF₃ is 2.0*10^{- 19} . Which expression listed below would be used to solve for the molar solubility of LaF 3 in water?

Accepted Answer

The answer of Exhibit 14-7 Consider dissolving LaF 3 in...

Question 14

Calculate the ion product value, Q _sp, for MgCO₃ when a 7.0×10^{- 5}M aqueous solution of Mg(NO₃)₂ is mixed with a 5.0×10^{- 4} M aqueous solution of Na₂CO₃. If the solubility product constant, K _sp, for MgCO₃ is 3.5×10^{- 8}, which statement below is true?

Accepted Answer

The answer of Calculate the ion product value, Q _sp,...

Question 15

What is the solubility of PbSO₄ ( K _sp = 2.0×10^{- 8}) in a solution that contains 100.0 g of Na₂SO₄ in 1.00 L?

Accepted Answer

The answer of What is the solubility of PbSO₄ (...

Question 16

What is the molar solubility of PbI₂ in 0.25 M NaI? K _sp(PbI₂) = 7.9×10^{- 9} (Hint: Assume "x" is small to make math easier.)

Accepted Answer

- Write the equilibrium equation and expression: 
PbIS1U1B12S1U1B0 (s) ⇌ PbS1U1B2S1U1B0 2+ (aq) + 3 I- (aq) 
K S1U1B1spS1U1B0(PbIS1U1B12S1U1B0) = [PbS1U1B2S1U1B0 2+ ][I - ]3 
- Substitute given values: 
K S1U1B1spS1U1B0(PbIS1U1B12S1U1B0) = 7.9×10S1U1P1 - 9S1S1P0 
[I-]=0.25 M 
- Assume "x" is small (compared to 0.25 M) and ignore its contribution to the initial [I-], and assume that [PbIS1U1B12S1U1B0] dissociates by "x". Then, at equilibrium: 
[PbIS1U1B12S1U1B0]= x M 
[PbS1U1B2S1U1B0 2+ ]= x M (Initially 0) 
[I-]= 0.25 M – 3x (since each PbIS1U1B12S1U1B0 molecule dissociates into 3 I-) 
K S1U1B1spS1U1B0(PbIS1U1B12S1U1B0) = [PbS1U1B2S1U1B0 2+ ][I - ]3 
7.9×10S1U1P1 - 9S1S1P0 = x * x * (0.25 M – 3x)3 
- Solve for "x" using approximation (or numerical methods): 
x ≈ 1.3×10S1U1P1 - 7S1S1P0 M 
- Choose the answer choice closest to the calculated value: 
D) [PbIS1U1B12S1U1B0] = 1.3×10S1U1P1 - 7S1S1P0 M (closest to calculated value)

Question 17

A mixture of NaCl, NaF, KOH, and KF is prepared. The common ion(s) is(are):

Accepted Answer

The common ions are those that are present in more than one of the compounds in the mixture. In this case, F- is present in both NaF and KF, and Na+ is present in both NaCl and NaF.

Question 18

What is the concentration of Pb 2+ ions in a solution resulting from the addition of 1.0 10 - 3 moles of Na₂ CrO₄ to 1.0 liter of solution containing solid PbCrO₄ ( K sp = 1.8*10^{- 14} ) in equilibrium with its ions? (Assume no volume change on addition of Na₂ CrO₄ .) PbCrO₄ (s) Pb 2+ (aq) + CrO₂ - (aq)

Accepted Answer

The answer of What is the concentration of Pb 2+...

Question 19

Exhibit 14-7 Consider dissolving LaF 3 in water as shown below to answer the following question(s):
LaFS1U1B13S1U1B0 (s)  LaS1U1P13+S1S1P0 (aq) + 3 F - (aq) K sp = 2.0*10S1U1P1- 19 S1S1P0
-Refer to Exhibit 14-7. What is the molar solubility of LaF 3 in a solution that contains 0.035 M KF?

Accepted Answer

The molar solubility of LaF₃ is affected by the common ion effect due to the presence of F⁻ ions from KF. The solubility product (Ksp) of LaF₃ is used to calculate the solubility in the presence of 0.035 M F⁻ ions.

Question 20

A mixture of Na₂CO₃, NaCl, and NiCO₃ is prepared. The common ion(s) is(are):

Accepted Answer

The common ion in the mixture is $CO_3^{2-}$ because it is present in both $Na_2CO_3$ and $NiCO_3$.

Quiz 14: Chemical Equilibrium

A solution containing AgNO₃ is prepared. Which of the following solutions would add common ions if added to this solution?

What is the solubility of PbI₂ ( K _sp = 7.1×10^{- 9}) in a 0.1 M solution of NaI?

What is the solubility of AgCl (s) ( K _sp = 1.7×10^{- 10}) in a 0.30 M NaCl solution?

What is the molar solubility of Ag₂SO₄ in the presence of 0.22 M aqueous solution of AgNO₃? K _sp(Ag₂SO₄) = 1.5×10^{- 5}

What is the concentration of F^- ions in a solution prepared by saturating a 0.0045 M Ba(NO₃) ₂ solution with BaF₂? ( K _sp = 1.0×10^{- 6})

The pH of a 0.050 M solution of CaCl₂ is adjusted such that it maintains a value of 8.0. Which statement below is true regarding whether a precipitate of Ca(OH) ₂ will form? K _sp[Ca(OH) ₂] = 6.5×10^{- 6}

Calculate the solubility of AgI in 0.10 M NaI solution at 25 ° C. K _sp for AgI is 8.5×10^{- 17} at 25 ° C.

What is the minimum concentration of Br^- ions necessary to precipitate PbBr₂ from a 0.080 M Pb(NO₃) ₂ solution, if K _sp = 4.0×10^{- 5} for PbBr₂?

Mg(OH) ₂ ( K _sp = 8.9×10^{- 12}) is precipitated by the addition of strong base to a 0.10 M solution of MgCl₂. At what [OH^-] will Mg(OH) ₂ begin to precipitate?

What is the molar solubility of PbBr₂ ( K _sp = 4.6×10^{- 6}) in 0.200 M NaBr?

What is the molar solubility of Ag₂SO₄ in 0.11 M Na₂SO₄? K _sp(Ag₂SO₄) = 1.5×10^{- 5}

Calculate the ion product value, Q _sp, for MgCO₃ when a 7.0×10^{- 5}M aqueous solution of Mg(NO₃) ₂ is mixed with a 5.0×10^{- 4} M aqueous solution of Na₂CO₃. If the solubility product constant, K _sp, for MgCO₃ is 3.5×10^{- 8}, which statement below is true?

What is the solubility of PbSO₄ ( K _sp = 2.0×10^{- 8}) in a solution that contains 100.0 g of Na₂SO₄ in 1.00 L?

What is the molar solubility of PbI₂ in 0.25 M NaI? K _sp(PbI₂) = 7.9×10^{- 9} (Hint: Assume "x" is small to make math easier.)

A mixture of NaCl, NaF, KOH, and KF is prepared. The common ion(s) is(are) :

Exhibit 14-7 Consider dissolving LaF 3 in water as shown below to answer the following question(s) : LaF₃ (s) La³⁺ (aq) + 3 F - (aq) K sp = 2.0*10^{- 19} -Refer to Exhibit 14-7. What is the molar solubility of LaF 3 in a solution that contains 0.035 M KF?

A mixture of Na₂CO₃, NaCl, and NiCO₃ is prepared. The common ion(s) is(are) :

Quiz 14: Chemical Equilibrium

A solution containing AgNO3 is prepared. Which of the following solutions would add common ions if added to this solution?

What is the solubility of PbI2 ( K sp = 7.1×10 - 9) in a 0.1 M solution of NaI?

What is the solubility of AgCl (s) ( K sp = 1.7×10 - 10) in a 0.30 M NaCl solution?

What is the molar solubility of Ag2SO4 in the presence of 0.22 M aqueous solution of AgNO3? K sp(Ag2SO4) = 1.5×10 - 5

What is the concentration of F - ions in a solution prepared by saturating a 0.0045 M Ba(NO3) 2 solution with BaF2? ( K sp = 1.0×10 - 6)

The pH of a 0.050 M solution of CaCl2 is adjusted such that it maintains a value of 8.0. Which statement below is true regarding whether a precipitate of Ca(OH) 2 will form? K sp[Ca(OH) 2] = 6.5×10 - 6

Calculate the solubility of AgI in 0.10 M NaI solution at 25 ° C. K sp for AgI is 8.5×10 - 17 at 25 ° C.

What is the minimum concentration of Br - ions necessary to precipitate PbBr2 from a 0.080 M Pb(NO3) 2 solution, if K sp = 4.0×10 - 5 for PbBr2?

Mg(OH) 2 ( K sp = 8.9×10 - 12) is precipitated by the addition of strong base to a 0.10 M solution of MgCl2. At what [OH - ] will Mg(OH) 2 begin to precipitate?

What is the molar solubility of PbBr2 ( K sp = 4.6×10 - 6) in 0.200 M NaBr?

What is the molar solubility of Ag2SO4 in 0.11 M Na2SO4? K sp(Ag2SO4) = 1.5×10 - 5

Calculate the ion product value, Q sp, for MgCO3 when a 7.0×10 - 5 M aqueous solution of Mg(NO3) 2 is mixed with a 5.0×10 - 4 M aqueous solution of Na2CO3. If the solubility product constant, K sp, for MgCO3 is 3.5×10 - 8, which statement below is true?

What is the solubility of PbSO4 ( K sp = 2.0×10 - 8) in a solution that contains 100.0 g of Na2SO4 in 1.00 L?

What is the molar solubility of PbI2 in 0.25 M NaI? K sp(PbI2) = 7.9×10 - 9 (Hint: Assume "x" is small to make math easier.)

A mixture of NaCl, NaF, KOH, and KF is prepared. The common ion(s) is(are) :

Exhibit 14-7 Consider dissolving LaF 3 in water as shown below to answer the following question(s) : LaF3 (s) La3+ (aq) + 3 F - (aq) K sp = 2.0*10- 19 -Refer to Exhibit 14-7. What is the molar solubility of LaF 3 in a solution that contains 0.035 M KF?

A mixture of Na2CO3, NaCl, and NiCO3 is prepared. The common ion(s) is(are) :

A solution containing AgNO₃ is prepared. Which of the following solutions would add common ions if added to this solution?

What is the solubility of PbI₂ ( K _sp = 7.1×10^{- 9}) in a 0.1 M solution of NaI?

What is the solubility of AgCl (s) ( K _sp = 1.7×10^{- 10}) in a 0.30 M NaCl solution?

What is the molar solubility of Ag₂SO₄ in the presence of 0.22 M aqueous solution of AgNO₃? K _sp(Ag₂SO₄) = 1.5×10^{- 5}

What is the concentration of F^- ions in a solution prepared by saturating a 0.0045 M Ba(NO₃) ₂ solution with BaF₂? ( K _sp = 1.0×10^{- 6})

The pH of a 0.050 M solution of CaCl₂ is adjusted such that it maintains a value of 8.0. Which statement below is true regarding whether a precipitate of Ca(OH) ₂ will form? K _sp[Ca(OH) ₂] = 6.5×10^{- 6}

Calculate the solubility of AgI in 0.10 M NaI solution at 25 ° C. K _sp for AgI is 8.5×10^{- 17} at 25 ° C.

What is the minimum concentration of Br^- ions necessary to precipitate PbBr₂ from a 0.080 M Pb(NO₃) ₂ solution, if K _sp = 4.0×10^{- 5} for PbBr₂?

Mg(OH) ₂ ( K _sp = 8.9×10^{- 12}) is precipitated by the addition of strong base to a 0.10 M solution of MgCl₂. At what [OH^-] will Mg(OH) ₂ begin to precipitate?

What is the molar solubility of PbBr₂ ( K _sp = 4.6×10^{- 6}) in 0.200 M NaBr?

What is the molar solubility of Ag₂SO₄ in 0.11 M Na₂SO₄? K _sp(Ag₂SO₄) = 1.5×10^{- 5}

Calculate the ion product value, Q _sp, for MgCO₃ when a 7.0×10^{- 5}M aqueous solution of Mg(NO₃) ₂ is mixed with a 5.0×10^{- 4} M aqueous solution of Na₂CO₃. If the solubility product constant, K _sp, for MgCO₃ is 3.5×10^{- 8}, which statement below is true?

What is the solubility of PbSO₄ ( K _sp = 2.0×10^{- 8}) in a solution that contains 100.0 g of Na₂SO₄ in 1.00 L?

What is the molar solubility of PbI₂ in 0.25 M NaI? K _sp(PbI₂) = 7.9×10^{- 9} (Hint: Assume "x" is small to make math easier.)

Exhibit 14-7 Consider dissolving LaF 3 in water as shown below to answer the following question(s) : LaF₃ (s) La³⁺ (aq) + 3 F - (aq) K sp = 2.0*10^{- 19} -Refer to Exhibit 14-7. What is the molar solubility of LaF 3 in a solution that contains 0.035 M KF?

A mixture of Na₂CO₃, NaCl, and NiCO₃ is prepared. The common ion(s) is(are) :