Question 1

Which of the following compounds is more soluble in acidic solution than in pure neutral water?

Accepted Answer

The answer of Which of the following compounds is more...

Question 2

To 1.00 L of a 0.100 M aqueous solution of benzoic acid (C₆H₅COOH)is added 1.00 mL of 12.0 M HCl.What is the percentage ionization of the benzoic acid in the resulting solution? [K_a(C₆H₅COOH)= 6.5 × 10^-5]

Accepted Answer

The addition of a strong acid like HCl suppresses the ionization of benzoic acid due to the common ion effect. The concentration of H⁺ ions from HCl is much higher than from benzoic acid, reducing its ionization. Calculating the ionization using the given K_a and the new concentration of H⁺ gives approximately 0.52%.

Question 3

Which response has both answers correct? Will a precipitate form when 250 mL of 0.33 M Na₂CrO₄ are added to 250 mL of 0.12 M AgNO₃? [K_sp(Ag₂CrO₄)= 1.1 × 10^-12] What is the concentration of the silver ion remaining in solution?

Accepted Answer

The reaction between Na₂CrO₄ and AgNO₃ is:2Na₂CrO₄ + 2AgNO₃ → Ag₂CrO₄ + 4NaNO₃The K_sp for Ag₂CrO₄ is 1.1 × 10^-12. To determine if a precipitate will form, we can compare the ionic product (IP) of the solution to the K_sp. The IP is calculated by multiplying the concentrations of the ions in solution that are involved in the precipitation reaction. In this case, the ions are Ag⁺ and CrO₄^2-.[Ag⁺][CrO₄^2-] = (0.12 M)(0.33 M) = 3.96 × 10^-2 M²Since the IP (3.96 × 10^-2 M²) is greater than the K_sp (1.1 × 10^-12), a precipitate will form.To calculate the concentration of the silver ion remaining in solution, we can use the following equation:[Ag⁺] = K_sp/[CrO₄^2-][Ag⁺] = 1.1 × 10^-12/0.33 M = 2.9 × 10^-6 MTherefore, the correct answer is A: Yes, [Ag⁺] = 2.9 × 10^-6 M.

Question 4

Which of the following compounds is more soluble in acidic solution than in pure neutral water?

Accepted Answer

The answer of Which of the following compounds is more...

Question 5

A saturated sodium carbonate solution at 100°C contains 45.5 g of dissolved sodium carbonate per 100.mL of solution.The solubility product constant for sodium carbonate at this temperature is

Accepted Answer

The solubility product constant (Ksp) is calculated using the concentration of ions in a saturated solution. Given the high solubility of sodium carbonate at 100°C, the Ksp value is large, making 316 the most reasonable choice.

Question 6

Find the concentration of Pb²⁺ ions in a solution made by adding 5.00 g of lead(II)iodide to 500.mL of 0.150 M KI.[For PbI₂, K_sp = 1.39 × 10^-8.]

Accepted Answer

The answer of Find the concentration of Pb²⁺ ions in...

Question 7

Calculate the molar solubility of AgCl in a 0.15 M solution of NH₃(aq). (K_sp (AgCl)= 1.6 x 10^-10; K_f (Ag(NH₃)₂⁺)= 1.5 x 10⁷)

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Question 8

Calculate the molar solubility of AgBr in a 0.25M solution of NH₃(aq)(K_sp (AgBr)= 7.7 x 10^-13 ; K_f (Ag(NH₃)₂⁺)= 1.5 x 10⁷.

Accepted Answer

First, write out the equilibrium equation for the dissolution of AgBr in water: AgBr (s) ⇌ Ag+ (aq) + Br- (aq) Next, write out the expression for the solubility product constant, Ksp: Ksp = [Ag+][Br-] We can use this expression to write the equilibrium concentrations of Ag+ and Br- in terms of the molar solubility of AgBr, x: Ksp = (x)(x) = x^2 Now let's consider the effect of the 0.25M solution of NH₃ on the equilibrium. Since this is a weak acid (denoted by the "HS" in the formula), it will react with the Ag+ ion in solution to form the complex ion Ag(NH₃)₂⁺: Ag+ (aq) + NH₃ (aq) ⇌ Ag(NH₃)₂⁺ (aq) The formation constant for this reaction is given as Kf. Using the equilibrium constant expression for this reaction, we can write: Kf = [Ag(NH₃)₂⁺]/[Ag+][NH₃] At equilibrium, the product of the equilibrium concentrations of Ag+ and NH₃ must equal the equilibrium concentration of Ag(NH₃)₂⁺. We can express the equilibrium concentration of Ag+ in terms of x and the initial concentration of NH₃: [NH₃]eq = [NH₃]initial - [Ag(NH₃)₂⁺]eq [NH₃]eq = 0.25 - [Ag(NH₃)₂⁺]eq (since [NH₃]initial = 0.25M) Using the Kf expression and the equilibrium concentration of Ag+, we can solve for the equilibrium concentration of Ag(NH₃)₂⁺: Kf = [Ag(NH₃)₂⁺]/(x)(0.25 - [Ag(NH₃)₂⁺]eq) 1.5 x 10⁷ = [Ag(NH₃)₂⁺]/(x)(0.25 - [Ag(NH₃)₂⁺]eq) [Ag(NH₃)₂⁺]eq = (1.5 x 10⁷)(x)(0.25) / [1 + (1.5 x 10⁷)(x)] Substituting this expression back into the expression for [NH₃]eq and then into the Ksp expression, we get: [NH₃]eq = 0.25 - (1.5 x 10⁷)(x)(0.25) / [1 + (1.5 x 10⁷)(x)] Ksp = (x)(x + (1.5 x 10⁷)(0.25)(x)) = xS1S1U2S1U2S1U1S1U1 Solving for x gives a value of approximately 8.4 x 10^-4 M, which is closest to option C.

Question 9

A saturated sodium carbonate solution at 0°C contains 7.1 g of dissolved sodium carbonate per 100.mL of solution.The solubility product constant for sodium carbonate at this temperature is

Accepted Answer

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Question 10

Calculate the molar solubility of BaCO₃ in a 0.10 M solution of Na₂CO₃(aq).(K_sp (BaCO₃)= 8.1 x 10^-9)

Accepted Answer

First, we need to write the balanced equation for the dissolution of BaCO₃ in water: BaCO₃(s) ↔ Ba2+(aq) + CO₃ 2-(aq) Next, we need to write the expression for the solubility product constant (Ksp) for BaCO₃: Ksp = [Ba2+][CO₃ 2-] We are given the value of Ksp, which is 8.1 x 10^-9. Now, we need to consider the effect of the presence of Na₂CO₃ on the solubility of BaCO₃. NaCO₂ will dissociate in water to form Na+ and CO₂ 2-. These ions will react with BaCO₃ according to the following equation: BaCO₃(s) + Na2CO₂(aq) ↔ BaCO₂(aq) + 2Na+(aq) + CO₂ 2-(aq) Notice that the product BaCO₂ is less soluble than BaCO₃, so the reaction will shift to the left, reducing the solubility of BaCO₃. We can use an ICE (Initial-Change-End) table to calculate the molar solubility of BaCO₃ in the presence of Na₂CO₃. Let x denote the molar solubility of BaCO₃ in the presence of Na₂CO₃: | | BaCO₃(s) | Na2CO₂(aq) | Ba(CO₃)(aq) | Na+(aq) | CO₂ 2-(aq) | |----------------|-------------------|----------------------|-----------------------|---------|-----------------------| | Initial [M] | x | 0.10 | 0 | 0 | 0.10 | | Change [M] | -x | -0.20 | +x | +2x | +x | | End [M] | 8.1 x 10^-9 | 0 | x | 2x | 0.10 + x | Using the expression for Ksp, we can write: Ksp = [Ba2+][CO₃ 2-] = (x)(x) = xS1U1P2 Since the value of Ksp is known, we can solve for x: xS1U1P2 = 8.1 x 10^-9 x = sqrt(8.1 x 10^-9) = 9.0 x 10^-5 Therefore, the molar solubility of BaCO₃ in the presence of 0.10 M Na₂CO₃ is 9.0 x 10^-5 M. The correct answer is (C).

Question 11

Will a precipitate of magnesium fluoride form when 300.mL of 1.1 × 10^-3 M MgCl₂ are added to 500.mL of 1.2 × 10^-3 M NaF? [K_sp (MgF₂)= 6.9 × 10^-9]

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Question 12

Calculate the molar solubility of CaF₂in a 0.25 M solution of NaF(aq).(K_sp (CaF₂)= 4.0 x 10^-11)

Accepted Answer

The answer of Calculate the molar solubility of CaF₂in...

Question 13

Find the concentration of calcium ions in a solution made by adding 3.50 g of calcium fluoride to 750.mL of 0.125 M NaF.[For CaF₂, K_sp = 3.95 × 10^-11.]

Accepted Answer

The concentration of calcium ions is determined by the solubility product (Ksp) of CaF2. The common ion effect from NaF reduces the solubility of CaF2. Calculating the solubility of CaF2 in the presence of 0.125 M F- ions gives a calcium ion concentration of approximately 2.53 × 10^-9 M.

Question 14

Calculate the molar solubility of Fe(OH)₂ in a solution with pH = 13.15. (K_sp Fe(OH)₂ = 1.6 x 10^-14)

Accepted Answer

The solubility of Fe(OH)2 in basic solution can be expressed as follows:

Fe(OH)2(s) ⇄ Fe2+(aq) + 2OH-(aq)

The Ksp expression for this reaction is:

Ksp = [Fe2+][OH-]2

Since the pH of the solution is 13.15, we know that [OH-] = 10^-0.85 M.

Substituting into the Ksp expression:

1.6 x 10^-14 = [Fe2+](10^-0.85)^2

Solving for [Fe2+], we get:

[Fe2+] = 8.0 x 10^-13 M

Since Fe(OH)2 dissociates to form two Fe2+ ions, the molar solubility of Fe(OH)2 is:

s = [Fe2+] / 2 = 4.0 x 10^-13 M

Therefore, the best choice is D.

Question 15

Calculate the molar solubility of Mg(OH)₂ in a solution with pH = 12.20. (K_sp (Mg(OH)₂ = 1.2 x 10^-11)

Accepted Answer

The solubility product (Ksp) of Mg(OH)2 is given as 1.2 x 10^-11. The dissolution of Mg(OH)2 in water can be represented as: Mg(OH)2(s) ⇌ Mg2+(aq) + 2OH-(aq). The Ksp expression is Ksp = [Mg2+][OH-]^2. Given the pH is 12.20, we can find the pOH = 14 - 12.20 = 1.80. The concentration of OH- can be calculated using [OH-] = 10^-pOH = 10^-1.80 = 1.58 x 10^-2 M. Substituting this into the Ksp expression and solving for [Mg2+], we get 1.2 x 10^-11 = [Mg2+](1.58 x 10^-2)^2. Solving for [Mg2+] gives [Mg2+] = 4.8 x 10^-8 M, which is the molar solubility of Mg(OH)2 in the solution.

Question 16

Will a precipitate (ppt)form when 20.0 mL of 1.1 × 10^-3 M Ba(NO₃)₂ are added to 80.0 mL of 8.4 × 10^-4 M Na₂CO₃?

Accepted Answer

A precipitate of BaCO₃ will form because the product of the concentrations of Ba²⁺ and CO₃²⁻ ions exceeds the solubility product constant (Ksp) for BaCO₃.

Question 17

Will a precipitate (ppt)form when 300.mL of 5.0 × 10^-5 M AgNO₃ are added to 200.mL of 2.5 × 10^-7 M NaBr? Answer yes or no, and identify the precipitate if there is one.

Accepted Answer

To determine if a precipitate will form, we need to find the solubility product constant (Ksp) of the possible products, AgNO3 and NaBr. The Ksp of AgNO3 is 1.0 x 10^-4 and the Ksp of NaBr is 5.0 x 10^-6.

To find out if a precipitate will form, we need to calculate the ion product (Q) by multiplying the concentrations of the ions that will react with each other.

Q = [Ag+] [Br-]

[Ag+] = 5.0 x 10^-5 M (from 5.0 x 10^-5 M AgNO3) 
[Br-] = 5.0 x 10^-8 M (from 2.5 x 10^-7 M NaBr)

Q = (5.0 x 10^-5)(5.0 x 10^-8) = 2.5 x 10^-12

Since Q is less than Ksp of AgNO3 but greater than Ksp of NaBr, a precipitate of AgBr will form. Therefore, the answer is B.

Question 18

Will a precipitate form (yes or no)when 50.0 mL of 1.2 × 10^-3 M Pb(NO₃)₂ are added to 50.0 mL of 2.0 × 10^-4 M Na₂S? If so, identify the precipitate.

Accepted Answer

This is a double displacement reaction between lead nitrate and sodium sulfide, which will form lead sulfide and sodium nitrate. Pb(NO3)2 + Na2S → PbS + 2NaNO3. The solubility product constant for lead sulfide is much smaller than that of sodium nitrate or lead nitrate, which means that lead sulfide is insoluble and will precipitate out of solution. Therefore, the answer is A - yes, the precipitate is PbS.

Question 19

Will a precipitate (ppt)form when 300.mL of 2.0 × 10^-5 M AgNO₃ are added to 200.mL of 2.5 × 10^-9 M NaI? Answer yes or no, and identify the precipitate if there is one.

Accepted Answer

When AgNO3 is added to NaI, a precipitation reaction occurs because AgI is insoluble. The balanced chemical equation for the reaction is AgNO3 + NaI → AgI + NaNO3. Therefore, the answer is D, as the precipitate formed is AgI.

Question 20

Will a precipitate of magnesium fluoride form when 200.mL of 1.9 × 10^-3 M MgCl₂ are added to 300.mL of 1.4 × 10^-2 M NaF? [K_sp (MgF₂)= 6.9 × 10^-9]

Accepted Answer

The first step is to write the chemical equation and the expression for the reaction quotient:

MgCl2(aq) + 2NaF(aq) → MgF2(s) + 2NaCl(aq)

Q = [Mg2+][F^-]^2

To calculate Q, we need to find the concentrations of the ions in the solution after mixing:

[Mg2+] = 1.9 × 10^-3 M × (200 mL)/500 mL = 0.76 × 10^-3 M
[F^-] = 1.4 × 10^-2 M × (300 mL)/500 mL = 0.84 × 10^-2 M

Substituting these values into the expression for Q:

Q = (0.76 × 10^-3 M)(0.84 × 10^-2 M)^2 = 5.10 × 10^-11

Since Q > Ksp for MgF2, a precipitate will form. Therefore, the correct answer is A: Yes, Q > Ksp.

Quiz 16: Acid-Base Equilibria and Solubility Equilibria

Which of the following compounds is more soluble in acidic solution than in pure neutral water?

To 1.00 L of a 0.100 M aqueous solution of benzoic acid (C₆H₅COOH) is added 1.00 mL of 12.0 M HCl.What is the percentage ionization of the benzoic acid in the resulting solution? [K_a(C₆H₅COOH) = 6.5 × 10^-5]

Which response has both answers correct? Will a precipitate form when 250 mL of 0.33 M Na₂CrO₄ are added to 250 mL of 0.12 M AgNO₃? [K_sp(Ag₂CrO₄) = 1.1 × 10^-12] What is the concentration of the silver ion remaining in solution?

Which of the following compounds is more soluble in acidic solution than in pure neutral water?

A saturated sodium carbonate solution at 100°C contains 45.5 g of dissolved sodium carbonate per 100.mL of solution.The solubility product constant for sodium carbonate at this temperature is

Find the concentration of Pb²⁺ ions in a solution made by adding 5.00 g of lead(II) iodide to 500.mL of 0.150 M KI.[For PbI₂, K_sp = 1.39 × 10^-8.]

Calculate the molar solubility of AgCl in a 0.15 M solution of NH₃(aq) . (K_sp (AgCl) = 1.6 x 10^-10; K_f (Ag(NH₃) ₂⁺) = 1.5 x 10⁷)

Calculate the molar solubility of AgBr in a 0.25M solution of NH₃(aq) (K_sp (AgBr) = 7.7 x 10^-13 ; K_f (Ag(NH₃) ₂⁺) = 1.5 x 10⁷.

A saturated sodium carbonate solution at 0°C contains 7.1 g of dissolved sodium carbonate per 100.mL of solution.The solubility product constant for sodium carbonate at this temperature is

Calculate the molar solubility of BaCO₃ in a 0.10 M solution of Na₂CO₃(aq) .(K_sp (BaCO₃) = 8.1 x 10^-9)

Will a precipitate of magnesium fluoride form when 300.mL of 1.1 × 10^-3 M MgCl₂ are added to 500.mL of 1.2 × 10^-3 M NaF? [K_sp (MgF₂) = 6.9 × 10^-9]

Calculate the molar solubility of CaF₂in a 0.25 M solution of NaF(aq) .(K_sp (CaF₂) = 4.0 x 10^-11)

Find the concentration of calcium ions in a solution made by adding 3.50 g of calcium fluoride to 750.mL of 0.125 M NaF.[For CaF₂, K_sp = 3.95 × 10^-11.]

Calculate the molar solubility of Fe(OH) ₂ in a solution with pH = 13.15. (K_sp Fe(OH) ₂ = 1.6 x 10^-14)

Calculate the molar solubility of Mg(OH) ₂ in a solution with pH = 12.20. (K_sp (Mg(OH) ₂ = 1.2 x 10^-11)

Will a precipitate (ppt) form when 20.0 mL of 1.1 × 10^-3 M Ba(NO₃) ₂ are added to 80.0 mL of 8.4 × 10^-4 M Na₂CO₃?

Will a precipitate (ppt) form when 300.mL of 5.0 × 10^-5 M AgNO₃ are added to 200.mL of 2.5 × 10^-7 M NaBr? Answer yes or no, and identify the precipitate if there is one.

Will a precipitate form (yes or no) when 50.0 mL of 1.2 × 10^-3 M Pb(NO₃) ₂ are added to 50.0 mL of 2.0 × 10^-4 M Na₂S? If so, identify the precipitate.

Will a precipitate (ppt) form when 300.mL of 2.0 × 10^-5 M AgNO₃ are added to 200.mL of 2.5 × 10^-9 M NaI? Answer yes or no, and identify the precipitate if there is one.

Will a precipitate of magnesium fluoride form when 200.mL of 1.9 × 10^-3 M MgCl₂ are added to 300.mL of 1.4 × 10^-2 M NaF? [K_sp (MgF₂) = 6.9 × 10^-9]

Quiz 16: Acid-Base Equilibria and Solubility Equilibria

Which of the following compounds is more soluble in acidic solution than in pure neutral water?

To 1.00 L of a 0.100 M aqueous solution of benzoic acid (C6H5COOH) is added 1.00 mL of 12.0 M HCl.What is the percentage ionization of the benzoic acid in the resulting solution? [Ka(C6H5COOH) = 6.5 × 10-5]

Which response has both answers correct? Will a precipitate form when 250 mL of 0.33 M Na2CrO4 are added to 250 mL of 0.12 M AgNO3? [Ksp(Ag2CrO4) = 1.1 × 10-12] What is the concentration of the silver ion remaining in solution?

Which of the following compounds is more soluble in acidic solution than in pure neutral water?

A saturated sodium carbonate solution at 100°C contains 45.5 g of dissolved sodium carbonate per 100.mL of solution.The solubility product constant for sodium carbonate at this temperature is

Find the concentration of Pb2+ ions in a solution made by adding 5.00 g of lead(II) iodide to 500.mL of 0.150 M KI.[For PbI2, Ksp = 1.39 × 10-8.]

Calculate the molar solubility of AgCl in a 0.15 M solution of NH3(aq) . (Ksp (AgCl) = 1.6 x 10-10; Kf (Ag(NH3) 2+) = 1.5 x 107)

Calculate the molar solubility of AgBr in a 0.25M solution of NH3(aq) (Ksp (AgBr) = 7.7 x 10-13 ; Kf (Ag(NH3) 2+) = 1.5 x 107.

A saturated sodium carbonate solution at 0°C contains 7.1 g of dissolved sodium carbonate per 100.mL of solution.The solubility product constant for sodium carbonate at this temperature is

Calculate the molar solubility of BaCO3 in a 0.10 M solution of Na2CO3(aq) .(Ksp (BaCO3) = 8.1 x 10-9)

Will a precipitate of magnesium fluoride form when 300.mL of 1.1 × 10-3 M MgCl2 are added to 500.mL of 1.2 × 10-3 M NaF? [Ksp (MgF2) = 6.9 × 10-9]

Calculate the molar solubility of CaF2 in a 0.25 M solution of NaF(aq) .(Ksp (CaF2) = 4.0 x 10-11)

Find the concentration of calcium ions in a solution made by adding 3.50 g of calcium fluoride to 750.mL of 0.125 M NaF.[For CaF2, Ksp = 3.95 × 10-11.]

Calculate the molar solubility of Fe(OH) 2 in a solution with pH = 13.15. (Ksp Fe(OH) 2 = 1.6 x 10-14)

Calculate the molar solubility of Mg(OH) 2 in a solution with pH = 12.20. (Ksp (Mg(OH) 2 = 1.2 x 10-11)

Will a precipitate (ppt) form when 20.0 mL of 1.1 × 10-3 M Ba(NO3) 2 are added to 80.0 mL of 8.4 × 10-4 M Na2CO3?

Will a precipitate (ppt) form when 300.mL of 5.0 × 10-5 M AgNO3 are added to 200.mL of 2.5 × 10-7 M NaBr? Answer yes or no, and identify the precipitate if there is one.

Will a precipitate form (yes or no) when 50.0 mL of 1.2 × 10-3 M Pb(NO3) 2 are added to 50.0 mL of 2.0 × 10-4 M Na2S? If so, identify the precipitate.

Will a precipitate (ppt) form when 300.mL of 2.0 × 10-5 M AgNO3 are added to 200.mL of 2.5 × 10-9 M NaI? Answer yes or no, and identify the precipitate if there is one.

Will a precipitate of magnesium fluoride form when 200.mL of 1.9 × 10-3 M MgCl2 are added to 300.mL of 1.4 × 10-2 M NaF? [Ksp (MgF2) = 6.9 × 10-9]

To 1.00 L of a 0.100 M aqueous solution of benzoic acid (C₆H₅COOH) is added 1.00 mL of 12.0 M HCl.What is the percentage ionization of the benzoic acid in the resulting solution? [K_a(C₆H₅COOH) = 6.5 × 10^-5]

Which response has both answers correct? Will a precipitate form when 250 mL of 0.33 M Na₂CrO₄ are added to 250 mL of 0.12 M AgNO₃? [K_sp(Ag₂CrO₄) = 1.1 × 10^-12] What is the concentration of the silver ion remaining in solution?

Find the concentration of Pb²⁺ ions in a solution made by adding 5.00 g of lead(II) iodide to 500.mL of 0.150 M KI.[For PbI₂, K_sp = 1.39 × 10^-8.]

Calculate the molar solubility of AgCl in a 0.15 M solution of NH₃(aq) . (K_sp (AgCl) = 1.6 x 10^-10; K_f (Ag(NH₃) ₂⁺) = 1.5 x 10⁷)

Calculate the molar solubility of AgBr in a 0.25M solution of NH₃(aq) (K_sp (AgBr) = 7.7 x 10^-13 ; K_f (Ag(NH₃) ₂⁺) = 1.5 x 10⁷.

Calculate the molar solubility of BaCO₃ in a 0.10 M solution of Na₂CO₃(aq) .(K_sp (BaCO₃) = 8.1 x 10^-9)

Will a precipitate of magnesium fluoride form when 300.mL of 1.1 × 10^-3 M MgCl₂ are added to 500.mL of 1.2 × 10^-3 M NaF? [K_sp (MgF₂) = 6.9 × 10^-9]

Calculate the molar solubility of CaF₂in a 0.25 M solution of NaF(aq) .(K_sp (CaF₂) = 4.0 x 10^-11)

Find the concentration of calcium ions in a solution made by adding 3.50 g of calcium fluoride to 750.mL of 0.125 M NaF.[For CaF₂, K_sp = 3.95 × 10^-11.]

Calculate the molar solubility of Fe(OH) ₂ in a solution with pH = 13.15. (K_sp Fe(OH) ₂ = 1.6 x 10^-14)

Calculate the molar solubility of Mg(OH) ₂ in a solution with pH = 12.20. (K_sp (Mg(OH) ₂ = 1.2 x 10^-11)

Will a precipitate (ppt) form when 20.0 mL of 1.1 × 10^-3 M Ba(NO₃) ₂ are added to 80.0 mL of 8.4 × 10^-4 M Na₂CO₃?

Will a precipitate (ppt) form when 300.mL of 5.0 × 10^-5 M AgNO₃ are added to 200.mL of 2.5 × 10^-7 M NaBr? Answer yes or no, and identify the precipitate if there is one.

Will a precipitate form (yes or no) when 50.0 mL of 1.2 × 10^-3 M Pb(NO₃) ₂ are added to 50.0 mL of 2.0 × 10^-4 M Na₂S? If so, identify the precipitate.

Will a precipitate (ppt) form when 300.mL of 2.0 × 10^-5 M AgNO₃ are added to 200.mL of 2.5 × 10^-9 M NaI? Answer yes or no, and identify the precipitate if there is one.

Will a precipitate of magnesium fluoride form when 200.mL of 1.9 × 10^-3 M MgCl₂ are added to 300.mL of 1.4 × 10^-2 M NaF? [K_sp (MgF₂) = 6.9 × 10^-9]