Question 1

The K_sp value for lead(II)chloride is 2.4 × 10^-4.What is the molar solubility of lead(II)chloride?

Accepted Answer

The answer of The K_sp value for lead(II)chloride is 2.4...

Question 2

The solubility of strontium carbonate is 0.0011 g/100 mL at 20ºC.Calculate the K_sp value for this compound.

Accepted Answer

The solubility product constant (Ksp) can be calculated using the solubility of the compound. For SrCO3, the dissolution is SrCO3(s) ⇌ Sr²⁺(aq) + CO₃²⁻(aq). The molar mass of SrCO3 is approximately 147.63 g/mol. The solubility in mol/L is 0.0011 g/100 mL × (1 mol/147.63 g) × (1000 mL/1 L) = 7.45 × 10⁻⁶ mol/L. Ksp = [Sr²⁺][CO₃²⁻] = (7.45 × 10⁻⁶)² = 5.6 × 10⁻⁹.

Question 3

The molar solubility of manganese(II)carbonate is 4.2 × 10^-6 M.What is K_sp for this compound?

Accepted Answer

The solubility product expression for manganese(II) carbonate is:

MnCO3(s) ⇌ Mn2+(aq) + CO32-(aq)

And its solubility product constant is defined as:

Ksp = [Mn2+][CO32-]

Using the given molar solubility value (4.2 × 10^-6 M), we can calculate the equilibrium concentrations of Mn2+ and CO32- :

[Mn2+] = 4.2 × 10^-6 M
[CO32-] = 4.2 × 10^-6 M

Substituting these values into the Ksp expression, we get:

Ksp = (4.2 × 10^-6 M)(4.2 × 10^-6 M) = 1.764 × 10^-11

Rounding to two significant figures, the answer is:

Ksp = 1.8 × 10^-11

Therefore, the correct answer is D.

Question 4

The solubility product for barium sulfate is 1.1 × 10^-10.Calculate the molar solubility of barium sulfate.

Accepted Answer

The solubility product (Ksp) for barium sulfate (BaSO4) is given by Ksp = [Ba²⁺][SO₄²⁻]. Since BaSO4 dissociates into one Ba²⁺ and one SO₄²⁻, let the molar solubility be 's'. Then, Ksp = s². Solving for 's', s = √(1.1 × 10⁻¹⁰) = 1.1 × 10⁻⁵ mol/L.

Question 5

The solubility product for calcium phosphate is K_sp = 1.3 × 10^-26.What is the molar solubility of calcium phosphate?

Accepted Answer

The solubility product expression for calcium phosphate is: K_sp = [Ca2+][PO43-] We need to find the molar solubility of calcium phosphate, which is the concentration of the ions in moles/L when the solid is at equilibrium with its saturated solution. Let's assume that the molar solubility of calcium phosphate is x mol/L. Then, at equilibrium, we have: [Ca2+] = x mol/L [PO43-] = x mol/L Substituting these values into the solubility product expression, we get: K_sp = x * x = x^2 Substituting the value of K_sp, we get: 1.3 × 10^-26 = x^2 Taking the square root of both sides, we get: x = 2.6 × 10^-6 M Therefore, the correct answer is (C).

Question 6

The K_sp for silver(I)phosphate is 1.8 × 10^-18.Calculate the molar solubility of silver(I)phosphate.

Accepted Answer

The dissolution of silver(I) phosphate, Ag₃PO₄, can be represented as: Ag₃PO₄(s) ⇌ 3Ag⁺(aq) + PO₄³⁻(aq). The Ksp expression is: Ksp = [Ag⁺]³[PO₄³⁻]. Let the molar solubility be 's'. Then, [Ag⁺] = 3s and [PO₄³⁻] = s. Substituting these into the Ksp expression gives: Ksp = (3s)³(s) = 27s⁴. Solving 27s⁴ = 1.8 × 10⁻¹⁸ gives s ≈ 1.6 × 10⁻⁵ M.

Question 7

Calculate the silver ion concentration in a saturated solution of silver(I)sulfate (K_sp = 1.4 × 10^-5).

Accepted Answer

The Ksp expression for silver(I)sulfate is:
Ksp = [Ag+][SO42-]
Since the solution is saturated, we can assume that the concentration of Ag+ ions in the solution is equal to the solubility of silver(I)sulfate. Therefore, we can write:
Ksp = [Ag+]2
1.4 × 10-5 = [Ag+]2
[Ag+] = √(1.4 × 10-5) = 3.0 × 10-3 M = 3.0 × 10-2 mM

Note that the answer is given in mM (millimolar) rather than M (molar) as stated in the answer choices.

Question 8

The K_sp for silver(I)phosphate is 1.8 × 10^-18.Determine the silver ion concentration in a saturated solution of silver(I)phosphate.

Accepted Answer

The answer of The K_sp for silver(I)phosphate is 1.8 ×...

Question 9

Calculate the concentration of chloride ions in a saturated lead(II)chloride (K_sp = 2.4 × 10^-4)solution.

Accepted Answer

The solubility product expression for lead(II) chloride is:

PbCl2 ⇋ Pb2+ + 2 Cl-

Ksp = [Pb2+][Cl-]2

At saturation, the concentrations of Pb2+ and Cl- are equal to the solubility product. Therefore:

Ksp = [Pb2+] [Cl-]2

Solving for [Cl-], we have:

[Cl-] = √(Ksp/[Pb2+])

Substituting the given values, we get:

[Cl-] = √(2.4 × 10-4/1) = 7.8 × 10-2 M = 7.8 × 10-2 mol/L

Therefore, the concentration of chloride ions in the saturated solution of lead(II) chloride is 7.8 × 10-2 M or 7.8 × 10-2 mol/L, which is answer choice E.

Question 10

Which of the following would decrease the K_sp for PbI₂?

Accepted Answer

The $K_{sp}$ of a compound, which stands for the solubility product constant, is indeed constant at a given temperature and does not change with changes in the concentration of reactants or products in the solution. It is a characteristic property of the substance. Therefore, none of the actions listed (lowering the pH, adding a solution of $Pb(NO_3)_2$, or adding a solution of KI) would change the $K_{sp}$ of $PbI_2$.

Question 11

For PbCl₂ (K_sp = 2.4 × 10^-4), will a precipitate of PbCl₂ form when 0.10 L of 3.0 × 10^-2 M Pb(NO₃)₂ is added to 400 mL of 9.0 × 10^-2 M NaCl?

Accepted Answer

The answer of For PbCl₂ (K_sp = 2.4 × 10^-4),...

Question 12

Calculate the silver ion concentration in a saturated solution of silver(I)carbonate (K_sp = 8.1 × 10^-12).

Accepted Answer

The dissolution of Ag2CO3 in water is represented by Ag2CO3(s) ⇌ 2Ag⁺(aq) + CO3²⁻(aq). Let the solubility of Ag2CO3 be s. Then, [Ag⁺] = 2s and [CO3²⁻] = s. The Ksp expression is Ksp = [Ag⁺]²[CO3²⁻] = (2s)²(s) = 4s³. Solving 4s³ = 8.1 × 10⁻¹² gives s = 1.25 × 10⁻⁴ M, so [Ag⁺] = 2s = 2.5 × 10⁻⁴ M.

Question 13

The molar solubility of magnesium carbonate is 1.8 × 10^-4 mol/L.What is K_sp for this compound?

Accepted Answer

The balanced chemical equation for the dissolving of magnesium carbonate is:

MgCO3(s) ⇌ Mg2+(aq) + CO32-(aq)

The solubility product constant expression for magnesium carbonate is:

Ksp = [Mg2+][CO32-]

If the molar solubility of MgCO3 is x, then [Mg2+] = x mol/L and [CO32-] = x mol/L. Therefore,

Ksp = x * x = x^2

Given the molar solubility of MgCO3 is 1.8 × 10^-4 mol/L, then

Ksp = (1.8 × 10^-4)^2 = 3.24 × 10^-8

Therefore, the answer is D) 3.2 × 10^-8.

Question 14

The solubility product for chromium(III)fluoride is K_sp = 6.6 × 10^-11.What is the molar solubility of chromium(III)fluoride?

Accepted Answer

The answer of The solubility product for chromium(III)fluoride is K_sp...

Question 15

Calculate the minimum concentration of Mg²⁺ that must be added to 0.10 M NaF in order to initiate a precipitate of magnesium fluoride.(For MgF₂ , K_sp = 6.9 × 10^-9.)

Accepted Answer

The balanced chemical equation for the precipitation reaction between Mg²⁺ and NaF is: Mg²⁺ + 2 NaF → MgF2 + Na2²⁺ The solubility product constant (Ksp) for MgF₂ is given as 6.9 × 10^-9. The equilibrium expression for the dissolution of MgF₂ is: Ksp = [Mg2+] [F-]2 At equilibrium, the concentration of Mg2+ ions and F- ions in the solution will be equal to the solubility of MgF₂: [Mg2+] = [F-] = s Substituting these values in the equation for Ksp, we get: Ksp = s (2s)2 = 4s3 Rearranging the equation, we can calculate the solubility (s) of MgF₂: s3 = Ksp/4 = 6.9 × 10^-9/4 s ≈ 1.09 × 10^-3 M To initiate precipitation of MgF2, we need to increase the concentration of Mg2+ ions and F- ions in the solution to a value greater than their solubility. Let x be the concentration of Mg²⁺ that must be added to the solution. After the reaction, the concentration of Mg2+ ions in the solution will be x M, and the concentration of F- ions will be (2x + 0.10) M (due to the presence of 0.10 M NaF initially). The concentration of Mg2+ ions and F- ions in the solution should be equal to the solubility of MgF2 to initiate precipitation. Therefore, we can write the following equation: s = [Mg2+] = [F-] = 2x + 0.10 Substituting the value of solubility (s), we get: 1.09 × 10^-3 = 2x + 0.10 x ≈ 6.95 × 10^-4 M Therefore, the minimum concentration of Mg²⁺ that must be added to 0.10 M NaF in order to initiate a precipitate of MgF2 is approximately 6.95 × 10^-4 M. This closest to answer E: 6.9 × 10^-7 M.

Question 16

Calculate the minimum concentration of Cr³⁺ that must be added to 0.095 M NaF in order to initiate a precipitate of chromium(III)fluoride.(For CrF₃, K_sp = 6.6 × 10^-11.)

Accepted Answer

To find the minimum concentration of Cr³⁺ needed to start precipitation, use the K_sp expression: K_sp = [Cr³⁺][F^-]³. Given [F^-] = 0.095 M, solve for [Cr³⁺]: [Cr³⁺] = K_sp / [F^-]³ = 6.6 × 10^-11 / (0.095)³ = 7.7 × 10^-8 M.

Question 17

The solubility of lead(II)iodide is 0.064 g/100 mL at 20ºC.What is the solubility product for lead(II)iodide?

Accepted Answer

The solubility product (Ksp) is calculated using the molar solubility. First, convert the solubility from g/100 mL to mol/L, then use the dissociation equation PbI2 ⇌ Pb²⁺ + 2I⁻ to find the concentrations of ions and calculate Ksp = [Pb²⁺][I⁻]².

Question 18

The molar solubility of tin(II)iodide is 1.28 × 10^-2 mol/L.What is K_sp for this compound?

Accepted Answer

The molar solubility of tin(II)iodide is 1.28 × 10^-2 mol/L. So, [Sn²⁺] = [I^-] = 1.28 × 10^-2 mol/L. K_sp = [Sn²⁺][I^-]² = (1.28 × 10^-2)(1.28 × 10^-2)² = 8.4 × 10^-6.

Question 19

Calculate the concentration of fluoride ions in a saturated barium fluoride (K_sp = 1.7 × 10^-6)solution.

Accepted Answer

The dissolution of barium fluoride (BaF2) in water can be represented as: BaF2(s) ⇌ Ba²⁺(aq) + 2F⁻(aq). The Ksp expression is Ksp = [Ba²⁺][F⁻]². Let the solubility of BaF2 be 's'. Then [Ba²⁺] = s and [F⁻] = 2s. Substituting into the Ksp expression: 1.7 × 10⁻⁶ = s(2s)² = 4s³. Solving for 's' gives s = 7.6 × 10⁻³ M, and [F⁻] = 2s = 1.5 × 10⁻² M.

Question 20

The molar solubility of lead(II)iodate in water is 4.0 × 10^-5 mol/L.Calculate K_sp for lead(II)iodate.

Accepted Answer

The solubility product constant, Ksp, expression for lead(II)iodate is:

Pb(IO3)2(s) ⇌ Pb2+(aq) + 2IO3-(aq)

The balanced equation shows that one mole of Pb(IO3)2 dissociates into one mole of lead ions and two moles of iodate ions. Thus, if we let x be the concentration of Pb2+ and 2x be the concentration of IO3-, then:

Ksp = [Pb2+][IO3-]^2

Substituting the molar solubility of Pb(IO3)2 into the above equation, we have:

Ksp = (4.0 × 10^-5)^1(2*4.0 × 10^-5)^2
     = 2.56 × 10^-14
     = 2.6 × 10^-13 (rounded to two significant figures)

Therefore, the correct answer is choice C.

Quiz 16: Acid-Base Equilibria and Solubility Equilibria

The K_sp value for lead(II) chloride is 2.4 × 10^-4.What is the molar solubility of lead(II) chloride?

The solubility of strontium carbonate is 0.0011 g/100 mL at 20ºC.Calculate the K_sp value for this compound.

The molar solubility of manganese(II) carbonate is 4.2 × 10^-6 M.What is K_sp for this compound?

The solubility product for barium sulfate is 1.1 × 10^-10.Calculate the molar solubility of barium sulfate.

The solubility product for calcium phosphate is K_sp = 1.3 × 10^-26.What is the molar solubility of calcium phosphate?

The K_sp for silver(I) phosphate is 1.8 × 10^-18.Calculate the molar solubility of silver(I) phosphate.

Calculate the silver ion concentration in a saturated solution of silver(I) sulfate (K_sp = 1.4 × 10^-5) .

The K_sp for silver(I) phosphate is 1.8 × 10^-18.Determine the silver ion concentration in a saturated solution of silver(I) phosphate.

Calculate the concentration of chloride ions in a saturated lead(II) chloride (K_sp = 2.4 × 10^-4) solution.

Which of the following would decrease the K_sp for PbI₂?

For PbCl₂ (K_sp = 2.4 × 10^-4) , will a precipitate of PbCl₂ form when 0.10 L of 3.0 × 10^-2 M Pb(NO₃) ₂ is added to 400 mL of 9.0 × 10^-2 M NaCl?

Calculate the silver ion concentration in a saturated solution of silver(I) carbonate (K_sp = 8.1 × 10^-12) .

The molar solubility of magnesium carbonate is 1.8 × 10^-4 mol/L.What is K_sp for this compound?

The solubility product for chromium(III) fluoride is K_sp = 6.6 × 10^-11.What is the molar solubility of chromium(III) fluoride?

Calculate the minimum concentration of Mg²⁺ that must be added to 0.10 M NaF in order to initiate a precipitate of magnesium fluoride.(For MgF₂ , K_sp = 6.9 × 10^-9.)

Calculate the minimum concentration of Cr³⁺ that must be added to 0.095 M NaF in order to initiate a precipitate of chromium(III) fluoride.(For CrF₃, K_sp = 6.6 × 10^-11.)

The solubility of lead(II) iodide is 0.064 g/100 mL at 20ºC.What is the solubility product for lead(II) iodide?

The molar solubility of tin(II) iodide is 1.28 × 10^-2 mol/L.What is K_sp for this compound?

Calculate the concentration of fluoride ions in a saturated barium fluoride (K_sp = 1.7 × 10^-6) solution.

The molar solubility of lead(II) iodate in water is 4.0 × 10^-5 mol/L.Calculate K_sp for lead(II) iodate.

Quiz 16: Acid-Base Equilibria and Solubility Equilibria

The Ksp value for lead(II) chloride is 2.4 × 10-4.What is the molar solubility of lead(II) chloride?

The solubility of strontium carbonate is 0.0011 g/100 mL at 20ºC.Calculate the Ksp value for this compound.

The molar solubility of manganese(II) carbonate is 4.2 × 10-6 M.What is Ksp for this compound?

The solubility product for barium sulfate is 1.1 × 10-10.Calculate the molar solubility of barium sulfate.

The solubility product for calcium phosphate is Ksp = 1.3 × 10-26.What is the molar solubility of calcium phosphate?

The Ksp for silver(I) phosphate is 1.8 × 10-18.Calculate the molar solubility of silver(I) phosphate.

Calculate the silver ion concentration in a saturated solution of silver(I) sulfate (Ksp = 1.4 × 10-5) .

The Ksp for silver(I) phosphate is 1.8 × 10-18.Determine the silver ion concentration in a saturated solution of silver(I) phosphate.

Calculate the concentration of chloride ions in a saturated lead(II) chloride (Ksp = 2.4 × 10-4) solution.

Which of the following would decrease the Ksp for PbI2?

For PbCl2 (Ksp = 2.4 × 10-4) , will a precipitate of PbCl2 form when 0.10 L of 3.0 × 10-2 M Pb(NO3) 2 is added to 400 mL of 9.0 × 10-2 M NaCl?

Calculate the silver ion concentration in a saturated solution of silver(I) carbonate (Ksp = 8.1 × 10-12) .

The molar solubility of magnesium carbonate is 1.8 × 10-4 mol/L.What is Ksp for this compound?

The solubility product for chromium(III) fluoride is Ksp = 6.6 × 10-11.What is the molar solubility of chromium(III) fluoride?

Calculate the minimum concentration of Mg2+ that must be added to 0.10 M NaF in order to initiate a precipitate of magnesium fluoride.(For MgF2 , Ksp = 6.9 × 10-9.)

Calculate the minimum concentration of Cr3+ that must be added to 0.095 M NaF in order to initiate a precipitate of chromium(III) fluoride.(For CrF3 , Ksp = 6.6 × 10-11.)

The solubility of lead(II) iodide is 0.064 g/100 mL at 20ºC.What is the solubility product for lead(II) iodide?

The molar solubility of tin(II) iodide is 1.28 × 10-2 mol/L.What is Ksp for this compound?

Calculate the concentration of fluoride ions in a saturated barium fluoride (Ksp = 1.7 × 10-6) solution.

The molar solubility of lead(II) iodate in water is 4.0 × 10-5 mol/L.Calculate Ksp for lead(II) iodate.

The K_sp value for lead(II) chloride is 2.4 × 10^-4.What is the molar solubility of lead(II) chloride?

The solubility of strontium carbonate is 0.0011 g/100 mL at 20ºC.Calculate the K_sp value for this compound.

The molar solubility of manganese(II) carbonate is 4.2 × 10^-6 M.What is K_sp for this compound?

The solubility product for barium sulfate is 1.1 × 10^-10.Calculate the molar solubility of barium sulfate.

The solubility product for calcium phosphate is K_sp = 1.3 × 10^-26.What is the molar solubility of calcium phosphate?

The K_sp for silver(I) phosphate is 1.8 × 10^-18.Calculate the molar solubility of silver(I) phosphate.

Calculate the silver ion concentration in a saturated solution of silver(I) sulfate (K_sp = 1.4 × 10^-5) .

The K_sp for silver(I) phosphate is 1.8 × 10^-18.Determine the silver ion concentration in a saturated solution of silver(I) phosphate.

Calculate the concentration of chloride ions in a saturated lead(II) chloride (K_sp = 2.4 × 10^-4) solution.

Which of the following would decrease the K_sp for PbI₂?

For PbCl₂ (K_sp = 2.4 × 10^-4) , will a precipitate of PbCl₂ form when 0.10 L of 3.0 × 10^-2 M Pb(NO₃) ₂ is added to 400 mL of 9.0 × 10^-2 M NaCl?

Calculate the silver ion concentration in a saturated solution of silver(I) carbonate (K_sp = 8.1 × 10^-12) .

The molar solubility of magnesium carbonate is 1.8 × 10^-4 mol/L.What is K_sp for this compound?

The solubility product for chromium(III) fluoride is K_sp = 6.6 × 10^-11.What is the molar solubility of chromium(III) fluoride?

Calculate the minimum concentration of Mg²⁺ that must be added to 0.10 M NaF in order to initiate a precipitate of magnesium fluoride.(For MgF₂ , K_sp = 6.9 × 10^-9.)

Calculate the minimum concentration of Cr³⁺ that must be added to 0.095 M NaF in order to initiate a precipitate of chromium(III) fluoride.(For CrF₃, K_sp = 6.6 × 10^-11.)

The molar solubility of tin(II) iodide is 1.28 × 10^-2 mol/L.What is K_sp for this compound?

Calculate the concentration of fluoride ions in a saturated barium fluoride (K_sp = 1.7 × 10^-6) solution.

The molar solubility of lead(II) iodate in water is 4.0 × 10^-5 mol/L.Calculate K_sp for lead(II) iodate.