Progress check 7.3 (22 --> 33)

For which of the following salts would the relationship between molar solubility, s, in mol/L, and the value of K_sp be represented by the equation K_sp⁢=4s³ ?

• PbCO₃

• Mg₃(PO₄)₂

• Ag₂SO₄

• MnS

Hg₂I₂(s)⇄Hg₂²⁺(aq)+2 I⁻(aq)

K_sp=[Hg₂²⁺][I⁻]²

A saturated solution of Hg₂I₂ is at equilibrium at 25°C as

represented by the equation above. If [I⁻]=4.6×10⁻¹⁰ M at equilibrium, which of the following gives the correct molar solubility, S, and K_sp for Hg₂I₂ ?

• S=4.6×10⁻¹⁰ M; K_sp =(2.3×10⁻¹⁰)(4.6×10⁻¹⁰)²

• S=4.6×10⁻¹⁰ M; K_sp =(4.6×10⁻¹⁰)(9.2×10⁻¹⁰)²

• S=2.3×10⁻¹⁰ M; K_sp =(2.3×10⁻¹⁰)(4.6×10⁻¹⁰)²

• S=2.3×10⁻¹⁰ M; K_sp=(4.6×10⁻¹⁰)(9.2×10⁻¹⁰)²

Tags

NGSS.HS-PS1-7

AgI(s)⇄ $Ag^+\left(aq\right)+I^-\left(aq\right)$   
Ksp= $8.3\times10^{-17}$   at 298K    

The dissolution of AgI is represented above. Which of the following shows the mathematical relationship between the molar solubility, S, of AgI and the Ksp at 298K?

• S=8.3x $10^{-17}$  mol/L

• S= $\frac{8.3\times10^{-17}}{2}$  mol/L

• S= $\sqrt{8.3\times10^{-17}}$  mol/L

• S=2 $\sqrt{8.3\times10^{-17}}$  mol/L

Ca(OH)₂(s)⇄Ca²⁺(aq)+2OH⁻(aq)

K_sp=5.5×10⁻⁶ at298K

The equilibrium in a saturated solution of Ca(OH)₂ is represented above. In an experiment, a student places 5.0 g of Ca(OH)₂(s) into 100.0mL of distilled water and stirs the mixture.

How would the results be affected if the student repeats the experiment but this time places 5.0g of Ca(OH)₂(s) into 100.0mL of 0.0010M NaOH(aq) instead of distilled water?

• Less solid will dissolve, because the larger value of [OH⁻] will cause the equilibrium position to lie farther to the right.

• Less solid will dissolve, because the larger value of [OH⁻] will cause the equilibrium position to lie farther to the left.

• More solid will dissolve, because the larger value of [OH⁻] will cause the equilibrium position to lie farther to the right.

• More solid will dissolve, because the smaller value of [OH⁻] will cause the equilibrium position to lie farther to the left.

Tags

NGSS.HS-PS1-5

AgCN(s)⇄Ag⁺(aq)+CN⁻(aq)

The dissolution of solid AgCN is represented by the chemical equation above. In pure water, the equilibrium concentration of Ag⁺ ions in a saturated solution is 7.7×10⁻⁹M. If a small amount of solid NaCN is added to the saturated AgCN solution, which of the following would be observed?

• The Ksp increases and more AgCN dissolves.

• The Ksp increases and some AgCN precipitates.

• The molar solubility of AgCN becomes smaller than 7.7×10⁻⁹M and some AgCN precipitates.

• The molar solubility of AgCN becomes larger than 7.7×10⁻⁹M and more AgCN dissolves.

Mg(OH)₂(s) ⇄ Mg²⁺(aq) + 2OH⁻(aq)

A student prepared a saturated aqueous solution of Mg(OH)₂ and measured its pH, as shown in Figure 1 above. Then the student added a few drops of an unknown solution to the test tube and observed cloudiness in the solutions as shown in Figure 2. On the basis of this information and the equilibrium represented above, which of the following is most likely the identity of the reagent added from the dropper?

• Distilled water

• NaNO₃(aq)

• HCl(aq)

• KOH(aq)

Tags

NGSS.HS-PS1-2

CaF₂(s) ⇄ Ca²⁺(aq)+2 F⁻(aq).

The value of K_sp for the dissolution is 3.5×10⁻¹¹. A student measures the concentration of Ca²⁺ ions in a saturated solution of CaF₂ at various pH values and uses those values to generate the graph above. Based on the data, which of the following observations about the solubility of CaF₂ is most valid?

• It does not depend on pH because [Ca²⁺] does not change between pH4 and pH12.

• It does not depend on pH because K_sp=[Ca²⁺][F⁻]², so as [Ca²⁺] decreases, [F⁻] increases to compensate, keeping K_sp constant.

• It is higher at a lower pH; there are more H⁺ ions in solution at low pH, so HF forms and shifts the equilibrium reaction above to the right.

• It is lower at a higher pH; there are more H⁺ ions in solution at high pH, so HF forms and shifts the equilibrium reaction above to the right.

Cu²⁺(aq) absorbs a certain frequency of visible light. Absorbance was measured for three saturated solutions of Cu(OH)₂, each at a different pH at 298K. Based on the data recorded in the table above, which of the following conclusions about the effect of pH on the solubility of Cu(OH)₂ can be made?

• When the concentration of H+ ions is increased, the solubility of Cu(OH)₂ increases.

• When the concentration of H+ ions is decreased, the solubility of Cu(OH)₂ increases.

• The solubility of Cu(OH)₂ is independent of pH.

• Cu(OH)2 is soluble only at a pH of 7.00.

Shown above are a chemical equation that represents the dissolution of PbBr₂ in pure water, a table of the changes in some thermodynamic properties for the process, and a particle diagram. Which of the following explains which relative change in a thermodynamic property is best illustrated by the particle diagram?

• The reorganization of the water molecules around the ions illustrates that ΔH°>0 because forming strong ion-dipole interactions releases energy.

• The very small amount of Pb²⁺ and Br⁻ ions illustrates that ΔS°>0 because entropy decreases when PbBr₂ dissolves.

• The very small amount of Pb²⁺ and Br⁻ ions illustrates that ΔG°>0 because the dissolution of PbBr₂ is not a favorable process.

• The very large amount of solid that remains undissolved illustrates that ΔG°>0 because the dissolution of PbBr₂ is a favorable process.

Tags

NGSS.HS-PS1-4

The particle diagram shown above represents the dissolution of CuCl(s) assuming an equilibrium concentration for Cu⁺ ions of about 4×10⁻⁴M in a saturated solution at 25°C. The equilibrium being represented is shown in the following chemical equation.

CuCl(s)⇄Cu+(aq)+Cl−(aq)

Which of the following changes to the particle diagram will best represent the effect of adding 1.0mL of 4MNaCl to the solution?

• Some of the Cu⁺ and Cl⁻ ions combine to form CuCl(s) because the K_sp will be lower than 1.6×10⁻⁷

• Some of the Cu⁺ and Cl⁻ ions combine to form CuCl(s) because the molar solubility will be lower than 4×10⁻⁴M.

• More Cu⁺ and Cl⁻ ions will be in solution because the molar solubility will be higher than 4×10⁻⁴M.

• More Cu⁺ and Cl⁻ ions will be in solution because the K_sp will be higher than 1.6×10⁻⁷.

Tags

NGSS.HS-PS1-5