THE SOLUBILITY CONSTANT

The equation for the reaction between Al(s) and Pb+2(aq) is: 2Al(s) + 3Pb+2(aq) 2Al+2(aq) + 3Pb(s). Therefore, for each 2 moles of Al(s) that are used, how many moles of Pb(s) are formed? From the moles of Al(s) consumed, 0.011 moles, calculate the number of moles of Pb(s) that should have been formed: moles

3 0.017

Which of the following apply to concentration? Select all that apply

Concentration of products can vary. As more solute dissolves, more ions form; but the volume in which they are contained does not change. Concentration of the liquid phase can change. Concentration of the solid phase does not change.

Using a value of Ksp = 1.8 x 10-2 for the reaction PbCl2(s) Pb+2(aq) + 2Cl-(aq). The concentration of the products yields a value of 2.1 x 10-2. Select the TWO statements that are true.

Ions are recombining to form a solid precipitate. The concentrations of Pb+2(aq) and Cl-(aq) are expected to decrease.

Select all that apply. When is Ksp determined for a system?

when equilibrium is established when the two phase system is saturated

Molarity is defined as per liter.

moles

The student placed 10 mL of PbCl2 (saturated solution) in the test tube and added a pinch of NaCl. A white precipitate of PbCl2 formed. What happens to the value of Ksp when NaCl was added?

remain constant

A saturated solution occurs when _____.

a solution can dissolve no more solute at a given temperature

Mass of Al wire before reaction = 3.96 g Mass of Al wire after reaction = 3.65 g Mass of Al lost = 0.31 g Moles of Al lost = 0.011 moles Mass of Pb + filter paper = 4.26 g Mass of filter paper = 0.92 g Mass of Pb = 3.34 g Moles of Pb formed = 0.016 moles Recall that there were 100 mL of solution. 0.016 moles of Pb were removed from the solution. What was the [Pb+2] for the saturated solution of PbCl2? Recall that there was 100 mL of solution. 0.016 moles of Pb were removed from the solution. Also recall the original equilibrium: PbCl2(s) Pb+2(aq) + 2(Cl-)(aq) Since there are 2 Cl ions formed for every Pb ion, what was the [Cl-] for the saturated solution of PbCl2? M

0.16 0.32

The equation for the reaction between Al(s) and Pb+2(aq) is: 2Al(s) + 3Pb+2(aq) 2Al+3(aq) + 3Pb(s) You know that 0.011 moles of Al(s) have been consumed and that 0.017 moles of Pb(s) have been formed. Calculate [Pb+2].

0.17M

Complete the Data Table: Use these values: The atomic mass of Al is 27.0 g/mole. The atomic mass of Pb is 207.2 g/mole. Use two significant digits for your answers: Mass of Al wire before reaction = 3.96 g Mass of Al wire after reaction = 3.65 g Mass of Al lost = g Moles of Al lost = moles Mass of Pb + filter paper = 4.26 g Mass of filter paper = 0.92 g Mass of Pb = g Moles of Pb formed = moles

0.31 0.011 3.34 0.016

You now know that [Pb+2] = 0.17 M. Calculate [Cl-]. Refer to equations given in the text to determine the relationship between these ions. 0.17 M

0.34M

A solution has a [Ag+(aq)] of 0.01 M. The chloride concentration in solution is 1 x 10-5 M. Based on the following reaction, answer the following questions: AgCl(s) Ag+(aq) + Cl-(aq), Ksp = 1.7 x 10-10 Calculate the value of the current solubility product. Is the K value obtained greater or less than Ksp? Will a precipitate of AgCl form?

1 x 10⁻⁷ greater yes

To verify the previous experiment, the chemistry student carried out the next experiment and recorded the data for you in the following Data Table. Directions: Place 100 mL of filtered, saturated PbCl2 in a 150 mL beaker and put it under a dryer to evaporate the water. Record the necessary data in the Data Table. Complete the Data Table. Since the data allows it, you should use 3 significant digits for all your answers. Molecular weight of PbCl2 = 278 g/mole Mass of beaker + PbCl2 = 62.19 g Mass of beaker = 57.77 g Mass of PbCl2 per 100 mL = g Moles PbCl2 per 100 mL = [Pb-2] based upon this investigation = M [Cl-] based upon this investigation = M This is close to our answer of 0.34 M above. The difference may be due to experimental error and to rounding numbers off because we were limited to 2 significant digits in the first part of this experiment. This illustrates the importance of using care with our procedures in the laboratory and to record instrument readings to as many significant digits as possible.

4.42 0.0159 0.159 0.318

Which of the following Gold compounds has the greatest solubility?

AuCl, Ksp = 2.0 x 10-13

Choose the Ksp expressions for the following reactions. Ca3(PO4)2(s) 3Ca+2 (aq) + 2PO4 -3(aq)

Ksp = [Ca+2]3[PO4-3] 2

When [ ] are around an ion, what does it represent?

The concentration of the ion is in moles/liter.

Using a value of Ksp = 1.8 x 10-2 for the reaction PbCl2(s) Pb+2(aq) + 2Cl-(aq). If a solution has a current value of Keq of 1.2 x 10-2, which statement is true?

The solution is unsaturated

Select the THREE answers that are correct concerning a saturated solution of PbCl2. PbCl2(s) Pb+2(aq) + 2Cl-(aq)

The system is in equilibrium. The rate of dissolving is equal to the rate of recrystallization. The concentration of the products only is used to determine the Ksp.

The student placed 10 mL of PbCl2 (saturated solution) in the test tube and added a pinch of NaCl. A white precipitate of PbCl2 formed. How will this affect the concentration of Pb+2?

decrease it

The student placed 10 mL of PbCl2 (saturated solution) in the test tube and added a pinch of lead acetate, Pb(C2H3O2)2. When the test tube was shaken, a white precipitate of PbCl2 formed. How will this affect the concentration of Cl-?

decrease it

Data Table 7 Compound Temperature, °C Ksp AgCl 4.7 0.21 x 10-10 AgCl 9.7 0.37 x 10-10 AgCl 25 1.56 x 10-10 AgCl 50 13.2 x 10-10 AgCl 100 215 x 10-10 From the data table, as the temperature increases, the Ksp value will . Does the change in solubility data mean that the AgCl becomes more or less soluble at higher temperature?

increase more

The student placed 10 mL of PbCl2 (saturated solution) in the test tube and added a pinch of NaCl. A white precipitate of PbCl2 formed. You have the Cl- concentration by adding NaCl.

increased

The student placed 10 mL of PbCl2 (saturated solution) in the test tube and added a pinch of lead acetate, Pb(C2H3O2)2. When the test tube was shaken, a white precipitate of PbCl2 formed. You have the Pb+2 concentration by adding lead acetate.

increased