aqueous equilibria entry task: feb 28 th thursday question: provide the k sp expression for calcium...

AqueousEquilibria

Entry Task: Feb 28th Thursday

Question:

Provide the Ksp expression for calcium phosphate, Ksp = 2.0 x 10-29.

From this expression, will there be a great deal of dissolved substance?

You have 5 minutes!

AqueousEquilibria

I can…• Define and explain the difference between:

SolubilitySolubility-productMolar solubility

• Write the Ksp expression for a given soluble ionic compound.

• Calculate the number of ions dissolved in solution.

AqueousEquilibria

Chapter 17Additional Aspects of Aqueous Equilibria

Sections 4

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Solubility Products

Define saturated solution.

Is a solution in which the solution is in contact with undissolved solute.

It’s the maximum amount of solute dissolved in solution at a specific temperature.

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Solubility Products

Consider the equilibrium that exists in a saturated solution of BaSO4 in water:

BaSO4(s) Ba2+(aq) + SO42-(aq)

Equilibrium shift to the left, because not a lot of product. Slightly souble.

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Solubility Products

The equilibrium constant expression for this equilibrium is

Ksp = [Ba2+] [SO42-]

where the equilibrium constant, Ksp, is called the solubility product.

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Solubility Products• Where can one find the Ksp values for a

particular ionic solid?

Appendix D on page 1023

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Solubility Products• What does a small Ksp value mean?

The smaller the Ksp value only a very small amount of solid would dissolve.

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Solubility Products• Write the Ksp constant for CaF2 including

the # value.

CaF2 (s) Ca+2 (aq) + 2F- (aq)

3.9 x 10-11 = [Ca+2] [F-]2

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Solubility Products• Write the Ksp constant for barium carbonate

include the # value.

BaCO3 (s) Ba+2 (aq) + CO3-2 (aq)

5.0 x 10-9 = [Ba+2] [CO3-2]

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Solubility Products• Write the Ksp constant for silver sulfate

include the # value.

Ag2SO4 (s) 2Ag+1 (aq) + SO4-2 (aq)

1.5 x 10-5 = [Ag+1]2 [SO4-2]

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Solubility and Ksp• It’s important to distinguish the difference

between solubility and solubility-product constant, Ksp.

• What is solubility?

Solubility of a substance is the quantity that dissolves to form a saturated solution

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• What is the molar solubility?

The molar solubility is the number of moles of the solute that dissolves in forming a liter of saturated solution of the solute (mol/L)

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• What is the solubility-product constant (Ksp)?

The solubility-product constant (Ksp) is the equilibrium constant for the equilibrium between ionic solid and its saturated solution.

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Solubility and Ksp• How can solubility of a solid change?

The solutions pH

Concentration of other ions in solution

Temperature

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Solubility Products

• Ksp is not the same as solubility.

• Solubility is generally expressed as the mass of solute dissolved in 1 L (g/L) or 100 mL (g/mL) of solution, or in mol/L (M).

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AqueousEquilibria

Calculating the Ksp

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Problem 1: Solid silver chromate is added to pure water at 25 C, and some of the solid remains undissolved. The mixture is stirred for several days to ensure that equilibrium is achieved between the undissolved Ag2CrO4(s) and the solution. Analysis of the equilibrated solution shows that its silver ion concentration is 1.3 104 M. Assuming that Ag2CrO4 dissociates completely in water and that there are no other important equilibria involving Ag+ or CrO4

2– ions in the solution, calculate Ksp for this compound.

First provide the equation:

Ag2CrO4(s) 2Ag+1(aq) + CrO42-(aq)

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What do we know:

Divide 1.3 x 10-4 by 2 = 6.5 x 10-5

We know that there is a 2 Ag1+ ions to 1 of CrO42-

Ag2CrO4(s) 2Ag+1(aq) + CrO42-(aq)

The Ag1+ ions 1.3 104 M.

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Provide the Ksp expression:

Ksp = [Ag+1]2 [CrO42-]

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Plug in the numbers:

Ksp = [1.3 x 10-4]2 [6.5 x 10-5]

Now solve:

1.1 x 10-12

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Problem 2: A saturated solution of Mg(OH)2 in contact with undissolved Mg(OH)2(s) is prepared at 25 C. The pH of the solution is found to be 10.17. Assuming that Mg(OH)2 dissociates completely in water and that there are no other simultaneous equilibria involving the Mg2+ or OH– ions, calculate Ksp for this compound.


Mg(OH)2(s) Mg+2(aq) + 2OH1-(aq)

AqueousEquilibria

Problem 2: A saturated solution of Mg(OH)2 in contact with undissolved Mg(OH)2(s) is prepared at 25 C. The pH of the solution is found to be 10.17. Assuming that Mg(OH)2 dissociates completely in water and that there are no other simultaneous equilibria involving the Mg2+ or OH– ions, calculate Ksp for this compound.What we know:pH is 10.17 , what can we get out of this information?

We can get OH- concentration.

pH = 10.17 – 14 = 3.83 is pOH

10-3.83 = 1.47 x 10-4 is the OH-1 concentrationAND there is twice the amount OH-1 to Mg+2 ions so divide OH by 2. 7.35 x 10-5 Mg+2 ions

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Ksp = [Mg+2] [OH-1]2

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Ksp = [7.35 x 10-5 ] [1.47 x 10-4 ]2

Ksp = 1.6 x 10-12

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Problem 3: Calculate the Ksp for CaCl2 if 200.0 g of CaCl2 are required to saturate 100.0 mL of solution.


CaCl2(s) Ca+2(aq) + 2Cl1-(aq)

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What we know:


200.0 g / 110.98 = 1.80 mol/0.1 L or 18.0M of CaCl2

There’s 1 Ca+2 ions to 2Cl-1 ions.

There’s 18.0M Ca+2 ions to 36.0M Cl-1 ions.

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Ksp = [Ca2+][F-]2

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Ksp = [18.0][36.0]2

There’s 18.0M Ca+2 ions to 36.0M Cl-1 ions.

Ksp = 2.33 x104

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Calculating Molar Solubility

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Problem 4: If the solubility product of HgSO4 is 6.4 x10 -5, then its molar solubility is:

Provide equation:

HgSO4(s) Hg+2(aq) + SO42-(aq)

AqueousEquilibria


What we know:

Initial

Change

Equilibrium

0 0-------

-------

-------

+x M +x M

x M x M

HgSO4(s) Hg+2(aq) + SO42-(aq)

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What we know:

Notice that the stoichiometry of the equilibrium for every mole of HgSO4 dissolves there is the same number of its ions.

Provide the Ksp expression: Ksp = [Hg2+][SO4-2]

Plug in the equilibrium [ ]: 6.3 x 10-5 = [x][x]

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Clean up and solve:

Plug in the equilibrium [ ]:

x2 = 6.3 x 10-5 x = 7.9 x 10-3

What does this number represent?

It’s the molar solubility of HgSO4 is 7.9 x 10-3

6.3 x 10-5 = [x][x]

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Problem 5: The solubility product of NaCl is 1.5x10-10 is at 298 K. its solubility in mol/L would be:

Provide equation:

NaCl(s) Na+1(aq) + Cl1-(aq)

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What we know:

Initial

Change

Equilibrium

0 0-------

-------

-------

+x M +x M

x M x M

NaCl(s) Na+1(aq) + Cl1-(aq)

AqueousEquilibria


What we know:

Notice that the stoichiometry of the equilibrium for every mole of NaCl dissolves there is the same number of its ions.

Provide the Ksp expression: Ksp = [Na1+][Cl-1]

Plug in the equilibrium [ ]: 1.5 x 10-10 = [x][x]

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Clean up and solve:

Plug in the equilibrium [ ]:

x2 = 1.5 x 10-10 x = 1.2 x 10-5


It’s the molar solubility of NaCl is 1.2 x 10-5

1.5 x 10-10 = [x][x]

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Calculating Solubility

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Problem 6: The Ksp for CaF2 is 3.9 1011 at 25 C . Assuming that CaF2 dissociates completely upon dissolving and that there are no other important equilibria affecting its solubility, calculate the solubility of CaF2 in grams per liter.


CaF2(s) Ca+2(aq) + 2F1-(aq)

AqueousEquilibria


What we know:

**Remember solubility is the amount of solute that can dissolve in a solvent, where as the solubility-product constant is an equilibrium constant.

**For this reason, we have to assume initially-that none of the salt has dissolved and allow x moles/liter of CaF2

to dissociate completely when equilibrium is achieved

AqueousEquilibria


What we know:

CaF2(s) Ca+2(aq) + 2F1-(aq)Initial

Change

Equilibrium

0 0-------

-------

-------

+x M +2x M

x M 2x M

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Problem 6: The Ksp for CaF2 is 3.9 1011 at 25 C . Assuming that CaF2 dissociates completely upon dissolving and that there are no other important equilibria affecting its solubility, calculate the solubility of CaF2 in grams per liter.What we know:

Notice that the stoichiometry of the equilibrium dictates that 2x moles/liter of F- ions are produced for each x moles/liter of CaF2 that dissolve. Provide the Ksp expression: Ksp = [Ca2+][F-]2

Plug in the equilibrium [ ]: 3.9 x 10-11 = [x][2x]2

AqueousEquilibria


Clean up and solve:

Provide the Ksp expression: Ksp = [Ca2+][F-]2


4x3 = 3.9 x 10-11

x = 3 3.9 x 10-11

4 x = 2.1 x 10-4


It’s the molar solubility of CaF2 is 2.1 x 10-4

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From the molarity solubility we can get to the number of grams dissolved in 1 liter of solvent

Now what:

2.1 x 10-4 mol CaF2

1 liter X

1.6 x 10-2 g CaF2/liter solution

78.1 g CaF2

1 mol

AqueousEquilibria

Problem 7: The Ksp for Cu(N3)2 is 6.3 1010 at 25 C . What is the solubility of Cu(N3)2 in grams per liter.

Provide equation:

Cu(N3)2(s) Cu+2(aq) + 2N31-(aq)

AqueousEquilibria


What we know:

**Remember solubility is the amount of solute that can dissolve in a solvent, where as the solubility-product constant is an equilibrium constant.

**For this reason, we have to assume initially-that none of the salt has dissolved and allow x moles/liter of Cu(N3)2 to dissociate completely when equilibrium is achieved

AqueousEquilibria


What we know:

Initial

Change

Equilibrium

0 0-------

-------

-------

+x M +2x M

x M 2x M

Cu(N3)2(s) Cu+2(aq) + 2N31-(aq)

AqueousEquilibria


What we know:

Notice that the stoichiometry of the equilibrium dictates that 2x moles/liter of N3- ions are produced for each x moles/liter of Cu(N3)2 that dissolve. Provide the Ksp expression: Ksp = [Cu2+][N3

-]2


AqueousEquilibria


What we know:

Clean up and solve:

Provide the Ksp expression: Ksp = [Cu2+][N3-]2


4x3 = 6.3 x 10-10

x = 3 6.3 x 10-10

4

x = 5.4 x 10-4


It’s the molar solubility of Cu(N3)2 is 5.4 x 10-4

AqueousEquilibria


What we know:

From the molarity solubility we can get to the number of grams dissolved in 1 liter of solvent

Now what:

5.4 x 10-4 mol Cu(N3)2

1 liter X

7.97 x 10-2 g Cu(N3)2 /liter solution

147.6 g Cu(N3)2 1 mol

aqueous equilibria entry task: feb 28 th thursday question: provide the k sp expression for calcium...

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