nc state universityfranzen/public_html/ch201/lecture/lecture_17.pdf · solubility equilibria...

Solubility equilibria

Chemistry 201

NC State University

Lecture 17

Solubility Equilibria

• What is the role of metal ions in solution pH?

• How do solution conditions affect solubility?

• Can precipitation be used to separate ions?

Text : Sections 8.1 - 8.3

Role of metal ions

Metal ions in solution can act as weak acids as

Shown in the Figure. These are called complex ions.

Hydrated metal ion Ka

Fe(H2O)63+ 6 x 10-3

Al(H2O)63+ 1 x 10-5

Cu(H2O)62+ 3 x 10-8

Zn(H2O)62+ 1 x 10-9

Ni(H2O)62+ 1 x 10-10

Role Hydrated metal ions Acidity constants


Consider the solubility of NaCl in water. The

ions Na+ and Cl- are in equilibrium with the solid

NaCl(s).

The equilibrium constant is K = 36. This applies

only to the ions and not to the solid. The solid

does not have concentration.


Since only the ions contribute to the equilibrium

constant, we can write,

The equilibrium constant is product of two

solubilities, and hence has the name solubility

product. Since K = 36, this means that we have

6 M [Na+] and [Cl-] in a saturated solution of

NaCl.

Given these Ksp’s:

– AgBr : 5.0 x 10-13

– Ag2S : 6.3 x 10-50

– Ag3PO4 : 2.6 x 10-18

What is the molar solubility of each compound?

Calculating Solubility Equilibria

Solution: the solubility equilibrium is

and the product is

The reaction leads to x moles of Ag+ and Br-.

Assuming a solid of AgBr is in equilibrium with

the ions in solution, calculate the concentration

of Ag+ and Br-. Ksp= 5.0 x 10-13


The concentrations are:



and the product is

The reaction leads to x moles of Ag+ and S2-.

Assuming a solid of Ag2S is in equilibrium with


of Ag+ and S2-. Ksp= 6.3 x 10-50


Assuming a solid of Ag3PO4 is in equilibrium with


of Ag+ and PO43-. Ksp= 2.6 x 10-18



and the product is

The reaction leads to x moles of Ag+ and PO43-.

Assuming a solid of Ag3PO4 is in equilibrium with


of Ag+ and PO43-. Ksp= 2.6 x 10-18


What is the solubility of AgBr in 0.02 M aqueous

NaBr?

Calculating Solubility Equilibria in the presence of a common ion


Make a reaction table

Substitute into the solubility product expression and

solve for x.


NaBr?


Species Ag+ Br-

Initial 0.0 0.02

Final x 0.02+x

Substitute into the solubility product expression and

solve for x.


NaBr?


Step 1. Write down the solubility equilibrium.

Step 2. Determine the hydroxide ion concentration.

Step 3. Substitute into the solubility product expression.

Step 4. Compare the value to the tablulated Ksp.

Will 0.01 M Mg2+ precipitate at pH 4?

Calculating Solubility Equilibria of metal hydroxides

Will 0.01 M Mg2+ precipitate at pH 4?


For this reaction Ksp = 5.61 x 10-12. Since the product of

the concentrations is less than this value the hydroxide will not precipitate. Here is a related, but harder problem.

At what pH will 0.01 M Mg2+ precipitate?

The comparison of concentrations in the solubility

product with the tabulated Ksp is an example of

a free energy relationship.

At equilibrium we have a saturated solution so

Thus, if Q > Ksp then DG > 0 and the solution

reaction,

is not spontaneous.

Justification for treatment of Ksp

Can 0.01 M Pb2+ be separated from 0.01 M Mg2+

at pH 9?


The answer to this question resides in the relative

Ksp for the hydroxides of each of these ions.

If one is completely insoluble then they can be

separated.


at pH 9?



at pH 9?


Step 1. Calculate the [OH-].


at pH 9?



Step2. Calculate the solubility products.


at pH 9?



Step2. Calculate the solubility products.

The answer in this case is yes, Pb(OH)2 can be

separated since it will precipitate, while Mg(OH) 2

remains in solution.

For polyprotic ions, the solubility product will depend

on the state of ionization of the ions.

For example, CaCO3 in the ocean is sparingly

soluble. However, if the CO32- concentration is

reduced by the equilibrium,

then CaCO3 will begin to dissolve. This equilibrium

is particularly important since coral reefs and diatoms

are composed of CaCO3.

pH-dependent equilibria

CaCO3 has a solubility product of Ksp = 3.36 x 10-9.

This is a moderately small Ksp, which means that

CaCO3 should precipitate if there is sufficient

Ca2+ and CO32-.

The equilibrium is

Precipitation of CaCO3

Step 1. Write the Ksp

Step2. Calculate the solubility.

What is the concentration of Ca2+ and CO32- in

a saturated solution at pH 7?

pH-dependent equilibria

The concentration of Ca2+ and CO32- in the world’s

oceans are 0.01 M and 3.1 x 10-4 M, respectively.

Calcium carbonate in the oceans

The concentration of Ca2+ and CO32- in the worlds


Will CaCO3 precipitate?


Step 1. Calculate the standard free energy

Step2. Calculate the free energy of CaCO3 in sea.





Step 2. Calculate the free energy (cont’d)

Since DG > 0 the reaction is NOT spontaneous as

written, formation of solution. Therefore, the

spontaneous process is precipitation.





Simpler approach. Just calculate the reaction

quotient and compare it to the solubility product.

The conclusion is the same. CaCO3 will precipitate.

But, wait a minute… why hasn’t it already done so?

Using solubility data to determine Ksp

Goals

• Quantify solubility of ionic compounds.

• Determine pH conditions for precipitation and separation of hydroxide compounds.

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