Humboldt State University ® Department of Chemistry

Richard A. Paselk

Chem 110	General Chemistry	Summer 2006
Lecture Notes::Lec3_1 June	© R. Paselk 2006
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Review of Aqueous Equilibria

Buffers, cont.

Buffer calculations, cont. The fact that "x" in buffer calculations is small is also why the Henderson-Hasselbalch equation is legitimate for buffer calculations (x is ignored).

Example: For the buffer from last time (0.0125 moles acetic acid and 0.0250 moles of sodium acetate in 1.000 L of solution), pK_a = -logK_a = 4.74.

pH = pK_a + log[A^-] / [HA]

pH = 4.74 + log(0.250) / (0.125)

= 4.74 + 0.301

pH = 5.04

Notice that only the ratio matters, the actual concentrations of the acid and salt are not that important for determining pH!

Buffer Preparation

Consider the preparation of a buffer using phosphate ions. The reactions involved and K_a and pK_a values are shown below:

Rxn   Ka  pKa  H3PO4 H+ + H2PO4-  7.5 x 10-3  2.2  H2PO4- H+ + HPO42-  6.2 x 10-8  7.2 HPO42-  H+ + PO43-  4.8 x 10-13  12.7

How would you make up a 0.10 molar phosphate buffer with a pH of 6.5 assuming all three sodium "phosphate" salts are available?

• First need to chose the appropriate salts. Looking at the pK_a values we would choose H₂PO₄^- as the acid and HPO₄^2- as the salt in the Henderson-Hasselbalch equation since the pH is between the pK_a's for these reactions:
  • pH = pK_a + log[A^-] / [HA]
  • pH = pK_a + log[HPO₄^2-] / [H₂PO₄^-]
  • and inserting values as appropriate, 6.5 = 7.2 + log[HPO₄^2-] / [H₂PO₄^-]
  • solving

  log[HPO₄^2-] / [H₂PO₄^-] = 6.5 - 7.2

  [H₂PO₄^-] / [HPO₄^2-] = 10^0.7 = 5.01

  [H₂PO₄^-] = 5.01*[HPO₄^2-]

  and [H₂PO₄^-] + [HPO₄^2-] = 0.1

  substituting, 5.01*[HPO₄^2-] + [HPO₄^2-] = 0.1

  [HPO₄^2-] = 0.017 M

  [H₂PO₄^-] = 0.084 M

Aqueous Ion Solubility

The solubility of ionic compounds in aqueous solutions has immense impact on the world. In particular ionic solubility determines the fate and deposition of vast mineral deposits on Earth's surface, it is used by organisms to create and model mineral components such as shell and bone, it impacts mineral accessibility in soils, and it is utilized and manipulated by humans in all manner of processes.

An example of a very important ionic solid is calcium carbonate, CaCO₃. CaCO₃ is used by organisms to make shells etc. Vast deposits of these shells make up limestone and (after heat and compression) marble deposits.

When looking at solubility situations it is most common to look at the dissolution of the solid:

CaCO₃ Ca²⁺ + CO₃^2-

The equilibrium expression for this reaction can then be written as:

But recall that the activity (the effective or apparent concentration of a substance in a chemical system) is 1 for any solid (remember, a solid is a crystalline substance, and therefore a pure phase). We can thus write the simpler expression for the equilibrium and give a new symbol, K_sp, the solubility product, for this expression:

Solubility from K_sp values:

Example: Using the solubility product we can then determine the solubility of calcium carbonate in water. Since for every calcium carbonate which dissolves one calcium ion and one carbonate ion goes into solution, we can use the concentration of calcium = the concentration of carbonate = the concentration of calcium carbonate in a pure saturated solution as the solubility:

[Ca²⁺] = [CO₃^2-] = (8.7 x 10^-9)^1/2 = 9.3 x 10^-5

So the solubility of calcium carbonate = 9.3 x 10^-5 M.

We can use this information to determine the conditions under which calcium carbonate will precipitate and dissolve. Of course the situations in nature are more complicated because the concentrations of calcium and carbonate are not generally equal. Further, the concentration of carbonate in natural environments depends on its equilibrium with bicarbonate which in turn depends on pH and carbon dioxide concentrations.

CO₃^2- + H⁺ HCO₃^-

HCO₃^- + H⁺ CO₂ + H₂O

Stories:

• Ocean shells and dissolution (Pacific more acidic than Atlantic).
• Limestone caves.

HCO₃^- + Ca²⁺ CaCO₃ + H⁺

HCO₃^- + H⁺ CO₂ + H₂O

K_sp values from solubility data: We can calculate K_sp values from solubility data, such as you are doing in lab.

Example: Calculate the the K_sp for lead chloride at 25°C if its solubility = 1.62 x 10^-2 M.

PbCl₂ Pb²⁺ + 2 Cl^-

K_sp = [Pb²⁺] [Cl^-]²

From the chemical equation we can see that twice as much chloride ion as lead ion results when lead chloride is dissolved, thus a saturated solution of lead chloride will have:

Pb²⁺ = 1.62 x 10^-2 M

Cl^- = 3.24 x 10^-2 M

Substituting these values into the solubility product expression then gives:

K_sp = (1.62 x 10^-2)(3.24 x 10^-2)² = 1.70 x 10^-5

 

Relative solubilities: Let's look at examples similar to your text.

As your text says, its easy to determine relative solubilities IF each compound being compared gives the same number of ions. The most soluble than has the largest K_sp, the least the smallest:

CaSO₄, K_sp = 2.4 x 10^-5 is more soluble than CaCO₃, K_sp = 8.7 x 10^-9

On the other hand, for Bi₂S₃, K_sp = 1.1 x 10^-73 compared to CuS, K_sp = 8.5 x 10^-45 we need to look at the equilibrium equation to calculate the solubility, x for Bi₂S₃

Bi2S3   Bi3+ + S2- amount dissolved x      in solution      2x    3x

To find the solubility, x, we can set up the solubility product:

Whereas for CuS:

K_sp = x² = 8.5 x 10^-45

Thus Bi₂S₃ is more soluble than CuS.

Common Ion Effects on Ionic Solubility: As we saw earlier, a common ion can greatly effect an equilibrium. Let's look at an example of a common ion effect on solubility.

Example: What is the solubility of ferric hydroxide in pH 8.0 buffered aqueous solution:

Fe(OH)₃ Fe³⁺ + 3 OH^-

K_sp = 4 x 10^-38

[Fe³⁺] [OH^-]³ = 4 x 10^-38

but the buffer fixes [OH^-] = 1 x 10^-6M

so [Fe³⁺] = (4 x 10^-38) / (10^-6)³ = 4 x 10^-20 M

© R A Paselk

Last modified 1 June 2006