Aqueous Equilibria

By: Chris Via

Common-Ion Effect

• C.I.E.- the dissociation of a weak electrolyte by adding to the solution a strong electrolyte that has an ion in common with the weak electrolyte.

• The addition of this strong electrolyte causes the equilibrium to shift as well as the pH to change.

Example

• The dissociation of HC2H3O2 is:

• HC2H3O2 H+1 + C2H3O2-1

• but if NaC2H3O2 is added to the solution, the addition of C2H3O2

- 1 from NaC2H3O2 causes the equilibrium to shift to the left and also reducing the concentration of H+1, therefore raising the pH of the solution.

Practice Problem

• Calculate the fluoride -ion concentration and pH of a solution containing 0.1 mol of HCl and 0.2 mol of HF in 1.0L solution. (Ka of HF = 6.8 x 10-4)

• Now you want to set up an ice box of the dissociation of HF.

Continued

Initial .2 M .1 M 0

Change -x M +x M +x M

Equilibrium (.2-x) M (.1+x) M x M

HF H+ F-

Ka=6.8*10-4 = [H+][F-]/[HF]=(.1+x)x/(.2-x)

Solution• In this case we can assume x is too small to

make a difference so (0.1+ x) becomes just 0.1

• So we are left with 0.1 = 6.8 x 10- 4

0.2• x =1.4 x 10-3 M= [F-1]• [H+1]= (0.1+ x) = 0.1M• -log ([H+]) = -log (0.1) = 1• The pH=1.00 and the [F-1]= 1.4 x 10-3 M

Buffered Solutions• Buffers are solutions that resist changes in

pH upon the addition of small amounts of acids or bases.

• A key example of a buffer is human blood which will stay around a pH of 7.4

• Buffers resist changes in pH because they contain both an acidic species to neutralize OH-1 ions and a basic species to neutralize H+1 ions.

Buffers• However, it is key that the acid and base

species of the buffer do not neutralize each other.

• To do this, buffers are often prepared by mixing a weak acid or base with a salt of that acid or base.

• Ex, the HC2H3O2 and C2H3O2-1 buffer can be

prepared by adding NaC2H3O2 to a solution of HC2H3O2.

Buffer Capacity• Buffer Capacity is the amount of an acid or

base the buffer can neutralize before the pH begins to change significantly.

• The greater the amounts of conjugate acid-base pair, the more resistant the solution is to change in pH.

• Henderson-Hasselbalch equation is used to calculate the pH of buffers:

• pH = pKa + log[base]/[acid]

Sample Exercise• What is the pH of a buffer that is 0.12 M

lactic acid and .1 M sodium lactate? (Ka for lactic acid = 1.4 x 10-4)

• You want to start out with an ice box of the dissociation of lactic acid

• lactic acid hydrogen lactateInitial .12 M 0 .1 M

Change -x M +x M +x M

Equilibrium (.12-x) M x M (.1+x) M

Problem Continued

• Ka = 1.4 x 10-4 = x (0.1+x)/(0.12-x)

• Once again in this case x will be too small so it can be ignored.

• X=1.7 x 10 -4 = [H+1]

• pH= -log (1.7 x 10-4) = 3.77

• With the Henderson-Hasselbalch equation:

• pH= pKa + log([base]/[acid])

• pH= 3.85 + (-0.08) =3.77

Titrations

• An acid-base titration is when an acid is added to a base or vice versa.

• Acid-Base indicators are usually used to determine the equivalence point which is the point at which stoichiometrically equivalent quantities of acid and base have been brought together.

Acid-Base Titrations

• When strong acids and bases are mixed the pH changes very dramatically near the equivalence point

• a single drop a this point could change the pH by a number of units.

Titrations

Polyprotic Acids

• When titrating with Polyprotic acids or bases the substance has multiple equivalence points.

• So for example, in a titration of Na2CO3 with HCl. There are two distinct equivalence points on the titration curve.

Solubility-Product Constant

• Ksp= Solubility Product Constant

• It expresses the degree to which the solid is soluble in water.

• The equation for Ksp is Ksp = [ion]a[ion]b

• An example is the expression of the Ksp of BaSO4:

• Ksp = [Ba2+][SO42-]

Factors that Affect Solubility

• They are:– the presence of common ions– the pH of the solution– presence of complexing agents

• The presence of common ions in a solution will reduce the solubility and make the equilibrium shift left.

pH and Solubility

• The general rule of solubility and pH is:

• The solubility of slightly-soluble salts containing basic anions increases as the pH of the solution is lowered.

• This is because the OH- ion is insoluble in water while the H+ ion is very soluble, therefore when a basic solution has a low concentration of OH- ions the salt will be easier to dissolve.

Continued

• The more basic the anion, the more the solubility is influenced by the pH of the solution.

• However, salts with anions of strong acids are unaffected by changes in pH.

Complex Ions

• A characteristic of most metal ions is their ability to act as a Lewis acid when interacting with water (Lewis base).

• However, when these metal ions interact with Lewis bases other than water, the solubility of that metal salt changes dramatically.

Complex Ion• Complex Ion-when a metal ion (Lewis acid)

is bonded together with a Lewis base other than water.– Ex. Ag(NH3)2

+, Fe(CN)6-4

• The stability of a complex ion can be judged by the size of the equilibrium constant for its formation.– This is called the formation constant, Kf

– the larger Kf the more stable the ion is.

Amphoterism

• Amphoteric-a metal hydroxide that is capable of being dissolved in strong acids or strong bases, but not in water because it can act like an acid or a base.– Ex. Al+3, Cr+3, Zn+2, and Sn+2

• However, these metal ions are more accurately expressed as Al(H2O)6

+3.

• This is a weak acid and as it is added to a strong base it loses protons and eventually forms the neutral and water-soluble Al(H2O)3(OH)3

Precipitation of Ions

• The reaction quotient, Q, can with the solubility product constant to determine if a precipitation will occur.

• If Q > Ksp-precipitation occurs until Q = Ksp

• If Q = Ksp-equilibrium exists (saturated solution)

• If Q < Ksp-solid dissolves until Q = Ksp

• Q is sometimes referred to as the ion product because there is no denominator.

Selective Precipitation

• Selective Precipitation-the separation of ions in an aqueous solution by using a reagent that forms a precipitate with one or more of the ions.

• The sulfide ion is widely used to separate metal ions because the solubilities of the sulfide salts span a wide range and are dependent on the pH of the solution.

Continued

• An example is a solution containing both Cu+2 and Zn+2 mixed with H2S gas.

• CuS ends up forming a precipitate, but ZnS does not.

• This is because CuS has a Ksp value of 6 x 10-37 making it less soluble than ZnS which has a Ksp value of 2 x 10-25

•

Qualitative Analysis

• Qualitative Analysis determines the presence or absence of a particular metal ion, whereas quantitative analysis determines how much of a certain substance is present or produced.

• There are 5 main groups of metal ions:– Insoluble Chlorides, Acid-insoluble sulfides, base-

insoluble sulfides and hydroxides, insoluble phosphates, and the alkali metal ions and NH4

+1

• These can be found on page 673 in your textbook