17.2-17.6: Solubility and the Common Ion Effect-Complex Ions and Solubility

Solubility and the Common Ion Effect:

If you have a salt in a solution of another salt having the same cation or anion, the salt will be less soluble in this solution than it would be in pure water. For example, when adding calcium oxalate, CaC₂O₄, to a solution of calcium chloride, CaCl₂, both salts will contribute a Ca²⁺ ion. This change in solubility is explained by Le Chatelier's principle:

CaC₂O₄ (s) ⇌ Ca²⁺ (aq) + C₂O₄^2- (aq)

Calcium chloride will also contribute Ca²⁺ ion, which will act as a stress on the right hand side of the equation and the equilibrium will shift to the left, and calcium oxalate will precipitate from the solution.

Example:

MgC₂O₄is added to a 0.437 M solution of Na₂C₂O₄. What is the solubility of MgC₂O₄ in this solution?

MgC₂O₄ ⇌ Mg²⁺ (aq) + C₂O₄^2- (aq)

Notice that there is an initial value for C₂O₄^2-, contributed by the Na₂C₂O₄.

K_sp = (x)(0.437 + x) = 8.5 x 10^-5
(always check, but the approximation should hold for K_sp problems)
x = 1.95 x 10^-4 M

Remember the question asked for solubility, not molar solubility. So we have to convert mols to grams using the periodic table:

( 1.95 x 10^-4 mol / 1 L) x (112 g MgC₂O₄ / 1 mol MgC₂O₄) = 0.022 g/L

Example:

Find the molar solubility for Ca₃(PO₄)₂ in 0.115 M solution of Ca(NO₃)₂.
(K_sp for Ca₃(PO₄)₂ = 1 x 10^-26)

Ca₃(PO₄)₂ (s) ⇌ 3Ca²⁺ (aq) + 2PO₄^3- (aq)



K_sp = (0.115 + 3x)³(2x)² → approximation holds
K_sp = (0.115)³(4x²) = 1 x 10^-26
x = 1.28 x 10^-12


Precipitation Reactions:

To try to find out whether or not precipitation will occur in a reaction, we need to compare the Q_c to the K_c, or in this case, the K_sp. (We went over Q_c and comparing it to K_c in Chapter 14). The Q_c, or ion product as it is called when considering solubility, is found the same way as the K_sp, except the K_sp is found at equilibrium and the Q_c is not.

Example:

AgCl (s) ⇌ Ag⁺ (aq) + Cl^- (aq)
[Ag⁺][Cl^-] = K_sp at equilibrium
[Ag⁺][Cl^-] = Q_c when not at equilibrium

When Q_c is less than K_sp, there is not enough product yet to be at equilibrium, so the reaction will shift to the right, and no ppt (precipitate) will form.

When Q_c is greater than K_sp, there is too much product, so the reaction will shift to the left, and a ppt will form.

When Q_c is equal to K_sp, the reaction is at equilibrium.

Example:

Will there be a ppt if 50.0mL of 0.450 M AgNO₃ is mixed with 50.0mL of 0.350 M NaCl?

The first step is to think about what the possible ppt might be from these compounds. Write out what happens when these two compounds break up:

AgNO₃ (s) ⇌ Ag⁺ (aq) + NO₃^-(aq)
NaCl (s) ⇌ Na⁺ (aq) + Cl^- (aq)

When these four ions are floating around in the solution, the only possible ppt is AgCl.

Because we are mixing, the first thing we need to do is convert to mols then divide by the new volume to find the new concentrations of AgNO₃ and NaCl:

(0.500 L)(0.450 mol / 1 L) = (0.0225 mol) / (0.1 L) = 0.225 M AgNO₃
(0.500 L)(0.350 mol / 1 L) = (0.0175 mol) / (0.1 L) = 0.175 M NaCl

Q_c = [Ag⁺][Cl^-] = (0.225)(0.175) = 0.0394
K_sp = 1.8 x 10^-10

Q_c is greater than K_sp, the reaction will shift to the left and a ppt will form.

Fractional Precipitation:

A technique of separating two or more ions from a solution by adding a reactant that precipitates first one ion, then another, and so forth.

Example:

Start with a solution of 0.200M each of Cd(NO₃)₂, Ca(NO₃)₂, Mg(NO₃)₂. Each will dissociate to give 0.200M each of Cd²⁺, Ca²⁺, and Mg²⁺. If Na₂C₂O₄ is added slowly, what is the order of ppt? Recognize that each of the three ions will react with the oxalate anion in Na₂C₂O₄, which is C₂O₄^2-.

Na₂C₂O₄ (s) ⇌ 2Na⁺ (aq) + C₂O₄^2- (aq)

Three compounds with low solubility will form: CdC₂O₄, CaC₂O₄, MgC₂O₄.

K_sp for CdC₂O₄ = 1.5 x 10^-8
K_sp for CaC₂O₄= 2.3 x 10^-9
K_sp for MgC₂O₄ = 8.5 x 10^-5

A precipitate will form when Q_c is greater than K_sp. In other words, just past the equilibrium. So an easy way to see when a ppt will form is to set Q_c = K_sp. To see which ppt will form first, look at which K_sp will be reached first, in other words, compare to see which K_sp is smaller.

In this case, CaC₂O₄ has the smallest K_sp, so it will be reached first, and CaC₂O₄ will form ppt first. The second largest is CdC₂O₄, so this will ppt out second, and MgC₂O₄ last.

So this question is extremely easy! When asked from a list of compounds which will precipitate out first, or the order of precipitation, look at the K_sp values and go from smallest to largest.

A harder question would be one that asks when each of the ions fall out of solution. In other words, at what [C₂O₄^2-] does each metal fall out?

First look at each compound and build the K_sp equation. Solve for the unknown, [C₂O₄^2-].

Example:

For CaC₂O₄:

K_sp = [Ca²⁺][C₂O₄^2-]
2.3 x 10^-9 = (0.200)[C₂O₄^2-]
[C₂O₄^2-] = 1.15 x 10^-8 M
Note that the concentration for Ca²⁺ was given at the beginning of the problem.

For CdC₂O₄:

K_sp = [Cd²⁺][C₂O₄^2-]
1.5 x 10^-8 = (0.200)[C₂O₄^2-]
[C₂O₄^2-] = 7.5 x 10^-8

For MgC₂O₄:

K_sp = [Mg²⁺][C₂O₄^2-]
8.5 x 10^-5 = (0.200)[C₂O₄^2-]
[C₂O₄^2-] = 4.25 x 10^-4

Example:

In this same situation, what is the percent Ca²⁺ ion left when Cd²⁺ begins to ppt?

Remember that we figured out earlier than Cd²⁺ begins to ppt when [C₂O₄^2-] = 7.5 x 10^-8. To see how much Ca²⁺ has fallen out at this point, substitute this value for [C₂O₄^2-] into the following equation and solve for Ca²⁺:

K_sp = [Ca²⁺][C₂O₄^2-]
2.3 x 10^-9 = [Ca²⁺](7.5 x 10^-8)
[Ca²⁺] = 0.0307 M

To find the percent Ca²⁺ that has precipitated out at this point, divide this concentration by the original concentration of Ca²⁺ and multiply by 100:

(0.0307 M) / (0.200 M) x 100 = 15.3 %

Effect of pH on Solubility:

What happens to the solubility of a compound when a strong acid is added?

Consider a solution of the solids CaSO₄ and CaCO₃ (according to solubility rules, these are insoluble, so imagine them as solids resting at the bottom of a solution while a tiny percent of their ions escape into solution).

CaSO₄ (s) ⇌ Ca²⁺ (aq) + SO₄^2-(aq)
CaCO₃ (s) ⇌ Ca²⁺ (aq) + CO₃^2- (aq)

If a strong acid is added, H₃O⁺ ion is added to the solution. The anions will react with it:

SO₄^2- (aq) + H₃O⁺ (aq) ⇌ HSO₄^- (aq) + H₂O (l)

The small amount of free SO₄^2- ions will react with H₃O⁺. According to Le Chatelier's, this will drive more SO₄^2- ion out of the CaSO₄, but only by a very small margin. The other compound will be more effected by the change in [H₃O⁺]:

CO₃^2- (aq) + H₃O⁺ (aq) ⇌ HCO₃^- (aq) + H₂O (l)

HCO₃^- reacts with more of the H₃O⁺:

HCO₃^- (aq) + H₃O⁺ (aq) ⇌ H₂CO₃ (aq) + H₂O (l)

H₂CO₃ is unstable and reacts further:

H₂CO₃ (aq) → H₂O (l) + CO₂ (g)

If the container is open, the CO₂ gas will escape, and according to Le Chatelier's, the HCO₃^- will continue to replenish out of the CaCO₃. This compound will be more soluble in the presence of a strong acid.

Complex Ion Formation:

In Chapter 15 we learned the definition of a Lewis Acid/Base:

Lewis Acid: a species that can form a covalent bond by accepting an electron pair from another species.

Lewis Base: a species that can form a covalent bond by donating an electron pair to another species.

And in previous classes we learned that a coordinate covalent bond is a form of bonding between two atoms in which both electrons shared in the bond come from the same atom.

Complex Ion: an ion formed from a metal ion with a Lewis base attached to it by a coordinate covalent bond. A complex is a compound containing complex ions.

Example:

Ligand: a Lewis base that bonds to a metal ion to form a complex ion.

Example:

The overall equation for when aqueous silver ions form a complex ion with ammonia:

Ag⁺ (aq) + 2NH₃ (aq) ⇌ Ag(NH₃)₂⁺ (aq)



Ag⁺ is the Lewis acid. NH₃ is the Lewis base, and ligand.

The formation constant, or stability constant: K_f, is yet another form of K_c. The equilibrium constant for the formation of the complex ion from aqueous metal ion and the ligands.

For Ag(NH₃)₂⁺:

K_f = [Ag(NH₃)₂⁺] / [Ag⁺][NH₃]²

K_f values are usually very large. For example,

K_f for Ag(NH₃)₂⁺ = 1.7 x 10⁷

When doing calculations with K_f values, it would be harder to work with these values, because the approximation method will never hold. An easy way around this is to remember the mathematical rules that apply to K_c.

When a reaction is reversed so that the reactants are products and the products are reactants, the new K_c is the inverse (1 / K_c) of the original K_c.

Dissociation Constant: K_d, the inverse of K_f.

Example:

Reverse the reaction and take the new K_f, which will be K_d:

Ag(NH₃)₂⁺ (aq) ⇌ Ag⁺ (aq) + 2NH₃ (aq)

K_d = [Ag⁺][NH₃]² / [Ag(NH₃)₂⁺] =( 1 / K_f)

Example:

For the formation of the complex ion:

Cu²⁺ (aq) + 4NH₃ (aq) ⇌ Cu(NH₃)₄²⁺ (aq), K_f = 4.8 x 10¹²

What is the [Cu²⁺] at equilibrium when you start with 0.100 M Cu(NO₃)₂ and 1.00 M NH₃?

Note that the approximation method will not hold for these numbers, and we'll be left to work through some unnecessarily tedious math. Or, we could use some shortcuts and make it much easier:

First, assume that the reaction goes to completion:

Cu²⁺ (aq) + 4NH₃ → Cu(NH₃)₄²⁺ (aq)


Note that only 0.100 M is subtracted in the second line of the table because there is an excess of NH₃, and less Cu²⁺.

Now reverse the reaction to let some of the complex ion fall apart and convert K_f to K_d:

Cu(NH₃)₄²⁺ (aq) ⇌ Cu²⁺ (aq) + 4NH₃ (aq)



K_d = 1 / K_f = 2.08 x 10^-13
2.08 x 10^-13 = (x)(0.600 + 4x)⁴ / (0.100 – x)
Approximation method holds,
2.08 x 10^-13 = (x)(0.600)⁴ / (0.100)
x = 1.60 x 10^-13

Complex Ions and Solubility:

Example:

Calculate the molar solubility of AgCl in 1.0 M NH₃.

This problem involves the solubility equilibrium for AgCl and the complex ion equilibrium:

AgCl (s) ⇌ Ag⁺ (aq) + Cl^- (aq), K_sp = 1.8 x 10^-10
Ag⁺ (aq) + 2NH₃ (aq) ⇌ Ag(NH₃)₂⁺ (aq), K_f = 1.7 x 10⁷

The overall equation for the process can be obtained by applying Hess's Law and adding them together:

AgCl (s) + Ag⁺ (aq) + 2NH₃ (aq) ⇌ Ag⁺ (aq) + Cl^- (aq) + Ag(NH₃)₂⁺ (aq)

The Ag⁺ will cancel on each side and the equation will be:

AgCl (s) + 2NH₃ (aq) ⇌ Cl^- (aq) + Ag(NH₃)₂⁺ (aq)



And remember from Chapter 14 that when you add two reactions, you obtain the new equilibrium constant by multiplying their equilibrium constants.

K_c = K_sp x K_f =
(1.8 x 10^-10)(1.7 x 10⁷) = 0.00306

0.00306 = (x²) / (1.00 – 2x)² → approximation holds
0.00306 = (x²) / (1.00)²
x = 0.0553 M = [Ag(NH₃)₂⁺] = [AgCl]

14.4-14.9: Qualitatively Interpreting the Equilibrium Constant-Effect of a Catalyst

Qualitatively Interpreting the Equilibrium Constant:

-If K_c is a large number (example: 4.1x10⁸) the products are favored at equilibrium, and there will be more product than reactant. To obtain K_c, the reactants are in the denominator, and when you divide by a smaller and smaller number (as the reactants dwindle and the products emerge), the value of K_c will get bigger and bigger.

-If K_c is a very small number (example: 4.6x10^-31) the reactants are favored at equilibrium, and there will be more reactant than product. Again, to obtain K_c, the products are in the numerator, and when you divide a small number (products) by a larger number (reactants), the value of K_c will be very small.

-If K_c is within an order of magnitude near 1 (example: 7.2), there is an even distribution of reactants and products.

-If K_c = 0, there has been no reaction.

Predicting the Direction of Reaction:

Q_c = reaction quotient
-Obtained the same exact way used to obtain K_c. Plug in the concentrations of the products raised to the power of their coefficients, divided by the concentrations of the reactants raised to the power of their coefficients.



-The only difference is K_c is at equilibrium and Q_c is not. So we can use Q_c to see how to get to K_c.

-If Q_c > K_c, numerator is too big, need more reactant, reaction will shift to the left
-If Q_c < K_c, denominator is too big, need more product, reaction will shift to the right -If Q_c = K_c, already at equilibrium

Calculating Equilibrium Concentrations:

1. Set up a table of concentrations (Initial, Change, and Equilibrium expressions in x)
2. Substitute expressions in x for equilibrium concentrations into the equilibrium-constant equation.
3. Solve the equilibrium-constant equation for the values of the equilibrium concentrations.

Note: If you cannot get rid of a squared x term easily by simply taking the square root of everything (which will happen if your initial concentrations are the same), you will have to use the quadratic equation.

This will yield two answers, only one answer will be correct. If one is positive and the other is negative, the positive answer is the correct answer. If both are positive, plug the answers into your x expressions to see which one is realistically plausible (doesn't yield a negative number for any concentrations).

Changing Reaction Conditions:

-Le Chatelier's Principle: when a system in chemical equilibrium is disturbed by a change in temperature, pressure, or a concentration, the system shifts in equilibrium composition in a way that tends to counteract this change variable.

-Three ways to alter equilibrium concentration of a gaseous reaction mixture:
1. Changing concentrations
2. Changing the partial pressures of the products and reactants by changing the volume
3. Changing the temperature

-Concentration:
-When more reactant is added or product removed from the reaction vessel, the reaction will shift to the right to compensate for this gain/loss.
-When more product is added or reactant removed, the reaction will shift to the left to compensate. More reactants will be produced and amount of product will be subsequently decreased.

-Pressures/Volume:
-Pressure is directly proportional to moles of gas, so a decrease in the volume of a reaction vessel will cause the reaction to move in the direction that has a smaller number of moles of gas.
-An increase in volume will cause the reaction to move in the direction that has a larger number of moles of gas.
-When there are equal numbers of moles of gas on either side of the reaction (example: 2 total moles of reactant yields 2 total moles of product), then a change in pressure causes no shift in the reaction.

-Temperature:
-Must know the change in enthalpy of a reaction (ΔH) to know how temperature will shift the reaction.
-If ΔH is negative, the reaction is exothermic. Heat can be "imagined" to be a product of the reaction, and when you increase a product, the reaction will shift to the left to compensate. So adding heat to an exothermic reaction will cause it to shift to the left. And a temperature decrease will cause the reaction to shift to the right.
-If ΔH is positive, the reaction is endothermic. Heat can now be imagined to be a reactant. When you increase a reactant, the reaction will shift to the right to compensate. So adding heat to an endothermic reaction will cause it to shift to the right. And a temperature decrease will cause the reaction to shift to the left.

Effect of a Catalyst:

K_c is unaffected by a catalyst. The rate of reaction increases, but the final composition at equilibrium will be the same!

14.1-14.2: Chemical Equilibrium-The Equlibrium Constant

Chemical Equilibrium:

At dynamic equilibrium, Rate_f = Rate_r, the forward rate equals the reverse rate. Both sides are always reacting at a microscopic level at the same rate.

Chemical Equilibrium: the state reached by a reaction mixture when the rates of forward and reverse reactions have become equal.

In a Moles vs Time graph, notice the amounts of substances become constant at equilibrium:

In a Rate vs Time graph, notice that the forward rate is large at first but steadily decreases, while the reverse rate starts at zero and steadily increases, until both rates are equal at equilibrium:


Molar Composition:

Example:

For the reaction

CO (g) + 3H₂ (g) ⇌ CH₄ (g) + H₂O (g)

When 1.000 mol CO and 3.000 mol H₂ are placed in a 10.00 L vessel at constant temperature and allowed to come to equilibrium, the mixture is found to contain 0.387 mol H₂O. How many moles of each substance are present in the equilibrium mixture?

Think about the starting amounts in moles, which are known. The change in the amount of moles as the reaction proceeds is an unknown, so we can label it "x". Each reactant will decrease by a factor of x (the factor is indicated by the coefficient of that particular reactant) and each product will increase by a factor of x (also found by the coefficient). The resulting amounts can be found by combining our known information into equations and solving for x.

It is easiest to do this by making a table. I, C, and E indicate Initial, Change, and Equilibrium.


If the initial amounts of products isn't stated, it is assumed to be 0. Also, if we know each product (with a coefficient of 1) increases by x, and we know the resulting amount of that product, then we know x. So here, x = 0.387 mol.

If the problem were to ask for concentrations, divide the results by the volume given, in this case 10.00L.

The Equilibrium Constant:

General Form of an Equation:
aA + bB ⇌ cC + dD

Where capital letters are reactants and products and lowercase letters are coefficients of the balanced chemical equation.

-The equilibrium constant can be obtained by multiplying the concentrations of products, dividing by the product of the concentrations of reactants, and raising each concentration term to a power equal to the coefficient in the chemical equation. This is much easier to see when it's written out:



-K_c can also be denoted as K_eq
-K_c has no units
-Concentrations must be entered in mols/Liter (M), otherwise convert to these units
-Must have an equilibrium to find K_c
-Ignore concentrations of pure liquids and solids, they are constant. Another way of writing concentrations is density (same dimensions of mass over volume) which will remain the same no matter how much of a solid or liquid you have.
-Just use the concentrations of gaseous and aqueous reactants and products.

Example:

CO (g) + 3H₂ (g) ⇌ CH₄ (g) + H₂O (g)

K_c = [CH₄][H₂O] / [CO][H₂]³

Example:

N₂ (g) + 3H₂ (g) ⇌ 2NH₃ (g)

K_c = [NH₃]² / [NH₂][H₂]³

Example:

2A (g) + B(s) + 7C (aq) ⇌ 14D (l) + 3/2E (g)

K_c = [E]^3/2 / [A]²[C]⁷

-Manipulating the balanced chemical equation will change K_c:

-If you multiply the coefficients of a reaction through by a number, A, then K_c = K_c^A
-Suppose you multiply a reaction through by 2, then you must square K_c
-If you flip a reaction around so the reactants are now the products and the products are now the reactants, then you get the reciprocal, K_c = 1 / K_c

The Pressure Equilibrium Constant, K_p:

-Won't use this until later when we study Thermodynamics, but it is relevant now.
-K_p is another version of K_c
-The values plugged into K_p are partial pressures converted to atm instead of concentrations.
-Related to K_c via the ideal gas law (PV = nRT → n/V = P/RT)
-In other words, the molar concentration of a gas equals its partial pressure divided by RT, which is constant at a given temperature.


Example:

For the reaction
CH₄ (g) + 2O₂ (g) ⇌ CO₂ (g) + 2H₂O (g)

K_c = [CO₂][H₂O]² / [CH₄][O₂]²

K_p = P_CO2 · P_H2O² / P_CH4 · P_O2²

-Sometimes K_c = K_p, but not always, it depends on the coefficients of the chemical equation.

K_c = K_p when the sum of reactant coefficients equals the sum of the product coefficients.

Example:

For the same reaction
CH₄ (g) + 2O₂ (g) ⇌ CO₂ (g) + 2H₂O (g)

1 + 2 = 1 + 2

K_c = K_p

Equation to relate K_p to K_c:

K_p = K_c (RT)^Δn

R = the ideal gas constant (0.0821 atm.L/mol.K)
T = Temperature in degrees Kelvin
Δn = sum of the coefficients of the products – the sum of the coefficients of the reactants.

Note: when the sum of the reactant coefficients equals the sum of product coefficients, Δn will be zero. Anything raised to the zero power is 1, so RT^Δn will equal 1, and K_p will equal K_c times 1, or just K_c.

Equilibrium Constant for the Sum of Reactions:

-If a given chemical equation can be obtained by taking the sum of other equations, the equilibrium constant for the given equation equals the product of the equilibrium constants of the other equations.
-Note: use Hess's Law

Example:

Equation 1:
CO (g) + 3H₂ (g) ⇌ CH₄ (g) + H₂O (g)
Equation 2:
CH₄ (g) + 2H₂S (g) ⇌ CS₂ (g) + 4H₂ (g)
Hess's Law:
CO + 3H₂ + CH₄ + 2H₂S ⇌ CH₄ + H₂O + CS₂ + 4H₂
Equation 3:
CO (g) + 2H₂S (g) ⇌ H₂O (g) + CS₂ (g) + H₂ (g)

K_c for Equation 1 = K₁
K₁ = [CH₄][H₂O] / [CO][H₂]³

K_c for Equation 2 = K₂
K₂ = [CS₂][H₂]⁴ / [CH₄][H₂S]²

K_c for Equation 3 = K₁ · K₂
K₃ = [CS₂][H₂O][H₂] / [CO][H₂S]²
(K₃ can be verified by looking at Equation 3 and writing K_c directly from it)