Study Guide: Big Idea 6: Any bond or intermolecular attraction that can be formed can be broken These

two processes are in a dynamic competition, sensitive to initial conditions and external perturbations.

6a: Chemical equilibrium is a dynamic, reversible state in which rates of opposing processes are equal.

● Chemical Equilibrium: Systems that have reached the point where the rates of the forward and reverse reactions are constant and equal.

● It is a dynamic process where reactants continuously form products and vice-versa but the net amount of each is stable and remains the same throughout.

● K is the constant that represents equilibrium

● ● K relates to all the constants and defines it a certain temperature ● Q: Describes relative amounts of products to reactants at any certain time ● K and Q only includes substances in aqueous solutions not as solids or gases

● ● Kinetics & Equilibrium

6b: Systems at equilibrium are responsive to external perturbations with the response leading to a change in the composition of the system.

Chemical equilibrium = dynamic state in which the rates of the forward and reverse reactions are equal Stress on the System = Change in conditions (ie temp or volume changed, change in chemical species), causes the rate of the forward and reverse reactions to fall out of balance

- Shift of the chemical reaction- stress on the system causes it to fall temporarily out of equilibrium

Le Chatelier's Principle- “Restoring Balance” When a change is imposed on a system that is at equilibrium, the equilibrium will shift to adjust to the change

- Used in predicting response of a system from stresses

6c:

Chemical and equilibrium plays an important role in acid-base chemistry and in solubility. Acid/Base Particulates:

Strong: Ka = >>1 at equilibrium strong acids are molecules that essentially ionize to[HA]

[H 0 ] [A ]3+ −

completion in aqueous solution, dissociating into H O ions and the additional anion. 3 + Weak: Ka = << 1 at equilibrium weak acids are molecules that partially ionize in[HA]

[H 0 ] [A ]2+ −

aqueous solutions, dissociating into few H O ions and the additional anion. 3 +

pH of weak or strong acids: Strong Acids: Strong Bases

● At the equivalence point of (OH- = H+) the pH=7 Strong Acid: Weak Base

● At the equivalence point, pH < 7 Weak Acid: Strong Base

● At the equivalence point, pH > 7

Weak Acid: Weak Base ● At the equivalence point, pH depends on the relative value of Ka and the Kb of the acid

and base. pH is a measure of the [H+] in solution. More moles of a weak acid are needed to achieve equivalent [H+] values of a strong acid of the same pH, since a weak acid only partially ionizes. If similar volumes of acids were titrated with the same strong base, the weak acid would require a larger volume of base to reach its equivalence point.

(Left) Shows titration curve of a polyprotic weak acid with a strong base. (Right) shows the titration curve of a weak base with a strong acid with indicator changes.

Shows the titration curve of a strong acid with a strong base. K : Depends on Temperature: w

Pure water is always neutral [H ]=[OH ]. This means that the pH value that is neutral ([H + − +

]=[OH ]) changes with temperature. − Most of the time the value for K is 1 10 w × −14 Calculating the initial pH:

Titration of 25 mL of 0.3 M HF with 0.3 M NaOH. The k value is 6.6×10−4 a HF+H2O⇌H3O++F−

ICE table is required to find the pH:

6.6 10 =× −4 X 2

0.3−X Manipulate the equation to get every thing on one side yields:

0=X +6.6 10 X -1.98 10 2 × −4 × −4 Now plug into quadratic formula:

X= 2−6.6×10 ±−4 √(6.6×10 ) − 4 (1) (−1.98×10 )−4 −4

Quadratic formula yields that X=0.013745 and X=-0.014405

Since there can not be a negative concentration, the pH= log(0.013745)

pH=1.86

Solving equilibrium problems involving weak acids:

Example: Consider the process by which we would calculate the H 3O+, OAc-, and HOAc concentrations at equilibrium in an 0.10 M solution of acetic acid in water.

What we know about the reaction:

We then compare the initial reaction quotient (Qa) with the equilibrium constant (Ka) for the reaction and reach the obvious conclusion that the reaction must shift to the right to reach equilibrium.

Recognizing that we get one H3O+ ion and one OAc- ion each time an HOAc molecule dissociates allows us to write equations for the equilibrium concentrations of the three components of the reaction.

Substituting what we know about the system at equilibrium into the Ka expression gives the following equation.

Use this approximate equation to solve for ΔC

ΔC can be ignored in this problem because it is less than 5% of the initial concentration of the acidic acid.

Can then use the value of ΔC to calculate the equilibrium concentrations of H3O+, OAc-, and HOAc.

Finally substitute these concentrations into the expression for Ka.

Acid/Base Reaction Species:

Is it a strong acid? If yes, it will completely dissociate in water.

Since it is a strong acid it dissociates completely, you will not have any HCl. Cl1- is not going to do anything in an acid base reaction. It is a spectator ion so leave it out. What you need is to figure out the [H3O1+]. Remember concentration is moles over liters.

Is it a strong base? If yes it will completely dissociate in water.

Since it is a strong base it dissociates completely. You will not have any NaOH. Na1+ is not going to do anything in an acid base reaction. It is a spectator ion so leave it out. What you need to do is figure out the [OH1-]

Is it a weak acid? If yes it will partially dissociate in water. Now you need to pay attention to the equilibrium reactions. First write out the balanced equilibrium reaction.

then write equilibrium expression:

You will need to use this expression to determine what concentration each species is at when the system is at equilibrium. Exactly how to solve this depends upon what other species are present.

Is it a salt? The salt will dissociate into ions.

After you write the salt out as ions, look at the ions to see if you recognize any of them as a weak acid or a weak base. Or the conjugate acid or base. Anything with a Ka or Kb. After you recognize it as an acid or base, write the appropriate reaction.

Buffer:

If you only have a weak acid. Determine the concentration of the acid (assuming that there is no dissociation). Then look up or determine Ka

If you have a weak acid AND the conjugate base. Solve for the buffer. Determine the concentration of the weak acid and the conjugate base (Ignoring any equilibrium effects at first ). Then look up or determine Ka.

If you only have the conjugate base. Solve for the pH of the base using Kb and the hydrolysis equation. Determine the concentration of the conjugate base. Then determine Kb (using Ka if necessary)

How to build a Buffer:

Getting the pH correct:

The pH of a buffer is primarily determined by the pKa of the weak acid in the conjugate acid-base pair.

When both species in the conjugate acid-base pair have equal concentrations, the pH of the buffer is equal to the pKa.

Choose a conjugate acid-base pair that has a pKa closest to the pH you desire and then adjust concentrations to fine tune from there.

Estimating Buffer Capacity:

A buffer is only effective as long as it has sufficient amounts of both members of the conjugate acid-base pair to allow equilibrium to shift during a stress.

Finding the major species:

pKa bears exactly the same relationship to Ka as pH does to the hydrogen ion concentration:

If you use your calculator on all the Ka values in the table above and convert them into pKa values, you get:

Notice that the weaker the acid, the larger the value of pKa. It is now easy to see the trend towards weaker acids as you go down the table.

The Buffer Mechanism:

A buffer is able to resist pH change because the two components (conjugate acid and conjugate base) are both present in appreciable amounts at equilibrium and are able to neutralize small amounts of other acids and bases (in the form of H3O+ and OH-) when they are added to the solution. Take, for example, a fluoride buffer made from hydrofluoric acid and NaF. A model fluoride buffer would contain equimolar concentrations of HF and NaF. Since they are a weak acid and a weak base, respectively, the amount of hydrolysis is minimal and both buffer species are present at, effectively, their initial supplied concentrations.

If a strong acid is added to the HF/F- buffer, then the the added acid will react completely with the available base, F-. This results in a nearly unchanged [H3O+] and a nearly unchanged pH.

If a strong base is added to the HF/F- buffer, then the added base will react completely with the available acid, HF. This results in a nearly unchanged [H3O+] and a nearly unchanged pH.

The slight shift in pH after challenge is governed by the hydrolysis equilibrium of HF, based on the new HF and F- concentrations:

K and Solubility Calculations: sp

The solubility product constant, , is the equilibrium constant for a solid substance dissolvingKsp in an aqueous solution. It represents the level at which a solute dissolves in solution. The more soluble a substance is, the higher the value it has.Ksp

General Dissolution reaction:

aA(s) ⇌ cC(aq) + dD(aq)

To solve for the it is necessary to take the molarities or concentrations of the products (cC K sp and dD) and multiply them. If there are coefficients in front of any of the products, it is necessary to raise the product to that coefficient power (and also multiply the concentration by that coefficient).

Ksp=[C]c[D]d

Note that the reactant, aA, is not included in the K equation. Solids are not included when sp calculating equilibrium constant expressions, because their concentrations do not change the expression; any change in their concentrations are insignificant, and therefore omitted. Hence,

represents the maximum extent that a solid that can dissolved in solution. K sp

Find a from solubility data:Ksp

The solubility product constant ( ) describes the equilibrium between a solid and itsKsp constituent ions in a solution. The value of the constant identifies the degree to which the compound can dissociate in water. For example, the higher the , the more soluble theKsp compound is. is defined in terms of activity rather than concentration because it is aKsp measure of a concentration that depends on certain conditions such as temperature, pressure, and composition. It is influenced by surroundings. is used to describe the saturated solution ofKsp ionic compounds.

(s) ⇌ (aq)+ (aq)BaCO3 Ba2+ CO32−

First, write down the equilibrium constant expression:

The activity of solid BaCO3 is 1, and considering that the concentrations of these ions are small, the activities of the ions are approximated to their molar concentrations. is therefore equal Ksp to the product of the ion concentrations:

Common ion effect:

The solubility of a sparingly soluble hydroxide can be greatly increased by the addition of an acid. For example, the hydroxide salt is relatively insoluble in water:Mg(OH)2

(s) ⇌ (aq) + 2OH−(aq)Mg(OH)2 Mg2+ With Ksp=5.61×10−12

When acid is added to a saturated solution that contains excess solid , the followingMg(OH)2 reaction occurs, removing OH− from solution:

H+(aq) + OH−(aq)→H2O(l)

The overall equation for the reaction of Mg(OH)2 with acid is:

Mg(OH)2(s) + 2H+(aq) ⇌ (aq) + 2H2O(l)Mg2+ (18.7.7)

As more acid is added to a suspension of , the equilibrium is driven to the right, soMg(OH)2 more dissolves.Mg(OH)2

In contrast, the solubility of a sparingly soluble salt may be decreased greatly by the addition of a common ion. For example, if is added to a saturated solution, additionalMgCl2 Mg(OH)2

will precipitate out. The additional ions will shift the original equilibrium toMg(OH)2 Mg2+ the left, thus reducing the solubility of the magnesium hydroxide.

Salt dissolution: ΔH and ΔS:

6d: The equilibrium constant is related to temperature and the difference in Gibbs free energy between reactants and products. Gibbs Equation: a thermodynamic equation used for calculating changes in the Gibbs energy of a system as a function of temperature Expressing the equilibrium constant in terms of delta G and RT

● ΔG° is the difference between the reactants and products.

● RT is the thermal energy within the reaction ● R = 8.314 J mol-1 K-1 or 0.008314 kJ mol-1 K-1. ● T is the temperature on the Kelvin scale.

Thermodynamic Favorability: A thermodynamically favorable reaction is one in which the energy state of reactants is higher than that of the products and will proceed spontaneously(without the need for added energy).

● Thermodynamically favored: reaction (ΔG < 0) is sometimes referred to as "spontaneous" In reactions in which enthalpy is favorable and entropy is unfavorable, the reaction becomes less

spontaneous ( G increases) until eventually the reaction is not spontaneous (when G >

0). As the magnitude of G changes, so does the equilibrium constant. K.