Chapter 18 - Aqueous Ionic Equilibrium

Buffer definition

a solution that resists pH changes when an acid or base is added

What is a buffer solution comprised of?

significant amounts of a weak acid and its conjugate base or significant amounts of a weak base and its conjugate acid. strong acids and bases never form a buffer solution.

What happens if you add an acid to a buffer solution?

the conjugate base neutralizes the added acid

What happens if you add a base to a buffer solution?

the weak acid neutralizes the added base

Henderson-Hasselbalch equation(s)

pH = pKa + log[conjugate base]/[weak acid]; pOH = pKb + log[conjugate acid]/[weak base] (only works if “x is small approximation“ is valid!), used to find the pH of a buffer system or to find the ratio of A to HA to make a buffer with a desired pH

Titration

technique used to determine the concentration of an unknown solution (analyte) by reacting it with a solution of known concentration (titrant)

equivalence point

[H3O+] = [OH-], n_acid = n_base (based on stoichiometry), pH = 7 (neutral salts), pH = >7 (basic salts), pH = <7 (acidic salts)

end point

when the indicator changes color, approximation of the eq. point

Titration of strong acid with strong base

1) initial pH is the pH of the strong acid solution

2) pH changes depending on the amount of added base (pH rises)

3) pH at equivalence point = 7

4) pH after equivalence point > 7, solution becomes basic

(when you swap acid and base, everything is the same but opposite)

Titration of weak acid with a strong base

1) find initial pH based on ICE table

2) after adding base you create a buffer system, find pH from Henderson-Hasselbalch equation

3) half volume of the equivalence point means pH=pKa

4) at equivalence point, you have a basic salt, find pH based on Kb of conjugate base (Ka=Kw/Kb)

5) after equivalence point, you have only strong base, find concentration of OH-, pOH, and pH

S

molar solubility

solubility product constant

Ksp = [aS]^a[bS]^b….

molar solubility

represents concentration of a solute in a saturated solution. max # of moles of a solute that can be dissolved in 1 L solution. the higher the molar solubility, the more soluble the salt.

comparing the solubility of two insoluble salts

if they both have the same # of ions, then compare the Ksp directly. larger Ksp = more soluble, smaller Ksp = less soluble. if they do not have the same # of ions, then molar solubility must be compared.

Q

defined the same as Ksp, but no coefficient inside the brackets

Q > Ksp

solution is very saturated, precipitate forms

Q = Ksp

solution is saturated and no additional salt can be dissolved

Q < Ksp

solution is not saturated, precipitate does not form (more insoluble salt can still be dissolved)

Buffer Capacity

The amount of acid or base a buffer can neutralize

Buffer range

The pH range in which the buffer can be effective

The effectiveness of a buffer depends on two factors:

1) the relative amounts of buffer acid and base; 2) the absolute concentrations of buffer acid and base

When is a buffer most effective?

When the ratio of [acid]:[base] = 1, and when the [acid] and [base] are large

Generally a buffer will be effective when:

0.1 < [base]:[acid] < 10

The effective pH range of a buffer is:

pK_a ± 1 (log(0.1), log(10))

When choosing an acid to make a buffer, choose one whose ___ is closest to the __ of the buffer.

pK_a, pH

As the [base]:[acid] ratio approaches 1,

the ability of the buffer to neutralize both acid and base improves.

Buffers that need to work mainly with added acid generally have

[base] > [acid]

Buffers that need to work mainly with added base generally have

[acid] > [base]

The buffer capacity increases with

increasing absolute concentration of the buffer components. (A concentrated buffer can neutralize more added acid or base than a dilute buffer.)

Indicator

Chemical that changes color when the pH changes

Titration curve

Plot of pH versus amount of added titrant (volume)

Inflection point

Equivalence point

Prior to the equivalence point

The original solution in the flask is in excess, so the pH will be closest to the original solution’s pH

Beyond the equivalence point

The solution in the burette is in excess, so the pH approaches the burette solution’s pH