Unit 8 - Acids and Bases

Question 1

Weak acid/base

Answer 1

Only partially ionized in solution. The majority still exists in the form of the original weak acid/base particles, not in the form of ions

*don’t split the ions up in the net ionic equation

Question 2

Bronsted Lowry model

Answer 2

An acid is a proton (H+ ion) donor and a base is a proton (H+ ion) acceptor

Question 3

Autoionization of water

Answer 3

Dissociation of H2O into H+ and OH-

Water autoionizes with an equilibrium constant Kw
Kw = [H3O+][OH-] = 1.0 x 10^-14 at 25 C
pKw = pH + pOH = 14.0 at 25 C
*both these values are temperature dependent

In pure water, [H3O+] = [OH-] = 1.0 x 10^-7 and pH = pOH = 7.0 → neutral solution

This process is endothermic because bonds are being broken which requires energy.

Question 4

What happens to the values of Kw, pKw, and pH of water as temperature increases?

Answer 4

Because the autoionization of water is an endothermic process, as temperature increases, equilibrium shifts right permanently. This increases the value of Kw (by favoring the products side), making it greater than 1.0 x 10-14 at higher temperatures. pKw is inversely proportional, so its value will decrease, making it smaller than 14.0 at higher temperatures.

The pH of pure water is 7.0 when the temperature is 25 C. At temperatures higher than 25 C, the pH is less than 7.0. At temperatures lower than 25 C, the pH is greater than 7.0. However, pure water is always neutral, with pH = pOH and [H3O+] = [OH-]

SUMMARY: Kw increases, pKw decreases, pH decreases

Question 5

Strong acid/base

Answer 5

Completely 100% ionizes in water. No original acid/base remains in solution.

The reaction goes to completion, they will never reach equilibrium with their conjugates - no tendency for reverse reaction to occur (use → instead of ⇌)

The initial concentration of the strong acid/base is equal to the final concentrations of the ions.

Question 6

Percent ionization

Answer 6

Percent ionization = [ions] / ( [ions] + [leftover acid/base]) = [ions] / [original acid/base]

EX: HF (aq) + H2O (l) → F+ (aq) + H3O+ (aq)
Percent ionization = [F+]/ ([leftover HF] + [F+])
Percent ionization = [F+]/ [original HF]

If there is no leftover acid/base and only ions in solution, there is 100% ionization

Question 7

How can you tell which acid is the stronger acid?

Answer 7

The larger the Ka value or the smaller the pKa value, the stronger the acid

The greater the percent ionization value, the stronger the acid

The lower the pH, the stronger the acid (higher pH means stronger base)

Fewer moles of strong acid/base are needed to the produce the same [H3O+] or [OH-] as a weak acid/base, which only partially ionize

Weak acids resist changes in pH more effectively than strong acids because so many molecules of weak acid are not dissociated in solution (that’s why weak acids make up a buffer solution)

The higher the electronegativity and the smaller the size of an anion (X) in an oxoacid, the lower the strength of the O-H bond (because O-X bond is stronger). Thus the stronger the acid (because H+ will more easily dissociate)

Question 8

Define a buffer

Answer 8

A solution that resists changes in pH when a small amount of strong acid or strong base is added

Consists of a weak acid and its conjugate base or a weak base and its conjugate acid

EX: 1 M HNO2 and 1 M NaNO2

Buffers are effective when the required pH of the solution is approximately equal to the pKa of the weak acid.

Question 9

How do you determine the pH of a buffer solution given the concentrations of the buffer components?

Answer 9

pH = pKa + log ([base]/[acid])

If [base] = [acid], so they are in the buffer in equal proportions, pH = pKa

If [base] > [acid], so there is a greater proportion of base, pH > pKa

If [base] < [acid], so there is a greater proportion of acid, pH < pKa

Diluting a buffer solution with distilled water does not change the pH because the concentration of each buffer component (weak acid and conjugate base) will decrease, keeping the ratio the same and therefore the pH of the solution as well

Question 10

Describe two ways to prepare a buffer solution.

Answer 10

1. Mix two solutions together, one that contains a weak acid (HA) and another that contains the conjugate base (A-).
2. Start with a solution that contains a weak acid (HA) and add a strong base until half of the weak acid has been neutralized. You don’t want to convert all the HA into A- because that would no longer be a buffer.

Question 11

Polyprotic acids

Answer 11

Polyprotic acids give up their first proton more easily than future protons. This is because after the first proton dissociates, there are a bunch of extra protons in solution → equilibrium shifts left to where there is no protons → reduces dissociation and weakens the acid

Concentration of each succeeding acid also decreases – [H2PO4 2-] > [HPO4-] > [PO4 3-]

The titration of a polyprotic acid has an equivalence point corresponding to the removal of each H+ ion (EX: H2SO4 has 2 equivalence points)

First dissociation: H2SO4 (aq) + OH- (aq) ⇌ HSO4 - (aq) + H2O (l)
After this equivalence point, the major species present in solution are [HSO4-] and [SO4 2-]. H2SO4 has been completely consumed.

Second dissociation: HSO4- (aq) + OH- (aq) ⇌ SO4 2- (aq) + H2O (l)
After this equivalence point, the major species present in solution is [SO4 2-]. HSO4- has been completely consumed.

Question 12

Equivalence point

EX: HA + NaOH ⇌ H2O + A-

Answer 12

Equivalence point = just enough titrant has been added to react with all the analyte

Initial # of moles for both the titrant (known acid/base) and analyte (unknown acid/base) are equal. The reaction goes to completion, meaning both are fully consumed until none is left

The major species present in solution (besides water) are Na+ and A- (HA and OH- have fully reacted)

This point is invisible, based on stoichiometric calculation

Question 13

Titration curve

Answer 13

Vertical line = equivalence point

Horizontal line where the curve levels out = buffer zone.

The center of the buffer zone line is the half equivalence point, where pH = pKa. You can find the pKa of the analyte acid by identifying the pH at this point – make sure it’s at half the volume of the equivalence point.

To the left of the center, [HA] > [A-] and pH < pKa. To the right of the center, [HA] < [A-] and pH > pKa

Question 14

Titrant

Answer 14

Standard solution of known concentration used in titration

Question 15

Analyte

Answer 15

Solution of unknown concentration used in titration

Question 16

Buffer capacity

Answer 16

The amount of H+ or OH- a buffer can absorb without a significant change in pH

Larger concentrations of components → absorb more H+ or OH- ions → greater capacity (but keeps pH of solution the same)

When a buffer has more conjugate acid than base, it has a greater capacity for addition of base. When a buffer has more conjugate base than acid, it has a greater capacity for addition of acid.

Question 17

End point

Answer 17

End point = when the pH of the equivalence point is reached, the indicator changes color

Visible, based on observation

Ideally, equivalence point and end point are the same

Question 18

Indicator

Answer 18

Indicators are weak acids that change colors in certain pH ranges (the point when pKa of indicator = pH of solution)

Find an indicator whose pKa matches the calculated pH at the equivalence point of the titration to find the most accurate endpoint

Will change color right around the point that the two species (HIn and In-) are present in equal amounts → values will cancel out in Ka expression → Ka = [H+]

Question 19

Half equivalence point

Answer 19

On the titration curve, this point is at the center of the buffer region which is where the curve levels out

Halfway to equivalence point, [HA] and [A-] are equal, so they cancel out and leave us with Ka = [H+] which is represented by pKa = pH

Question 20

Explain the effect of changing the concentration of the titrant (known acid/base)

Answer 20

Higher concentration of known base = final pH of solution will be slightly higher where OH- dictates pH (after equivalence point)

Higher concentration of known base = less volume needed to reach equivalence point (double the concentration = half the volume)

Question 21

How does a titration curve show different equivalence points?

Answer 21

1. Strong acid + strong base

pH of 7 at equivalence point (H2O is the only major species)
pH changes slowly initially, then rapidly passes through pH = 7, then changes slowly again

2. Weak acid + strong base

pH > 7 at equivalence point. The weaker the acid, the larger the pH. (The conjugate base is a major species present and reacts with water to product OH-)

For weak acids, the initial pH of the titration curve is higher than strong acids

3. Weak base + strong acid

pH < 7 at equivalence point. The weaker the base, the smaller the pH. (The conjugate acid is a major species present and reacts with water to product H3O+)

For weak bases, the initial pH of the titration curve is lower than strong bases

Question 22

Why can’t you form a buffer with a strong acid and its conjugate base?

Answer 22

Because the base will be very weak, essentially neutral, and won’t readily accept H+ ions. In a buffer, you need both the acid and base components

Question 23

How do you determine the relative strengths of acids or bases given an equation and the value of K?

Answer 23

Strong acid + strong base → weak acid + weak base

If K > 1, that means the strong acid and base are on the left side because the forward reaction is favored.

If K < 1, that means the strong acid and base on the right side because the reverse reaction is favored.