Ch.16 Flashcards

Question 1

Q

Buffers

Answer

A

Recall that the equilibrium concentrations of H3O+ and OHin pure water (or a solution of a
neutral salt) are very small (1.0 x 10-7 M)
Therefore, it takes very little strong acid or strong base to dramatically affect the [H3O+] and [OH-]
If one wishes to produce a neutral or near-neutral aqueous solution with a stable pH, something
must be present that can neutralize any added H3O+ or OH-
In other words, the solution must contain an acid (to react with OH-) and a base (to react with H+)
Strong acids and bases are obviously unsuitable, as they would immediately react with each other

Question 2

Q

Buffers continued

Answer

A

A mixture of a weak acid and a weak base that do not immediately react with each other is possible if the Ka and the Kb are properly matched
The simplest and easiest way to achieve this is to mix a weak acid (HA) with its own conjugate base (A-)
This mixture is called a buffer
Buffer solutions can react with both acids and bases, minimizing the effect that added H3O+ or OH- has on the pH

– Weak acid neutralizes any added base

– Conjugate base neutralizes any added acid

Question 3

Q

Example – Acetic acid – Acetate Buffer

CH3COO- + H3O+ ⇌ CH3COOH + H2O

CH3COOH + OH- ⇌ CH3COO- + H2O

Answer

A

If you add acid:
Additional H3O+ will react with CH3COOto form CH3COOH
[CH3COO-] decreases, [CH3COOH] increases
The overall [H3O+] will increase only slightly
If you add base:
Additional OH- will react with CH3COOH to form CH3COO-
[CH3COOH] decreases, [CH3COO-] increases
The overall [H3O+] will decreases only slightly

Question 4

Q

Buffer pH

Answer

A

CH3COOH + H2O ⇌ CH3COO- + H3O+

Ka =[CH3COO-][H3O+] / [CH3OOH]
or
[H3O+] = Ka x [CH3OOH] / [CH3COO-] (buffer ratio)

The pH of a buffer solution depends on the Ka and the HA to A ratio
Ka depends on the acid chosen, and so long as there is a substantial amount of both HA and A-, added acid or base will have only a small effect on the ratio
– We will have an effective buffer solution
If the HA or A- concentration becomes small, converting any HA into A- or A- into HA will affect the ratio significantly

– The buffer will be much less effective at resisting changes in pH

Question 5

Q

Henderson-Hasselbalch Equation

Answer

A

Rearranging the Ka equation and taking the log of both sides results in something very useful for calculations involving buffer solutions, the Henderson-Hasselbalch equation

pH = pKa + log ([A-]/[HA]) or pH = pKa + log ([base]/[acid])

The Henderson-Hasselbalch equation can be used to:

– calculate concentration of a buffer
– calculate the pH of a buffer after addition of acid or base

Major Limitation:
[H3O+] must be small relative to [HA] (acid dissociation must be minimal)

– The H-H equation is making use of the “x is small” approximation, and will fail when that approximation is not valid

Question 6

Q

Base/Conjugate acid buffers

Answer

A

We can create a buffer system based on a weak base and its conjugate acid

– for example: NH3/NH4Cl

Will act the same as an acid/conjugate base buffer
– Additional H3O+ will be neutralized by the weak base (NH3)
– Additional OH- will be neutralized by the conjugate acid (NH4+)

– In order to use the Henderson-Hasselbalch equation, we must convert the Kb of the base into a Ka

pKa + pKb= 14

Question 7

Q

Buffer Range and Capacity

Answer

A

Buffers do not have an unlimited ability to neutralize additions of acid or base
Two characteristics of buffers can be defined:
Buffer capacity

– The amount of acid or base that can be added without dramatically changing the pH

– Units of mol/L (moles of acid or base per liter of buffer)

– A “significant change” is usually taken to mean +/- 1 pH unit

Buffer range

– The pH range over which the buffer capacity is significant

Buffers are most effective when the concentrations of acid and conjugate base are equal
As the concentrations begin to differ the buffer’s ability to resist pH change decreases

– Converting HA to A- or vise-versa starts to have a large effect on the A-/HA ratio

Buffers are also more effective when their concentrations are high

– More HA and more Ato react

Question 8

Q

Buffer Capacity

Answer

A

Buffers are most effective when
their concentrations are high
As the [HA] + [A-] increases, the
buffering capacity increases
because there is more HA and Ato
react with the added acid or base

Question 9

Q

Buffer Range

Answer

A

Buffers are most effective
when the concentrations of
acid and conjugate base are
equal
If either the HA or Aconcentration is small,mconverting HA to A- or vise versa will have a larger effect
on the A-/HA ratio
The buffer range is the range of pH over which the buffer is effective
Both A- and HA need to be present in substantial amounts for effective buffering, so the [A-]/[HA] ratio needs to be close to 1
If [A-]/[HA] is greater than 10 or less than 0.1, the buffering ability is poor
The buffering capacity is exhausted

pH = pKa + log (1/10) = pKa - 1
pH = pKa + log (10/1) = pKa + 1

The most useable pH range for a buffer is ±1 pH unit of the pKa of the acid

Question 10

Q

Preparing a Buffer

Answer

A

Suppose you needed to make a buffer with a pH of 4.00. What would you do?

Choose the appropriate acid
* pKa of acid must be within ± 1 of the desired pH

Closer is better - Choices include:

Benzoic acid: Ka = 6.31x10-5, pKa = 4.20

Formic acid: Ka = 1.77x10-4, pKa = 3.75
Lactic acid: Ka = 1.38x10-4, pKa = 3.86

Lactic acid is the best choice

Calculate the acid to conjugate base ratio
* Use Henderson-Hasselbalch equation
* pH = pKa + log ([A-]/[HA]), 4.00 = 3.86 + log ([A-]/[HA]), log ([A-]/[HA]) = 0.14, ([A-]/[HA]) = 1.3804
Decide on a buffer concentration

Concentration of a buffer is [HA] + [A-], typically in the range of 0.1 M to 1.0 M
Let’s assume 1.0 M: [HA] + [A-] = 1.0 M

Calculate amounts of chemicals needed

[HA] + [A-] = 1.0 M and [A-]/[HA] = 1.3804, so [HA] = 1.0 - [A-] and [A-] / 1.0 - [A-] = 1.3804
2.3804[A-] = 1.3804, [A-] = 0.58 M, [HA] = 0.42 M
Verify: pH = pKa + log([A-]/[HA]) = 3.86 + log (0.58/0.42) = 4.00

Question 11

Q

Preparing a Buffer – Alternate method

Answer

A

We can also create a buffer by partial neutralizing a solution of weak acid or weak base with strong base or strong acid, respectively

Choose the acid or base to make pKa ≈pH
Calculate the amount of HA (or B) needed to produce the desired volume and concentration of buffer
Use the Henderson-Hasselbalch equation to determine how much HA must be converted into A- (or how much B must be converted into HB+)
For every mole of A- needed, add one mole of strong base (or one mole of strong acid for every mole of HB+ needed)

Question 12

Q

Titrations and pH curves

Answer

A

In Chem 101, you briefly looked at acid-base titration reactions (neutralization)
The pH of the solution is monitored electronically or with a pH indicator until the solution
has been neutralized
The equivalence point is reached when the moles of base in solution is stoichiometrically equivalent to the number of moles of acid (moles OH-=moles H3O+)
Plot of the pH of solution during titration is called a titration curve or pH curve
The appearance of the curve depends on the nature of the titration

– Starting with acid or starting with base? Strong or weak?

Question 13

Q

Titrating a Strong Acid with a Strong Base

Answer

A

Initially:
[H3O+] = [HA]
Strong acid dissociates quantitatively
After addition of some base, but before equivalence:
H3O+ + OH- → 2 H2O
Amount of H3O+ = (initial amount of H3O+) – (amount of OH- added)
Addition of titrant has increased the total volume (remaining H3O+ has been diluted)
pH is becoming less acidic
At the equivalence point:
Amount of H3O+ = amount of OH-
[H3O+] = [OH-] (pH is neutral)
Conjugate base and conjugate acid are spectators
Past the equivalence point:
[OH-] > [H3O+] (pH has become > 7)
Amount of OH- = (amount of OH- added) – (initial amount of HA)
Volume increase continues

Question 14

Q

Titrating a Weak Acid with a Strong Base

Answer

A

Initially:
Ka = [H3O+][A-] / [HA]. [H3O]≈ √(Ka x [HA]) if [HA] / Ka > ~1000
Calculate pH of weak acid solution using [HA] and Ka
Use “x is small” approximation or solve quadratic equation as appropriate
After addition of some base, but before equivalence:
HA + OH- → A- + H2O
Amount of HA = (initial amount of HA) – (amount of OH- added)
Amount of A- = amount of OH- added
Mixture of HA and A- → you have a buffer
Use Henderson-Hasselbalch equation to find pH
The mole ratio is the same as the concentration ratio
The increase in total volume can be ignored
At the equivalence point:
Exactly enough OH- has been added to convert all HA to A-
You have a solution of a weak base (A-, the conjugate base of HA)
Use Kb to find [OH-] and subsequently pH
Kb = Kw/Ka, Kb = [OH-][HA] /[A-],[OH−] ≈ √(Kb x [A-]) if [A-] / Kb > ~1000
pH is basic
Do not forget to account for dilution
Use “x is small” approximation or solve quadratic equation as appropriate
Past the equivalence point:
pH continues to increase
Amount of OH- = (amount of OH- added) – (initial amount of HA)
Volume increase continues (dilution)
Identical to strong acid – strong base titration

Question 15

Q

What to do if the “x is small” approximation is invalid

Answer

A

At the beginning:
Set up an ICE for the reaction HA + H2O ⇌ H3O+ + A-
Ka = [H3O+][A-] / [HA]
You will get a quadratic polynomial, solve for x using the quadratic equation
After addition of a small volume of titrant:
Amount of A- = (amount of OH- added) + (A- due to acid dissociation)
Set up an ICE for the reaction HA + H2O ⇌ H3O+ + A- (You are still using the Ka equation)
When determining initial conditions, you need to account for both dilution and reaction of HA with OH to form A-
Initial [A-] =moles of OH− added/total volume and initial [HA] = original moles of HA − moles of OH− added/total volume
Solve resulting quadratic polynomial for x

Question 16

Q

What to do if the “x is small” approximation is invalid

Just before the equivalence point:

Answer

A

Amount of HA = (amount of excess HA) + (HA due to hydrolysis of A-)

Set up an ICE for the reaction A- + H2O ⇌ OH- + HA (You are now using the Kb equation)
When determining initial conditions, you need to account for both dilution and reaction of HA with OH to form A-
Initial [A-] =moles of OH− added/total volume and initial [HA] = original moles of HA − moles of OH− added/total volume
Solve resulting quadratic polynomial for x

Question 17

Q

What to do if the “x is small” approximation is invalid

At the equivalence point:

Answer

A

Set up an ICE for the reaction A- + H2O ⇌ OH- + HA
Kb = Kw/Ka, kb = [OH-][HA] / [A-]
You will get a quadratic polynomial, solve for x using the quadratic equation

Question 18

Q

What to do if the “x is small” approximation is invalid

Just after the equivalence point:

Answer

A

All HA has been converted to A-, slight excess of OH-, some hydrolysis may still occur
Set up an ICE for the reaction A- + H2O ⇌ OH- + HA (You are still using the Kb equation)
When determining initial conditions, you need to account for both dilution and reaction of HA with OHto form A-
Initial [A-] = original moles of HA /total volume and initial [OH] = moles of OH− original moles of HA/ total volume
Solve resulting quadratic polynomial for x

Question 19

Q

Titrating a Strong Base with a Strong Acid

Answer

A

Initially:
[OH-] = [B], [H3O+] = Kw / [OH-]
Strong base dissociates quantitatively
After addition of some acid, but before equivalence:
H3O+ + OH- → 2 H2O
Amount of OH- = (initial amount of OH-) – (amount of H3O+ added)
Addition of titrant has increased the total volume (remaining OH- has been diluted)
pH is becoming less basic
At the equivalence point:
Amount of H3O+ = amount of OH-
[H3O+] = [OH-] (pH is neutral)
Conjugate base and conjugate acid are spectators
Past the equivalence point:
[OH-] < [H3O+] (pH has become < 7)
Amount of H3O+ = (amount of H3O+ added) – (initial amount of B)
Volume increase continues

Question 20

Q

Titrating a Weak Base with a Strong Acid

Answer

A

Initially:

Kb = [OH−][HB+] / [B], [OH−] ≈ √(Kb x [B]) if [B] / Kb > ~1000
Calculate pH of weak base solution using [B] and Kb
Use “x is small” approximation or solve quadratic equation as appropriate

After addition of some acid, but before equivalence:

B + H3O+ → HB+ + H2O
Amount of B = (initial amount of B) – (amount of H3O+ added)
Amount of HB+ = amount of H3O+ added
Mixture of B and HB+ → you have a buffer
Use Henderson-Hasselbalch equation to find pH
The mole ratio is the same as the concentration ratio
The increase in total volume can be ignored

At the equivalence point:

Exactly enough H3O+ has been added to convert all B to HB+
You have a solution of a weak acid (HB+), the conjugate acid of B)
Use Ka to find [H3O+] and subsequently pH
Ka = Kw/Kb, Ka = [H3O+][A−]/[HA],[H3O+] ≈ √(Ka x [HB+]) if [HB+] / Ka > ~1000
pH is acidic
Do not forget to account for dilution
Use “x is small” approximation or solve quadratic equation as appropriate

Past the equivalence point:

pH continues to drop
Amount of H3O+ = (amount of H3O+ added) – (initial amount of B)
Volume increase continues (dilution)
Identical to strong base – strong acid titration

Question 21

Q

Titration of Polyprotic Acid with Strong Base

Answer

A

If Ka1 and Ka2 are sufficiently different the pH curve will have two equivalence points
If Ka1 and Ka2 are quite similar (pKa within 2), the first equivalence point will be indistinct, as the buffer regions for the first and second ionization will overlap

– You will not need to calculate pH for such buffers, but you should be able to recognize that they will have an extended buffer range

– First equivalence point does not become easily visible until DpKa ~4

The volume required to reach the first equivalence point is identical to the volume required to reach the second one

– Amount of HA- at the first equivalence point is the same as the amount of H2A at the beginning

Question 22

Q

Indicators

Answer

A

Indicators are weak acids or bases
Protonation or deprotonation results in a color change

– HA is a different color than A

HIn(aq)_{colour 1} + H2O(l) ⇌ H3O+ (aq) + In- (aq)_{coulour 2}

Ka= [H3O+][In-]/[Hln]

[Hln]/[In-] = [H3O+]/ka

We see the HIn color if [HIn]/[In-] >10
We see the In- color if [HIn]/[In-]<0.1
In between, we see both colors

Question 23

Q

Indicators in Titrations

Answer

A

Indicators are used to provide a visual cue that the equivalence point has been reached in
an acid-base titration

– The visible color change is called the end point

The indicator should thus change color very near to the equivalence point
For strong acid – strong base titrations, the pH change through the equivalence point is very large, and a variety of indicators are suitable

– Indicator pKa of ~4 to ~10 are all equally good

For weak acid – strong base titrations, indicator pKa should be ~8 to ~10

– Phenolphthalein is a common choice, pKa = 9.3

For weak base – strong acid titrations, indicator pKa should be ~4 to ~6

Question 24

Q

Solubility Equilibria

Answer

A

In Chem 101, you considered solubility qualitatively
Materials were classified as soluble or insoluble, and the outcome of mixing ionic solutions was predicted by which combination of ions formed an insoluble salt
In reality, there is a continuum between highly soluble and highly insoluble
We will now consider precipitation reactions as equilibria
Equilibrium constant is designated Ksp
Consider the dissolution of calcium phosphate:
Ca3 (PO4)2 (s) → 3 Ca2+ (aq) + 2 PO43-(aq)
As dissolution proceeds, the concentrations of PO4 3-(aq) and Ca2+(aq) steadily increase, and the reverse reaction also becomes possible:
– 3 Ca2+ (aq) + 2 PO43-(aq) → Ca3 (PO4)2 (s) (decreases)
Eventually, a dynamic equilibrium is reached:
Ca3 (PO4)2 (s) → 3 Ca2+ (aq) + 2 PO43-(aq)
The equilibrium constant, Ksp = [Ca2+]3[PO43-]2– Products over reactants, coefficients become exponents, separate phases have activity = 1
Ksp is called the solubility product constant

Question 25

Q

Solubility Products

Answer

A

We can generalize the expression for the Ksp of any ionic compound, MpXq, as:

Ksp=[Mn+]p[Xz-]q (the [] of each species is raised to the power of its subscripts in compounds formula

Remember that the solid is in a separate phase and thus has activity = 1, regardless of the
actual amount that is present

– Salt to a saturated solution won’t dissolve

Ksp is most useful for slightly soluble salts

– Highly soluble salts have very high ion concentrations when saturated, and concentration is no longer a good approximation of activity

Question 26

Q

Solubility and Solubility Product

Answer

A

Solubility Product ≠ Solubility

Solubility Product (Ksp ) is an equilibrium constant, and as such, has just one value for a given
ionic compound at a given temperature.
Solubility is the extent to which the compound in questions dissolves in a given solution.
In pure water at a specified temperature, a given salt has a particular solubility, expressed
usually in g/L or moles/L of saturated solution.
If a common ion is present, however, the solubility will vary with the concentration of the common ion (i.e. the common ion represses dissolution).

– This changes the solubility, but does not change the Ksp

In both cases, Ksp can be used to calculate solubility

Question 27

Q

Molar Solubility

Answer

A

The solubility of a compound is the quantity of the compound that dissolves in a certain amount of liquid
Molar solubility is the number of moles of substance that dissolves in 1 litre of water
We can calculate the molar solubility from the Ksp values of a solid

AgBr(s) ⇌ Ag+(aq) + Br-(aq) Ksp=5.35 x 10-13

Ksp = [Ag+ ][Br-], so molar solubility = Ksp = 7.31 x 10-7 mol/L

Question 28

Q

Molar Solubility

Be careful with salts that do not contain a 1:1 ratio of ions – eg.

Answer

A

CaF2(s) ⇌ Ca2+(aq) + 2 F-(aq) Ksp = 1.5 x 10-10

Ksp = [Ca2+][F-]2, but [Ca2+] ≠ [F-], so molar solubility ≠ 3 Ksp
The F- concentration is twice the molar concentration of CaF2
(aq)
The Ksp equation contains [F-]2, so fluoride’s coefficient gets squared as well
Ksp = [Ca2+][F-]2 = (x)(2x)2 = 4x3 Need to divide Ksp by 4 before cube rooting
Molar solubility = x = (cubed root)
(Ksp/4) = 3.35 x 10-4 mol/L
Generally, for any salt MnXm: molar solubility = (n+m)squareroot ksp/n^n x m^m

Question 29

Q

Relative Solubilities from Ksp

Answer

A

For salts that produce the same number of ions in the solubility reaction, the solubility decreases with decreasing Ksp:
NiCO3(s) ⇌ Ni2+(aq) + CO3
2-(aq) Ksp = 1.4 x 10-7 Solubility = 3.7x10-4 M
CaCO3(s) ⇌ Ca2+(aq) + CO3
2-(aq) Ksp = 5.0 x 10-9 Solubility = 7.1x10-5 M
CuCO3(s) ⇌ Cu2+(aq) + CO32-(aq) Ksp = 1.4 x 10-10 Solubility = 1.2x10-5 M
For salt pairs with different numbers of ions, one cannot do a direct qualitative comparison of
Ksp due to differing exponents:
PbSO4(s) ⇌ Pb2+(aq) + SO42-(aq) Ksp = 1.8 x 10-8 Solubility = 1.3x10-4 M
SrF2(s) ⇌ Sr2+(aq) + 2 F-(aq) Ksp = 4.3 x 10-9 Solubility = 1.0x10-3 M
Li3PO4(s) ⇌ 3 Li+(aq) + PO43-(aq) Ksp = 3.2 x 10-9 Solubility =3.3x10-3 M

Question 30

Q

Determining Ksp from solubilities

Answer

A

The method by which Ksp values are obtained from measured solubilities is the opposite of the solubility calculations procedure we have just considered.
e.g. The solubility of PbI2 is 0.060 grams per 100. mL of pure water at 25oC. Use thisinformation to calculate the Ksp for PbI2 at 25oC.
0.060 g / (461.0 g/mol) = 1.3 x 10-4 mol/100 mL = 1.3 x 10-3 mol/L

– A saturated solution of PbI2
thus contains 1.3 x 10-3 M of Pb2+(aq)

– and 2.6 x 10-3 M of I-(aq)

Ksp = [Pb2+][I-]2 = (1.3 x 10-3)(2.6 x 10-3)2 = 8.8 x 10-9

Question 31

Q

Common Ion Effect

Answer

A

What happens to solubility if the solvent is not pure water, but water that already containsone or more of the ions in the salt?
Consider the silver(I) acetate dissolution reaction:

AgCH3COO(s) ⇌ Ag+(aq) + CH3COO-(aq)

What happens if we add Ag+ by adding a highly soluble salt such as AgNO3?
Le Chatelier’s principle suggests the solubility will decrease because we are adding product
The new molar solubility can be calculated using Ksp and an ICE table

– Initial [Ag+] = [AgNO3]

Question 32

Q

Effect of pH on Solubility

Answer

A

For salts with ions that have acidic or basic properties the pH of the solution can affect the solubility
We can explain the effect with the common ion effect and Le Chatelier’s principle
For hydroxide salts, the solubility increases with increasing solution acidity
[OH-] = 10-pOH = 10-(14-pH)
H3O+(aq) + OH-(aq) → 2H2O(l)
Salts containing basic anions also increase in solubility in acidic solutions
Protonation reduces the concentration of anion and drives the solvation reaction to the right

Question 33

Q

Precipitation Reactions

Answer

A

Precipitation reactions occur when the concentration of ions exceed the solubility of the ionic compound
To determine whether a precipitate will form you can calculate the reaction quotient Qsp and compare it to the Ksp
Remember that Ksp refers to the
dissolving process (aqueous ions are products), so calculate Qsp the same way

Question 34

Q

Reaction Quotient, Qsp

Answer

A

If we compare the reaction quotient, Qsp, for the current solution concentrations to the value of Ksp, we can determine if precipitation will occur
Qsp = Ksp, the solution is saturated, no precipitation
Qsp < Ksp, the solution is unsaturated, no precipitation
Qsp > Ksp, the solution would be above saturation, the salt above saturation will precipitate
The solvation process proceeds in reverse because Q > K
The Qsp differs from the Ksp in that the concentrations used in the Qsp calculation are the initial values and not necessarily those at equilibrium.

Question 35

Q

Complex Ion Equilibria

Answer

A

A complex ion contains a central metal ion bound to one or more ligands
Ligands are neutral molecules or ions that act as Lewis base with the central metal ion
Complex ions form as a result of a Lewis acid/base reaction
each ligand donates an electron pair to the central metal

Question 36

Q

Formation Constant

Answer

A

The reaction between an ion and a ligand to form a complex ion is called a complex ion formation reaction
The equilibrium constant for the reaction is called the formation constant, Kf
Example:
Ag+(aq) + 2 NH3(aq) ⇌ Ag(NH3)2+(aq)
kf = [Ag(NH3)2+] / [Ag+][NH3]2
Kf = 1.7 x 107
A large Kf indicates a stable complex, which can have a significant effect on other equilibria
Example: Silver salts readily dissolve in concentrated aqueous ammonia (pH ~12) , despite
the very low Ksp for AgOH (2.0 x 10-8)

Question 37

Q

Effect of Complex Ion Equilibria on Solubility

Answer

A

The solubility of an ionic compound containing a metal cation that forms complex ions increases in the presence of Lewis bases that complex with the cation
Recall that adding reactions requires multiplying equilibrium constants