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Chem 102 > Ch.16 > Flashcards

Ch.16 Flashcards

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1
Q

Buffers

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
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2
Q

Buffers continued

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

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3
Q

Example – Acetic acid – Acetate Buffer

CH3COO- + H3O+ ⇌ CH3COOH + H2O

CH3COOH + OH- ⇌ CH3COO- + H2O

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
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4
Q

Buffer pH

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

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5
Q

Henderson-Hasselbalch Equation

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

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6
Q

Base/Conjugate acid buffers

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

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7
Q

Buffer Range and Capacity

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

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8
Q

Buffer Capacity

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
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9
Q

Buffer Range

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

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10
Q

Preparing a Buffer

A
  • Suppose you needed to make a buffer with a pH of 4.00. What would you do?
  1. 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

  1. 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
  2. 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
  1. 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
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11
Q

Preparing a Buffer – Alternate method

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
  1. Choose the acid or base to make pKa ≈pH
  2. Calculate the amount of HA (or B) needed to produce the desired volume and concentration of buffer
  3. Use the Henderson-Hasselbalch equation to determine how much HA must be converted into A- (or how much B must be converted into HB+)
  4. For every mole of A- needed, add one mole of strong base (or one mole of strong acid for every mole of HB+ needed)
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12
Q

Titrations and pH curves

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?

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13
Q

Titrating a Strong Acid with a Strong Base

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
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14
Q

Titrating a Weak Acid with a Strong Base

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
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15
Q

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

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
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16
Q

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

Just before the equivalence point:

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
17
Q

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

At the equivalence point:

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
18
Q

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

Just after the equivalence point:

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
19
Q

Titrating a Strong Base with a Strong Acid

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
20
Q

Titrating a Weak Base with a Strong Acid

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
21
Q

Titration of Polyprotic Acid with Strong Base

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

22
Q

Indicators

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
23
Q

Indicators in Titrations

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
24
Q

Solubility Equilibria

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
25
Q

Solubility Products

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

26
Q

Solubility and Solubility Product

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
27
Q

Molar Solubility

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
28
Q

Molar Solubility

  • Be careful with salts that do not contain a 1:1 ratio of ions – eg.
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
29
Q

Relative Solubilities from Ksp

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
30
Q

Determining Ksp from solubilities

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
31
Q

Common Ion Effect

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]

32
Q

Effect of pH on Solubility

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
33
Q

Precipitation Reactions

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
34
Q

Reaction Quotient, Qsp

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.
35
Q

Complex Ion Equilibria

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
36
Q

Formation Constant

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)
37
Q

Effect of Complex Ion Equilibria on Solubility

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

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