Ch.16 Flashcards
Buffers
- 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
Buffers continued
- 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
Example – Acetic acid – Acetate Buffer
CH3COO- + H3O+ ⇌ CH3COOH + H2O
CH3COOH + OH- ⇌ CH3COO- + H2O
- 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
Buffer pH
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
Henderson-Hasselbalch Equation
- 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
Base/Conjugate acid buffers
- 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
Buffer Range and Capacity
- 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
Buffer Capacity
- 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
Buffer Range
- 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
Preparing a Buffer
- 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
Preparing a Buffer – Alternate method
- 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)
Titrations and pH curves
- 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?
Titrating a Strong Acid with a Strong Base
- 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
Titrating a Weak Acid with a Strong Base
- 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
What to do if the “x is small” approximation is invalid
- 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
What to do if the “x is small” approximation is invalid
Just before the equivalence point:
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
What to do if the “x is small” approximation is invalid
At the equivalence point:
- 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
What to do if the “x is small” approximation is invalid
Just after the equivalence point:
- 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
Titrating a Strong Base with a Strong Acid
- 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
Titrating a Weak Base with a Strong Acid
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
Titration of Polyprotic Acid with Strong Base
- 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
Indicators
- 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
Indicators in Titrations
- 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
Solubility Equilibria
- 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
Solubility Products
- 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
Solubility and Solubility Product
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
Molar Solubility
- 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
Molar Solubility
- Be careful with salts that do not contain a 1:1 ratio of ions – eg.
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
Relative Solubilities from Ksp
- 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
Determining Ksp from solubilities
- 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
Common Ion Effect
- 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]
Effect of pH on Solubility
- 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
Precipitation Reactions
- 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
Reaction Quotient, Qsp
- 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.
Complex Ion Equilibria
- 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
Formation Constant
- 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)
Effect of Complex Ion Equilibria on Solubility
- 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