Chapter 16: Aqueous Ionic Equilibrium. Buffers Solutions that resist changes in pH when an acid or...
-
Upload
samson-ryan -
Category
Documents
-
view
223 -
download
0
Transcript of Chapter 16: Aqueous Ionic Equilibrium. Buffers Solutions that resist changes in pH when an acid or...
Chapter 16: Aqueous Ionic Equilibrium
Buffers• Solutions that resist changes in pH when an acid or base
is added• Act by neutralizing acid or base that is added to the
buffered solution• There is a limit to buffering capacity eventually the pH
changes• Standard Buffer Solutions =
– solution of a weak acid + solution of soluble salt containing the conjugate base anion
Making an Acid Buffer
How Acid Buffers Work:Addition of Base
HA(aq) + H2O(l) A−(aq) + H3O+
(aq)
• Buffers = Application of Le Châtelier’s Principle to weak acid equilibrium
• Buffer solutions contain significant amounts of the weak acid molecules, HA
• HA react with added A- to neutralize itHA(aq) + OH−(aq) → A−(aq) + H2O(l)
– you can also think of the H3O+ combining with the OH− to make H2O; the H3O+ is then replaced by the shifting equilibrium
H2O
HA
How Buffers Work
HA + H3O+
A−
AddedHO−
newA−
A−
How Acid Buffers Work:Addition of Acid
HA(aq) + H2O(l) A−(aq) + H3O+
(aq)
• The buffer solution also contains significant amounts of the conjugate base anion, A−
• These ions combine with added acid to make more HA
H+(aq) + A−(aq) → HA(aq)• After the equilibrium shifts, the concentration
of H3O+ is kept constant
H2O
How Buffers Work
HA + H3O+A−A−
AddedH3O+
newHA
HA
Common Ion Effect HA(aq) + H2O(l) A−
(aq) + H3O+(aq)
• Adding NaA, a salt containing the A−
– A− = conjugate base of the acid (the common ion)– This shifts the equilibrium to the left
• The new pH is higher than the pH of the acid solution– Lower H3O+ ion concentration
Common Ion Effect
Practice − What is the pH of a buffer that is 0.14 M HF (pKa = 3.15) and 0.071 M KF?
write the reaction for the acid with water
construct an ICE table for the reaction
enter the initial concentrations – assuming the [H3O+] from water is ≈ 0
HF + H2O F + H3O+
[HA] [A−] [H3O+]
initial 0.14 0.071 ≈ 0
change
equilibrium
[HA] [A−] [H3O+]
initial 0.14 0.071 0
change
equilibrium
Practice – What is the pH of a buffer that is 0.14 M HF (pKa = 3.15) and 0.071 M KF?
represent the change in the concentrations in terms of x
sum the columns to find the equilibrium concentrations in terms of x
substitute into the equilibrium constant expression
+x+xx0.14 x 0.071 + x x
HF + H2O F + H3O+
pKa for HF = 3.15Ka for HF = 7.0 x 10−4
Practice – What is the pH of a buffer that is 0.14 M and 0.071 M KF?
determine the value of Ka
because Ka is very small, approximate the [HA]eq = [HA]init and [A−]eq = [A−]init solve for x
[HA] [A−] [H3O+]
initial 0.14 0.071 ≈ 0
change −x +x +x
equilibrium 0.012 0.100 x0.14 x 0.071 +x
Practice – What is the pH of a buffer that is 0.14 M HF (pKa = 3.15) and 0.071 M KF?
check if the approximation is valid by seeing if x < 5% of [HC2H3O2]init
Ka for HF = 7.0 x 10−4
the approximation is valid
x = 1.4 x 10−3
[HA] [A−] [H3O+]
initial 0.14 0.071 ≈ 0
change −x +x +x
equilibrium 0.14 0.071 x
Practice – What is the pH of a buffer that is 0.14 M HF (pKa = 3.15) and 0.071 M KF?
substitute x into the equilibrium concentration definitions and solve
x = 1.4 x 10−3
[HA] [A−] [H3O+]
initial 0.14 0.071 ≈ 0
change −x +x +x
equilibrium 0.14 0.072 1.4E-30.071 + x x 0.14 x
Practice – What is the pH of a buffer that is 0.14 M HF (pKa = 3.15) and 0.071 M KF?
substitute [H3O+] into the formula for pH and solve
[HA] [A−] [H3O+]
initial 0.14 0.071 ≈ 0
change −x +x +x
equilibrium 0.14 0.072 1.4E−3
Practice – What is the pH of a buffer that is 0.14 M HF (pKa = 3.15) and 0.071 M KF?
check by substituting the equilibrium concentrations back into the equilibrium constant expression and comparing the calculated Ka to the given Ka
the values are close enough
[HA] [A−] [H3O+]
initial 0.14 0.071 ≈ 0
change −x +x +x
equilibrium 0.14 0.072 1.4E−3
Ka for HF = 7.0 x 10−4
Henderson-Hasselbalch Equation• Calculating the pH of a buffer solution can be
simplified by using an equation derived from the Ka expression called the Henderson-Hasselbalch Equation
• The equation calculates the pH of a buffer from the pKa and initial concentrations of the weak acid and salt of the conjugate base– as long as the “x is small” approximation is valid
Practice – What is the pH of a buffer that is 0.14 M HF (pKa = 3.15) and 0.071 M KF?
find the pKa from the given Ka
assume the [HA] and [A−] equilibrium concentrations are the same as the initial
substitute into the Henderson-Hasselbalch equation
check the “x is small” approximation
HF + H2O F + H3O+
When should you use the “Full Equilibrium Analysis” or the “Henderson-Hasselbalch Equation”?
• The Henderson-Hasselbalch equation is generally good enough when the “x is small” approximation is applicable
• Generally, the “x is small” approximation will work when both of the following are true:
a) the initial concentrations of acid and salt are not very dilute
b) the Ka is fairly small• For most problems, this means that the initial acid
and salt concentrations should be over 100 to 1000x larger than the value of Ka
How Much Does the pH of a Buffer Change When an Acid or Base Is Added?
• While buffers resist change in pH when acid or base is added to them, their pH does change
• Calculating the new pH after adding acid or base requires breaking the problem into two parts
1. a stoichiometry calculation for the reaction of the added chemical with one of the ingredients of the buffer to reduce its initial concentration and increase the concentration of the other added acid reacts with the A− to make more HA added base reacts with the HA to make more A−
2. an equilibrium calculation of [H3O+] using the new initial values of [HA] and [A−]
Practice – What is the pH of a buffer that has 0.140 moles HF and 0.071 moles KF in 1.00 L of solution when 0.020 moles of HCl is
added?
If the added chemical is a base, write a reaction for OH− with HA. If the added chemical is an acid, write a reaction for H3O+ with A−.
construct a stoichiometry table for the reaction
F− + H3O+ HF + H2O
F− H3O+ HF
mols before 0.071 0 0.140
mols added – 0.020 –
mols after
Practice – What is the pH of a buffer that has 0.140 moles HF and 0.071 moles KF in 1.00 L of solution when 0.020 moles of HCl is
added?
fill in the table – tracking the changes in the number of moles for each component
F− + H3O+ HF + H2O
F− H3O+ HF
mols before 0.071 0 0.140
mols added – 0.020 –
mols after 0.051 0 0.160
Practice – What is the pH of a buffer that has 0.140 moles HF and 0.071 moles KF in 1.00 L of solution when 0.020 moles of
HCl is added?
If the added chemical is a base, write a reaction for OH− with HA. If the added chemical is an acid, write a reaction for H3O+ with A−.
construct a stoichiometry table for the reaction
enter the initial number of moles for each
F− + H3O+ HF + H2O
F− H3O+ HF
mols before 0.071 0.020 0.140
mols change
mols after
new molarity
F− H3O+ HFmols before 0.071 0.020 0.140
mols change
mols after
new molarity
Practice – What is the pH of a buffer that has 0.140 moles HF and 0.071 moles KF in 1.00 L of solution
when 0.020 moles of HCl is added?
using the added chemical as the limiting reactant, determine how the moles of the other chemicals change
add the change to the initial number of moles to find the moles after reaction
divide by the liters of solution to find the new molarities
0.16000.051
F− + H3O+ HF + H2O
+0.020−0.020 −0.020
0.16000.051
Practice – What is the pH of a buffer that has 0.140 moles HF and 0.071 moles KF in 1.00 L of solution when 0.020 moles of
HCl is added?
write the reaction for the acid with water
construct an ICE table.
assume the [HA] and [A−] equilibrium concentrations are the same as the initial
substitute into the Henderson-Hasselbalch equation
HF + H2O F + H3O+
[HF] [F−] [H3O+]
initial 0.160 0.051 ≈ 0
change −x +x +x
equilibrium 0.160 0.051 x
Basic BuffersB:(aq) + H2O(l) H:B+
(aq) + OH−(aq)
• Buffers can also be made by mixing a weak base, (B:), with a soluble salt of its conjugate acid, H:B+Cl−
H2O(l) + NH3 (aq) NH4+
(aq) + OH−(aq)
• The Henderson-Hasselbalch equation is written for a chemical reaction with a weak acid reactant and its conjugate base as a product
• The chemical equation of a basic buffer is written with a weak base as a reactant and its conjugate acid as a product
B: + H2O H:B+ + OH−
• To apply the Henderson-Hasselbalch equation, the chemical equation of the basic buffer must be looked at like an acid reactionH:B+ + H2O B: + H3O+
this does not affect the concentrations, just the way we are looking at the reaction
Henderson-Hasselbalch Equation for Basic Buffers
Relationship between pKa and pKb
• Just as there is a relationship between the Ka of a weak acid and Kb of its conjugate base, there is also a relationship between the pKa of a weak acid and the pKb of its conjugate base
Ka Kb = Kw = 1.0 x 10−14
−log(Ka Kb) = −log(Kw) = 14
−log(Ka) + −log(Kb) = 14
pKa + pKb = 14
Buffering Effectiveness• A good buffer should be able to neutralize moderate
amounts of added acid or base• However, there is a limit to how much can be added
before the pH changes significantly• The buffering capacity is the amount of acid or base a
buffer can neutralize• The buffering range is the pH range the buffer can be
effective• The effectiveness of a buffer depends on two factors
(1) the relative amounts of acid and base, and (2) the absolute concentrations of acid and base
HA A− OH−
mols before 0.18 0.020 0
mols added ─ ─ 0.010
mols after 0.17 0.030 ≈ 0
Effect of Relative Amounts of Acid and Conjugate Base
Buffer 10.100 mol HA & 0.100 mol A−
Initial pH = 5.00
Buffer 20.18 mol HA & 0.020 mol A−
Initial pH = 4.05pKa (HA) = 5.00
after adding 0.010 mol NaOHpH = 5.09
HA + OH− A + H2O
HA A− OH−
mols before 0.100 0.100 0
mols added ─ ─ 0.010
mols after 0.090 0.110 ≈ 0
after adding 0.010 mol NaOHpH = 4.25
A buffer is most effective with equal concentrations of acid and base
HA A− OH−
mols before 0.50 0.500 0
mols added ─ ─ 0.010
mols after 0.49 0.51 ≈ 0
HA A− OH−
mols before 0.050 0.050 0
mols added ─ ─ 0.010
mols after 0.040 0.060 ≈ 0
Effect of Absolute Concentrations of Acid and Conjugate Base
Buffer 10.50 mol HA & 0.50 mol A−
Initial pH = 5.00
Buffer 20.050 mol HA & 0.050 mol A−
Initial pH = 5.00pKa (HA) = 5.00
after adding 0.010 mol NaOHpH = 5.02
HA + OH− A + H2O
after adding 0.010 mol NaOHpH = 5.18
A buffer is most effective when the concentrations of acid and base are largest
Buffering Capacity
a concentrated buffer can neutralize more added acid or base than a dilute buffer
Effectiveness of Buffers
• A buffer will be most effective when the [base]:[acid] = 1– equal concentrations of acid and base
• A buffer will be effective when 0.1 < [base]:[acid] < 10
• A buffer will be most effective when the [acid] and the [base] are large
Buffering Range• We have said that a buffer will be effective when
0.1 < [base]:[acid] < 10• Substituting into the Henderson-Hasselbalch
equation we can calculate the maximum and minimum pH at which the buffer will be effective
Lowest pH Highest pH
Therefore, the effective pH range of a buffer is pKa ± 1When choosing an acid to make a buffer, choose one whose is pKa closest to the pH of the buffer
Practice – What ratio of NaC7H5O2 : HC7H5O2 would be required to make a buffer with pH 3.75?Benzoic acid, HC7H5O2, pKa = 4.19
to make a buffer with pH 3.75, you would use 0.363 times as much NaC7H5O2 as HC7H5O2
Buffering Capacity• Buffering capacity = the amount of acid or base that
can be added to a buffer without causing a large change in pH
• The buffering capacity increases with increasing absolute concentration of the buffer components
• As the [base]:[acid] ratio approaches 1, the ability of the buffer to neutralize both added 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]
Titration• In an acid-base titration, a solution of unknown
concentration (titrant) is slowly added to a solution of known concentration from a burette until the reaction is complete– when the reaction is complete we have reached the
endpoint of the titration• An indicator may be added to determine the endpoint
– an indicator is a chemical that changes color when the pH changes
• When the moles of H3O+ = moles of OH−, the titration has reached its equivalence point
Titration
Titration Curve: pH vs Amount of Titrant Added
• Prior to the equivalence point, the known solution in the flask is in excess, so the pH is closest to its pH
• The inflection point of the curve is the equivalence point of the titration– The pH of the equivalence point depends on the pH of
the salt solution• equivalence point of neutral salt, pH = 7 • equivalence point of acidic salt, pH < 7 • equivalence point of basic salt, pH > 7
• Beyond the equivalence point, the unknown solution in the burette is in excess, so the pH approaches its pH
Titration Curve:Unknown Strong Base Added to Strong Acid
Before Equivalence(excess acid)
After Equivalence(excess base)
Titration of 25 mL of 0.100 M HCl with 0.100 M NaOH
Equivalence Point equal moles of HCl and NaOH
pH = 7.00
Because the solutions are equal concentration, and 1:1 stoichiometry, the equivalence point is at equal volumes
HCl NaCl NaOH
mols before 2.50E-3 0 5.0E-4
mols change −5.0E-4 +5.0E-4 −5.0E-4
mols end 2.00E-3 5.0E-4 0
molarity, new 0.0667 0.017 0
HCl NaCl NaOH
mols before 2.50E-3 0 5.0E-4
mols change
mols end
molarity, new
HCl NaCl NaOH
mols before 2.50E-3 0 5.0E-4
mols change −5.0E-4 +5.0E-4 −5.0E-4
mols end 2.00E-3 5.0E-4 0
molarity, new
5.0 x 10−4 mole NaOH added
Titration of 25 mL of 0.100 M HCl with 0.100 M NaOH
• HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l)
• Initial pH = −log(0.100) = 1.00• Initial mol of HCl = 0.0250 L x 0.100 mol/L = 2.50 x 10−3
• Before equivalence point added 5.0 mL NaOH
42
Titration of 25 mL of 0.100 M HCl with 0.100 M NaOH
• HCl(aq) + NaOH(aq) NaCl(aq) + H2O(aq)
• To reach equivalence, the added moles NaOH = initial moles of HCl = 2.50 x 10−3 moles
• At equivalence, we have 0.00 mol HCl and 0.00 mol NaOH left over
• Because the NaCl is a neutral salt, the pH at equivalence = 7.00
HCl NaCl NaOH
mols before 2.50E-3 0 2.5E-3
mols change
mols end
molarity, new
HCl NaCl NaOH
mols before 2.50E-3 0 2.5E-3
mols change −2.5E-3 +2.5E-3 −2.5E-3
mols end 0 2.5E-3 0
molarity, new
HCl NaCl NaOH
mols before 2.50E-3 0 2.5E-3
mols change −2.5E-3 +2.5E-3 −2.5E-3
mols end 0 2.5E-3 0
molarity, new 0 0.050 0
2.5 x 10−3 mole NaOH added
Titration of 25 mL of 0.100 M HCl with 0.100 M NaOH
• HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l)
• Initial pH = −log(0.100) = 1.00• Initial mol of HCl = 0.0250 L x 0.100 mol/L = 2.50 x 10−3
• At equivalence point added 25.0 mL NaOH
HCl NaCl NaOH
mols before 2.50E-3 0 3.0E-3
mols change −2.5E-3 +2.5E-3 −2.5E-3
mols end 0 2.5E-3 5.0E-4
molarity, new 0 0.045 0.0091
HCl NaCl NaOH
mols before 2.50E-3 0 3.0E-3
mols change
mols end
molarity, new
HCl NaCl NaOH
mols before 2.50E-3 0 3.0E-3
mols change −2.5E-3 +2.5E-3 −2.5E-3
mols end 0 2.5E-3 5.0E-4
molarity, new
Titration of 25 mL of 0.100 M HCl with 0.100 M NaOH
• HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l)
• Initial pH = −log(0.100) = 1.00• Initial mol of HCl = 0.0250 L x 0.100 mol/L = 2.50 x 10−3
• After equivalence point added 30.0 mL NaOH
3.0 x 10−3 mole NaOH added
added 5.0 mL NaOH0.00200 mol HClpH = 1.18
added 10.0 mL NaOH0.00150 mol HClpH = 1.37
added 25.0 mL NaOHequivalence pointpH = 7.00
added 30.0 mL NaOH0.00050 mol NaOHpH = 11.96
added 40.0 mL NaOH0.00150 mol NaOHpH = 12.36
added 50.0 mL NaOH0.00250 mol NaOHpH = 12.52
added 35.0 mL NaOH0.00100 mol NaOHpH = 12.22
added 15.0 mL NaOH0.00100 mol HClpH = 1.60
added 20.0 mL NaOH0.00050 mol HClpH = 1.95
Adding 0.100 M NaOH to 0.100 M HCl
25.0 mL 0.100 M HCl0.00250 mol HClpH = 1.00
HNO3 NaNO3 NaOH
mols before 1.25E-2 0 1.5E-3
mols change −1.5E-3 +1.5E-3 −1.5E-3
mols end 1.1E-3 1.5E-3 0
molarity, new
HNO3 NaNO3 NaOH
mols before 1.25E-2 0 1.5E-3
mols change −1.5E-3 +1.5E-3 −1.5E-3
mols end 1.1E-3 1.5E-3 0
molarity, new 0.018 0.025 0
Practice – Calculate the pH of the solution that results when 10.0 mL of 0.15 M NaOH is added to 50.0 mL of 0.25
M HNO3
• HNO3(aq) + NaOH(aq) NaNO3(aq) + H2O(l)• Initial pH = −log(0.250) = 0.60• Initial mol of HNO3= 0.0500 L x 0.25 mol/L=1.25 x 10−2
• Before equivalence point added 10.0 mL NaOH
Practice – Calculate the amount of 0.15 M NaOH solution that must be added to 50.0 mL of 0.25 M HNO3 to reach equivalence
• HNO3(aq) + NaOH(aq) NaNO3(aq) + H2O(l)
• Initial pH = −log(0.250) = 0.60• Initial mol of HNO3= 0.0500 L x 0.25 mol/L=1.25 x 10−2
• At equivalence point: moles of NaOH = 1.25 x 10−2
HNO3 NaNO3 NaOH
mols before 1.25E-2 0 1.5E-2
mols change −1.25E-2 +1.25E-2 −1.25E-2
mols end 0 1.25E-2 0.0025
molarity, new 0 0.0833 0.017
HNO3 NaNO3 NaOH
mols before 1.25E-2 0 1.5E-2
mols change −1.25E-2 +1.25E-2 −1.25E-2
mols end 0 1.25E-2 0.0025
molarity, new
Practice – Calculate the pH of the solution that results when 100.0 mL of 0.15 M NaOH is added to 50.0 mL of
0.25 M HNO3
• HNO3(aq) + NaOH(aq) NaNO3(aq) + H2O(l)• Initial pH = −log(0.250) = 0.60• Initial mol of HNO3= 0.0500 L x 0.25 mol/L=1.25 x 10−2
• After equivalence point added 100.0 mL NaOH
HNO3 NaNO3 NaOH
mols before 1.25E-2 0 1.5E-2
mols change −1.25E-2 +1.25E-2 −1.25E-2
mols end 0 1.25E-2 0.0025
molarity, new 0 0.0833 0.017
HNO3 NaNO3 NaOH
mols before 1.25E-2 0 1.5E-2
mols change −1.25E-2 +1.25E-2 −1.25E-2
mols end 0 1.25E-2 0.0025
molarity, new
Practice – Calculate the pH of the solution that results when 100.0 mL of 0.15 M NaOH is added to 50.0 mL of
0.25 M HNO3
• HNO3(aq) + NaOH(aq) NaNO3(aq) + H2O(l)• Initial pH = −log(0.250) = 0.60• initial mol of HNO3= 0.0500 L x 0.25 mol/L=1.25 x 10−2
• After equivalence point added 100.0 mL NaOH
Titration of a Strong Base with a Strong Acid
• If the titration is run so that the acid is in the burette and the base is in the flask, the titration curve will be the reflection of the one just shown
Titration of a Weak Acid with a Strong Base• Titrating a weak acid with a strong base results in differences
in the titration curve at the equivalence point and excess acid region
• The initial pH is determined using the Ka of the weak acid• The pH in the excess acid region is determined as you would
determine the pH of a buffer • The pH at the equivalence point is determined using the Kb of
the conjugate base of the weak acid• The pH after equivalence is dominated by the excess strong
base– the basicity from the conjugate base anion is negligible
Titration of 25 mL of 0.100 M HCHO2 with 0.100 M NaOH
• HCHO2(aq) + NaOH(aq) NaCHO2(aq) + H2O(aq)
• Initial pH
[HCHO2] [CHO2−] [H3O+]
initial 0.100 0.000 ≈ 0
change −x +x +x
equilibrium 0.100 − x x x
Ka = 1.8 x 10−4
Titration of 25 mL of 0.100 M HCHO2 with 0.100 M NaOH
• HCHO2(aq) + NaOH(aq) NaCHO2 (aq) + H2O(aq)
• Initial mol of HCHO2 = 0.0250 L x 0.100 mol/L = 2.50 x 10−3
• Before equivalence added 5.0 mL NaOH
HA A− OH−
mols before 2.50E-3 0 0
mols added – – 5.0E-4
mols after 2.00E-3 5.0E-4 ≈ 0
HA A− OH−
mols before 2.50E-3 0 0
mols added – – 2.50E-3
mols after 0 2.50E-3 ≈ 0
[HCHO2] [CHO2−] [OH−]
initial 0 0.0500 ≈ 0
change +x −x +x
equilibrium x 5.00E-2-x x
Titration of 25 mL of 0.100 M HCHO2 with 0.100 M NaOH
added 25.0 mL NaOHCHO2
−(aq) + H2O(l) HCHO2(aq) + OH−
(aq)
Kb = 5.6 x 10−11
[OH−] = 1.7 x 10−6 M
• HCHO2(aq) + NaOH(aq) NaCHO2 (aq) + H2O(aq)
• Initial mol of HCHO2 = 0.0250 L x 0.100 mol/L = 2.50 x 10−3
• At equivalence
HA A− NaOH
mols before 2.50E-3 0 3.0E-3
mols change −2.5E-3 +2.5E-3 −2.5E-3
mols end 0 2.5E-3 5.0E-4
molarity, new
HA A− NaOH
mols before 2.50E-3 0 3.0E-3
mols change −2.5E-3 +2.5E-3 −2.5E-3
mols end 0 2.5E-3 5.0E-4
molarity, new 0 0.045 0.0091
HA A− NaOH
mols before 2.50E-3 0 3.0E-3
mols change
mols end
molarity, new
Titration of 25 mL of 0.100 M HCHO2 with 0.100 M NaOH
56
added 30.0 mL NaOH
3.0 x 10−3 mole NaOH added
• HCHO2(aq) + NaOH(aq) NaCHO2(aq) + H2O(aq)
• Initial mol of HCHO2 = 0.0250 L x 0.100 mol/L = 2.50 x 10−3
• After equivalence
added 35.0 mL NaOH0.00100 mol NaOH xspH = 12.22
initial HCHO2 solution0.00250 mol HCHO2
pH = 2.37
added 5.0 mL NaOH0.00200 mol HCHO2
pH = 3.14
added 10.0 mL NaOH0.00150 mol HCHO2
pH = 3.56
added 25.0 mL NaOHequivalence point0.00250 mol CHO2
−
[CHO2−]init = 0.0500 M
[OH−]eq = 1.7 x 10−6
pH = 8.23
added 30.0 mL NaOH0.00050 mol NaOH xspH = 11.96
added 20.0 mL NaOH0.00050 mol HCHO2
pH = 4.34
added 15.0 mL NaOH0.00100 mol HCHO2
pH = 3.92
added 12.5 mL NaOH0.00125 mol HCHO2
pH = 3.74 = pKa
half-neutralization
Adding NaOH to HCHO2
added 40.0 mL NaOH0.00150 mol NaOH xspH = 12.36
added 50.0 mL NaOH0.00250 mol NaOH xspH = 12.52
Titrating Weak Acid with a Strong Base
• The initial pH is that of the weak acid solution– calculate like a weak acid equilibrium problem
• e.g., 15.5 and 15.6
• Before the equivalence point, the solution becomes a buffer– calculate mol HAinit and mol A−
init using reaction stoichiometry
– calculate pH with Henderson-Hasselbalch using mol HAinit and mol A−
init • Half-neutralization pH = pKa
Titrating Weak Acid with a Strong Base
• At the equivalence point, the mole HA = mol Base, so the resulting solution has only the conjugate base anion in it before equilibrium is established– mol A− = original mole HA
• calculate the volume of added base as you did in Example 4.8– [A−]init = mol A−/total liters– calculate like a weak base equilibrium problem
• e.g., 15.14
• Beyond equivalence point, the OH is in excess– [OH−] = mol MOH xs/total liters– [H3O+][OH−]=1 x 10−14
Titration Curve of a Weak Base with a Strong Acid
Practice – Titration of 25.0 mL of 0.10 M NH3 with 0.10 M HCl. Calculate the initial pH of the NH3(aq)
• NH3(aq) + HCl(aq) NH4Cl(aq)
• Initial: NH3(aq) + H2O(l) NH4+
(aq) + OH−(aq)
[HCl] [NH4+] [NH3]
initial 0 0 0.10
change +x +x −x
equilibrium x x 0.10−x
pKb = 4.75Kb = 10−4.75 = 1.8 x 10−5
Practice – Titration of 25.0 mL of 0.10 M NH3 with 0.10 M HCl. Calculate the initial pH of the NH3(aq)
• NH3(aq) + HCl(aq) NH4Cl(aq)
• Initial: NH3(aq) + H2O(l) NH4+
(aq) + OH−(aq)
[HCl] [NH4+] [NH3]
initial 0 0 0.10
change +x +x −x
equilibrium x x 0.10−x
pKb = 4.75Kb = 10−4.75 = 1.8 x 10−5
Practice – Titration of 25.0 mL of 0.10 M NH3 (pKb = 4.75) with 0.10 M HCl. Calculate the pH of the solution after adding 5.0 mL of HCl.
• NH3(aq) + HCl(aq) NH4Cl(aq)
• Initial mol of NH3 = 0.0250 L x 0.100 mol/L = 2.50 x 10−3
• Before equivalence: after adding 5.0 mL of HCl
NH3 NH4Cl HCl
mols before 2.50E-3 0 5.0E-4
mols change −5.0E-4 −5.0E-4 −5.0E-4
mols end 2.00E-3 5.0E-4 0
molarity, new 0.0667 0.017 0
NH4+
(aq) + H2O(l) NH4+
(aq) + H2O(l) pKb = 4.75pKa = 14.00 − 4.75 = 9.25
Practice – Titration of 25.0 mL of 0.10 M NH3 (pKb = 4.75) with 0.10 M HCl. Calculate the pH of the solution after adding 5.0 mL of HCl.
• NH3(aq) + HCl(aq) NH4Cl(aq)
• Initial mol of NH3 = 0.0250 L x 0.100 mol/L = 2.50 x 10−3
• Before equivalence: after adding 5.0 mL of HCl
NH3 NH4Cl HCl
mols before 2.50E-3 0 5.0E-4
mols change −5.0E-4 −5.0E-4 −5.0E-4
mols end 2.00E-3 5.0E-4 0
molarity, new 0.0667 0.017 0
Practice – Titration of 25.0 mL of 0.10 M NH3 (pKb = 4.75) with 0.10 M HCl. Calculate the pH of the solution at equivalence.
• NH3(aq) + HCl(aq) NH4Cl(aq)
• Initial mol of NH3 = 0.0250 L x 0.100 mol/L = 2.50 x 10−3
• At equivalence mol NH3 = mol HCl = 2.50 x 10−3
added 25.0 mL HClNH3 NH4Cl HCl
mols before 2.50E-3 0 2.5E-3
mols change −2.5E-3 +2.5E-3 −2.5E-3
mols end 0 2.5E-3 0
molarity, new 0 0.050 0
Practice – Titration of 25.0 mL of 0.10 M NH3 (pKb = 4.75) with 0.10 M HCl. Calculate the pH of the solution at equivalence.
NH3(aq) + HCl(aq) NH4Cl(aq) at equivalence [NH4Cl] = 0.050 M
[NH3] [NH4+] [H3O+]
initial 0 0.050 ≈ 0
change +x −x +x
equilibrium x 0.050−x x
NH4+
(aq) + H2O(l) NH3(aq) + H3O+(aq)
Practice – Titration of 25.0 mL of 0.10 M NH3 (pKb = 4.75) with 0.10 M HCl. Calculate the pH of the solution after adding 30.0 mL of HCl.
• NH3(aq) + HCl(aq) NH4Cl(aq)
• Initial mol of NH3 = 0.0250 L x 0.100 mol/L = 2.50 x 10−3
• After equivalence: after adding 30.0 mL HClNH3 NH4Cl HCl
mols before 2.50E-3 0 3.0E-3
mols change −2.5E-3 +2.5E-3 −2.5E-3
mols end 0 2.5E-3 5.0E-4
molarity, new 0 0.045 0.0091
when you mix a strong acid, HCl, with a weak acid, NH4
+, you only need to consider the strong acid
Titration of a Polyprotic Acid• If Ka1 >> Ka2, there will be two equivalence points
in the titration– the closer the Ka’s are to each other, the less
distinguishable the equivalence points are
titration of 25.0 mL of 0.100 M H2SO3 with 0.100 M NaOH
Monitoring pH During a Titration• The general method for monitoring the pH during the
course of a titration is to measure the conductivity of the solution due to the [H3O+]– using a probe that specifically measures just H3O+
• The endpoint of the titration is reached at the equivalence point in the titration – at the inflection point of the titration curve
• If you just need to know the amount of titrant added to reach the endpoint, we often monitor the titration with an indicator
Monitoring pH During a Titration
Indicators• Many dyes change color depending on the pH of the
solution• These dyes are weak acids, establishing an equilibrium with
the H2O and H3O+ in the solutionHInd(aq) + H2O(l) Ind
(aq) + H3O+(aq)
• The color of the solution depends on the relative concentrations of Ind:HInd– when Ind:HInd ≈ 1, the color will be mix of the colors of Ind
and HInd – when Ind:HInd > 10, the color will be mix of the colors of Ind
– when Ind:HInd < 0.1, the color will be mix of the colors of HInd
Phenolphthalein
Methyl Red
Monitoring a Titration with an Indicator
• For most titrations, the titration curve shows a very large change in pH for very small additions of titrant near the equivalence point
• An indicator can therefore be used to determine the endpoint of the titration if it changes color within the same range as the rapid change in pH– pKa of HInd ≈ pH at equivalence point
Acid-Base Indicators
Solubility Equilibria
• All ionic compounds dissolve in water to some degree – however, many compounds have such low
solubility in water that we classify them as insoluble
• We can apply the concepts of equilibrium to salts dissolving, and use the equilibrium constant for the process to measure relative solubilities in water
Solubility Product• The equilibrium constant for the dissociation of a solid
salt into its aqueous ions is called the solubility product, Ksp
• For an ionic solid MnXm, the dissociation reaction is:MnXm(s) nMm+(aq) + mXn−(aq)
• The solubility product would be Ksp = [Mm+]n[Xn−]m
• For example, the dissociation reaction for PbCl2 isPbCl2(s) Pb2+(aq) + 2 Cl−(aq)
• And its equilibrium constant is Ksp = [Pb2+][Cl−]2
Molar Solubility• Solubility is the amount of solute that will dissolve in
a given amount of solution– at a particular temperature
• The molar solubility is the number of moles of solute that will dissolve in a liter of solution– the molarity of the dissolved solute in a saturated solution
for the general reaction MnXm(s) nMm+(aq) + mXn−(aq)
Practice – Determine the Ksp of PbBr2 if its molar solubility in water at 25 C is 1.05 x 10−2 M
write the dissociation reaction and Ksp expression
create an ICE table defining the change in terms of the solubility of the solid
[Pb2+] [Br−]
initial 0 0
change +(1.05 x 10−2) +2(1.05 x 10−2)
equilibrium (1.05 x 10−2) (2.10 x 10−2)
PbBr2(s) Pb2+(aq) + 2 Br−(aq)
Ksp = [Pb2+][Br−]2
Practice – Determine the Ksp of PbBr2 if its molar solubility in water at 25 C is 1.05 x 10−2 M
substitute into the Ksp expression
plug into the equation and solve
Ksp = [Pb2+][Br−]2
Ksp = (1.05 x 10−2)(2.10 x 10−2)2
[Pb2+] [Br−]
initial 0 0
change +(1.05 x 10−2) +2(1.05 x 10−2)
equilibrium (1.05 x 10−2) (2.10 x 10−2)
Ksp and Relative Solubility
• Molar solubility is related to Ksp
• But you cannot always compare solubilities of compounds by comparing their Ksps
• To compare Ksps, the compounds must have the same dissociation stoichiometry
The Effect of Common Ion on Solubility• Addition of a soluble salt that contains one of
the ions of the “insoluble” salt, decreases the solubility of the “insoluble” salt
• For example, addition of NaCl to the solubility equilibrium of solid PbCl2 decreases the solubility of PbCl2
PbCl2(s) Pb2+(aq) + 2 Cl−(aq)addition of Cl− shifts the
equilibrium to the left
Practice – Determine the concentration of Ag+ ions in seawater that has a [Cl−] of 0.55 M
write the dissociation reaction and Ksp expression
create an ICE table defining the change in terms of the solubility of the solid
[Ag+] [Cl−]
initial 0 0.55
change +S +S
equilibrium S 0.55 + S
AgCl(s) Ag+(aq) + Cl−(aq)
Ksp = [Ag+][Cl−]
Practice – Determine the concentration of Ag+ ions in seawater that has a [Cl−] of 0.55 M
substitute into the Ksp expression,assume S is small
find the value of Ksp from Table 16.2, plug into the equation, and solve for S
[Ag+] [Cl−]
Initial 0 0.55
Change +S +S
Equilibrium S 0.55 + S
Ksp = [Ag+][Cl−]
Ksp = (S)(0.55 + S)
Ksp = (S)(0.55)
The Effect of pH on Solubility• For insoluble ionic hydroxides, the higher the pH, the
lower the solubility of the ionic hydroxide– and the lower the pH, the higher the solubility– higher pH = increased [OH−]
M(OH)n(s) Mn+(aq) + nOH−(aq)• For insoluble ionic compounds that contain anions of
weak acids, the lower the pH, the higher the solubilityM2(CO3)n(s) 2 Mn+(aq) + nCO3
2−(aq)
H3O+(aq) + CO32− (aq) HCO3
− (aq) + H2O(l)
Precipitation• Precipitation will occur when the concentrations of the
ions exceed the solubility of the ionic compound• If we compare the reaction quotient, Q, for the current
solution concentrations to the value of Ksp, we can determine if precipitation will occur– Q = Ksp, the solution is saturated, no precipitation– Q < Ksp, the solution is unsaturated, no precipitation– Q > Ksp, the solution would be above saturation, the salt
above saturation will precipitate• Some solutions with Q > Ksp will not precipitate unless
disturbed – these are called supersaturated solutions
precipitation occurs if Q > Ksp
a supersaturated solution will precipitate if a seed crystal is added
Selective Precipitation
• A solution containing several different cations can often be separated by addition of a reagent that will form an insoluble salt with one of the ions, but not the others
• A successful reagent can precipitate with more than one of the cations, as long as their Ksp values are significantly different
Practice – Will a precipitate form when we mix Ca(NO3)2(aq) with NaOH(aq) if the concentrations after mixing are both 0.0175 M?
write the equation for the reaction
determine the ion concentrations of the original salts
determine the Ksp for any “insoluble” product
write the dissociation reaction for the insoluble product
calculate Q, using the ion concentrations
compare Q to Ksp. If Q > Ksp, precipitation
Ca(NO3)2(aq) + 2 NaOH(aq) → Ca(OH)2(s) + 2 NaNO3(aq)
Ksp of Ca(OH)2 = 4.68 x 10–6
Ca(OH)2(s) Ca2+(aq) + 2 OH−
(aq)
Ca(NO3)2 = 0.0175 M Ca2+ = 0.0175 M, NO3
− = 2(0.0175 M)
NaOH = 0.0175 M Na+ = 0.0175 M, OH− = 0.0175 M
Q > Ksp, so precipitation
Practice – What is the minimum concentration of Ca(NO3)2(aq) that will precipitate Ca(OH)2 from 0.0175 M NaOH(aq)?
precipitating may just occur when Q = Ksp
[Ca(NO3)2] = [Ca2+] = 0.0153 M
Ca(NO3)2(aq) + 2 NaOH(aq) → Ca(OH)2(s) + 2 NaNO3(aq)
Ca(OH)2(s) Ca2+(aq) + 2 OH−
(aq)
Practice – A solution is made by mixing Pb(NO3)2(aq) with AgNO3(aq) so both compounds have a concentration of 0.0010 M. NaCl(s) is added to precipitate out both AgCl(s) and
PbCl2(aq). What is the [Ag+] concentration when the Pb2+ just begins to precipitate?
Practice – What is the [Ag+] concentration when the Pb2+(0.0010 M) just begins to precipitate?
precipitating may just occur when Q = Ksp
AgNO3(aq) + NaCl(aq) → AgCl(s) + NaNO3(aq)
AgCl(s) Ag+(aq) + Cl−(aq)
Practice – What is the [Ag+] concentration when the Pb2+(0.0010 M) just begins to precipitate?
precipitating may just occur when Q = Ksp
Pb(NO3)2(aq) + 2 NaCl(aq) → PbCl2(s) + 2 NaNO3(aq)
PbCl2(s) Pb2+(aq) + 2 Cl−(aq)
Practice – What is the [Ag+] concentration when the Pb2+(0.0010 M) just begins to precipitate
precipitating Ag+ begins when [Cl−] = 1.77 x 10−7 M
precipitating Pb2+ begins when [Cl−] = 1.08 x 10−1 M
when Pb2+ just begins to precipitate out, the [Ag+] has dropped from 0.0010 M to 1.6 x 10−9 M
AgCl(s) Ag+(aq) + Cl−(aq)
Qualitative Analysis• An analytical scheme that utilizes selective
precipitation to identify the ions present in a solution is called a qualitative analysis scheme– wet chemistry
• A sample containing several ions is subjected to the addition of several precipitating agents
• Addition of each reagent causes one of the ions present to precipitate out
Qualitative Analysis
Group 1
• Group one cations are Ag+, Pb2+ and Hg22+
• All these cations form compounds with Cl− that are insoluble in water– as long as the concentration is large enough– PbCl2 may be borderline
• molar solubility of PbCl2 = 1.43 x 10−2 M
• Precipitated by the addition of HCl
Group 2• Group two cations are Cd2+, Cu2+, Bi3+, Sn4+, As3+,
Pb2+, Sb3+, and Hg2+
• All these cations form compounds with HS− and S2− that are insoluble in water at low pH
• Precipitated by the addition of H2S in HCl
Group 3
• Group three cations are Fe2+, Co2+, Zn2+, Mn2+, Ni2+ precipitated as sulfides; as well as Cr3+, Fe3+, and Al3+ precipitated as hydroxides
• All these cations form compounds with S2− that are insoluble in water at high pH
• Precipitated by the addition of H2S in NaOH
Group 4
• Group four cations are Mg2+, Ca2+, Ba2+ • All these cations form compounds with PO4
3− that are insoluble in water at high pH
• Precipitated by the addition of (NH4)2HPO4
Group 5• Group five cations are Na+, K+, NH4
+ • All these cations form compounds that are
soluble in water – they do not precipitate• Identified by the color of their flame
Complex Ion Formation
• Transition metals tend to be good Lewis acids• They often bond to one or more H2O molecules to
form a hydrated ion– H2O is the Lewis base, donating electron pairs to form
coordinate covalent bondsAg+(aq) + 2 H2O(l) Ag(H2O)2
+(aq)• Ions that form by combining a cation with several
anions or neutral molecules are called complex ions– e.g., Ag(H2O)2
+
• The attached ions or molecules are called ligands– e.g., H2O
Complex Ion Equilibria
• If a ligand is added to a solution that forms a stronger bond than the current ligand, it will replace the current ligand
Ag(H2O)2+
(aq) + 2 NH3(aq) Ag(NH3)2+
(aq) + 2 H2O(l) – generally H2O is not included, because its complex ion
is always present in aqueous solution
Ag+(aq) + 2 NH3(aq) Ag(NH3)2
+(aq)
Formation Constant
• The reaction between an ion and ligands to form a complex ion is called a complex ion formation reaction
Ag+(aq) + 2 NH3(aq) Ag(NH3)2
+(aq)
• The equilibrium constant for the formation reaction is called the formation constant, Kf
Practice – What is [HgI42−] when 125 mL of 0.0010 M KI is reacted with 75
mL of 0.0010 M HgCl2?4 KI(aq) + HgCl2(aq) 2 KCl(aq) + K2HgI4(aq)
Practice – What is [HgI42−] when 125 mL of 0.0010 M KI is reacted with 75 mL of
0.0010 M HgCl2?4 KI(aq) + HgCl2(aq) 2 KCl(aq) + K2HgI4(aq)
Write the formation reaction and Kf expression.Look up Kf value
determine the concentration of ions in the diluted solutions
Hg2+(aq) + 4 I−(aq) HgI42−(aq)
Practice – What is [HgI42−] when 125 mL of 0.0010 M KI is reacted with 75 mL of
0.0010 M HgCl2?4 KI(aq) + HgCl2(aq) 2 KCl(aq) + K2HgI4(aq)
Create an ICE table. Because Kf is large, assume all the lim. rgt. is converted into complex ion, then the system returns to equilibrium.
[Hg2+] [I−] [HgI42−]
initial 3.75E-4 6.25E-4 0
change ≈¼(−6.25E-4) ≈−(6.25E-4) ≈¼(+6.25E-4)
equilibrium 2.19E-4 x 1.56E-4
Hg2+(aq) + 4 I−(aq) HgI42−(aq)
I− is the limiting reagent
Practice – What is [HgI42−] when 125 mL of 0.0010 M KI is reacted with 75 mL of
0.0010 M HgCl2?4 KI(aq) + HgCl2(aq) 2 KCl(aq) + K2HgI4(aq)
substitute in and solve for x
confirm the “x is small” approximation
2 x 10−8 << 1.6 x 10−4, so the approximation is valid
Hg2+(aq) + 4 I−(aq) HgI42−(aq)
[HgI42−] = 1.6 x 10−4
[Hg2+] [I−] [HgI42−]
initial 3.75E-4 6.25E-4 0
change ≈¼(−6.25E-4) ≈−(6.25E-4) ≈¼(+6.25E-4)
equilibrium 2.19E-4 x 1.56E-4
The Effect of Complex Ion Formation on Solubility
• The solubility of an ionic compound that contains a metal cation that forms a complex ion increases in the presence of aqueous ligands
AgCl(s) Ag+(aq) + Cl−
(aq) Ksp = 1.77 x 10−10
Ag+(aq) + 2 NH3(aq) Ag(NH3)2
+(aq) Kf = 1.7 x 107
• Adding NH3 to a solution in equilibrium with AgCl(s) increases the solubility of Ag+
Solubility of Amphoteric Metal Hydroxides
• Many metal hydroxides are insoluble• All metal hydroxides become more soluble in acidic
solution– shifting the equilibrium to the right by removing OH−
• Some metal hydroxides also become more soluble in basic solution– acting as a Lewis base forming a complex ion
• Substances that behave as both an acid and base are said to be amphoteric
• Some cations that form amphoteric hydroxides include Al3+, Cr3+, Zn2+, Pb2+, and Sb2+
Al3+
• Al3+ is hydrated in water to form an acidic solutionAl(H2O)6
3+(aq) + H2O(l) Al(H2O)5(OH)2+
(aq) + H3O+(aq)
• Addition of OH− drives the equilibrium to the right and continues to remove H from the molecules
Al(H2O)5(OH)2+(aq) + OH−
(aq) Al(H2O)4(OH)2+
(aq) + H2O (l)
Al(H2O)4(OH)2+
(aq) + OH−(aq) Al(H2O)3(OH)3(s) + H2O
(l)