chapter 16 aqueous ionic equilibrium. ph / poh calculations strong acids strong bases weak acids...
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Chapter 16
Aqueous Ionic Equilibrium
pH / pOH Calculations
• Strong acids
• Strong bases
• Weak acids
• Weak bases
• Salts
NaAc(aq) → Na+(aq) + Ac−(aq)
Calculate pH for a mixture of HAc and Ac−
HAc(aq) ⇌ H+(aq) + Ac−(aq)acid
Ac−(aq) + H2O(l) ⇌ HAc(aq) + OH−(aq)base
NH4Cl(aq) → NH4+(aq) + Cl−(aq)
Calculate pH for a mixture of NH3 and NH4+
NH4+(aq) H⇌ +(aq) + NH3(aq)
acid
NH3(aq) + H2O(l) NH⇌ 4+(aq) + OH−(aq)
base
Calculate the pH of a solution that contains 0.100 M HAc
and 0.100 M NaAc. Ka for HAc is 1.8 x 10−5.
Example 16.1, page 716
pH = 4.74
weak acid + its conjugate base = buffer solution
weak base + its conjugate acid = buffer solution
Adding H+ or OH−, pH does not change too much
Calculate the change in pH that occurs when 0.010 mol solid
NaOH is added to 1.0 L of the buffered solution described in
Example 16.1. Compare this pH change with that which
occurs when 0.010 mol solid NaOH is added to 1.0 L of water.
Assume there is no volume change after solid NaOH is added.
Example 16.3, page 722
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[A ]pH = pK + log
[HA]
Henderson-Hasselbalch equationfor buffer solutions
pH for buffer solutions: ICE → exact answer
pH for buffer solutions: approximation
Calculate the pH of a solution that contains 0.50 mol/L HAc
and 0.50 mol/L NaAc. Ka for HAc is 1.8 x 10−5.
pH = 4.74
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[A ]pH = pK + log
[HA]
Example 16.1, page 716. revisited
Calculate the pH of a buffer solution that is 0.050 M in
benzoic acid (HC7H5O2) and 0.150 M in sodium benzoate (NaC7H5O2).
For benzoic acid, Ka = 6.5 10–5.
Example 16.2, page 718.
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[A ]pH = pK + log
[HA]
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[A ]pH = pK + log
[HA]
Use the Henderson–Hasselbalch equation to calculate
the pH of a buffer solution that is 0.50 M in NH3 and
0.20 M in NH4Cl. For ammonia, pKb = 4.75.
Example 16.4, page 725.
Calculate the change in pH that occurs when 0.010 mol gaseous
HCl is added to 1.0 L of each of the following solutions:
Solution A: 5.00 mol/L HAc and 5.00 mol/L NaAc
Solution B: 0.050 mol/L HAc and 0.050 mol/L NaAc
Ka for HAc is 1.8 x 10−5.
HAc Ac− HAc Ac−
H+
HAc Ac−
H+
Solution A
HAc Ac−
Solution B
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[A ]pH = pK + log
[HA]
Buffer capacity
To maximize buffer capacity:
1) High [HA] and [A−]
2) [HA] = [A−] (example page 725-726)
pH = pKa
Titration
Titrate strong acid with strong base
Calculate the pH when the following quantities of
0.100 mol/L NaOH solution have been added to
50.0 mL of 0.100 mol/L HCl solution
a)0 mL; b) 49.0 mL;c) 50.0 mL; d) 51.0 mL;e) 60.0 mL.
a)1.000; b) 3.00;c) 7.00; d) 11.00;e) 11.96
0 10 20 30 40 50 60 700
1
2
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4
5
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8
9
10
11
12
13
14
pH
mL of NaOH
equivalence point
Titrate weak acid with strong base
Calculate the pH when the following quantities of
0.100 mol/L NaOH solution have been added to
50.0 mL of 0.100 mol/L HAc solution. Ka of HAc
is 1.8 x 10−5.
a)0 mL; b) 49.0 mL; c) 50.0 mL;d) 51.0 mL; e) 60.0 mL.
A very similar example: page 733 -- 738
a)2.87; b) 6.43;c) 8.72; d) 11.00;e) 12.00
Chapter 17
The pH Curves for the Titrations of 50.0-mL Samples of 0.10 M Acids with Various Ka Values
with 0.10 M NaOH
Solubility Equilibrium
CaF2 (s) Ca2+ (aq) + 2F− (aq)H2O
Solubility equilibrium is established
Aqueous solution of CaF2 is saturated
Ksp = [Ca2+][F−]2
solubility product constant
“insoluble” salts
Write down the dissociation reactions and the expression of Ksp
Solid silver chromate is added to pure water at 25 °C.
Equilibrium is achieved between undissolved Ag2CrO4(s)
and its aqueous solution. Silver ion concentration is
1.3 x 10−4 mol/L. Calculate Ksp for this compound.
CaF2 (s) Ca2+ (aq) + 2F− (aq)H2O
Solubility equilibrium is established
Aqueous solution of CaF2 is saturated
Solubility: the concentration of a saturated solution.
Unit: mol/L or ∙ ∙ ∙
solubility ≠ solubility product constant Ksp
Solubility ↔ Ksp
value of Kc or Kp
equilibrium concentrations or pressures
Copper(I) bromide has a measured solubility of 2.0 x 10−4 mol/L
at 25 °C. Calculate its Ksp.
Ksp = 4.0 x 10−8
Bismuth sulfide (Bi2S3) has a measured solubility of
1.0 x 10−15 mol/L at 25 °C. Calculate its Ksp.
Ksp = 1.1 x 10−73
Try example 16.9 on page 746
The Ksp for copper(II) iodate, Cu(IO3)2, is 1.4 x 10−7 at 25 °C.
Calculate its solubility at 25 °C.
3.3 x 10−3 M
Try example 16.8 on page 745
Calculate the solubility of CaF2 (Ksp = 1.46 x 10−10) in a
0.100 mol/L NaF solution.
Example 16.10, page 747
CaF2 (s) Ca2+ (aq) + 2F− (aq)H2O
Increase [Ca2+] or [F−] → equilibrium shifts to left → solubility ↓
common ion effect
For Mg(OH)2 Ksp = 1.8 x 10−11. What is the pH of a saturated
solution of Mg(OH)2? What is its solubility?
Suppose that solid Mg(OH)2 is equilibrated with a solution
buffered at a more acidic pH of 9.00. Ksp of Mg(OH)2 is 1.8 x 10−11
What are the [Mg2+] and solubility?
CaF2 (s) Ca2+ (aq) + 2F− (aq)H2O
Q = [Ca2+][F−]2
If Q > Ksp, precipitate will form
If Q < Ksp, precipitate will not form
A solution is prepared by adding 750.0 mL of 4.00 x 10−3 mol/L
Ce(NO3)3 to 300.0 mL of 2.00 x 10−2 mol/L KIO3. Will Ce(IO3)3
(Ksp = 1.9 x 10−10) precipitate from this solution?
Try example 16.12 on page 750
1 Ka for acetic acid (CH3COOH) is 1.8 x 10-5 while Ka for hypochlorous (HClO) ion is 3.0 x 10-8. A. Which acid is the stronger acid?B. Which is the stronger conjugate base? Acetate ion (CH3COO-) or chlorous (ClO-) ion?C. Calculate kb values for CH3COO- and ClO-. 2. a. Calculate the pH of a 1.50 L solution containing 0.750 mole of HCN and 0.62 mole of KCN. Ka = 4.0 x 10-10
b. If 0.015 mole of KOH was added, calculate the pH of the solution.
c. If 0.015 mole of HBr was added, calculate the pH of the solution.