Transcript file
Acid-base equilibria & common ions
Consider solution containing HF (weak acid)
and salt NaF
What effect does presence of NaF have on
dissociation equilibrium of HF?
HF(aq) H+(aq) + F-(aq)
The Common Ion Effect
Equilibrium calculations involving
common ions
34.6 g of NH4Cl is added to 3.98 L of a 0.0145 M solution of
NH3. Kb(NH3) = 1.8 x 10-5.
What is the pH of the original solution before the addition of
NH4Cl?
NH3(aq) + H2O(l) NH4+(aq) + OH-(aq)
So – what’s the pH of the solution after the addition of the
NH4Cl (assume that the volume stays constant)?
Buffered solutions
Buffer: A solution that resists a change in its pH
when either H+ or OH- ions are added
Buffers can contain weak acids & salts or weak
bases & salts (solutions can be buffered at almost
any pH)
How does buffering work?
Buffer contains relatively large concs. of weak acid HA and its
conjugate base A-.
Add OH- ions into solution, what happens? (OH- is a strong base
– it will look for H+) Weak acid HA is best source of H+.
OH- + HA H2O + A-
OH- ions cannot accumulate (replaced by A- ions)
Original buffered
solution pH
OH- added
added OH- ions
replaced by A- ions
Final pH of buffer
close to original pH
How does buffering work?
If OH- ions converted to A- ions, how does the pH stay so
stable? Look at eq. expression for [H+]; dissociation equilibrium
for HA
+
Ka = [H
][A
______]
[HA]
or
[H+] = Ka[HA]
___
[A-]
The essence of buffering: [HA] and [A-]
are very large relative to [OH-] added.
The essence of buffering
Original [HA]/[A]
ratio
OH- added
Final [HA]/[A]
ratio close to original
Added OH- changes HA to A-,
but [HA] and [A-] are large
compared to [OH-] added
Buffer solution pH calculation
Buffered solution contains 0.50 M HC2H3O2 (Ka = 1.8x10-5) and
0.50 M NaC2H3O2. Calculate pH of solution.
HC2H3O2(aq) C2H3O2-(aq) + H+(aq)
pH changes in buffered solutions
Calculate change in pH that occurs when 0.010 mol solid
NaOH added to 1.0 L of buffered solution from last
question.Compare this with pH change that occurs when 0.010
mol solid NaOH added to 1.0 L of H2O.
Expected: pH of buffered solution will change very little; pH of
water will change by much larger amount.
How to handle these problems
Original buffered
solution pH
Step 1: Do stoichiometry
calculations; assume reaction
with H+/OH- goes to completion
Modified pH
Step 2: Do equilibrium
calculations
Adding H+, rather than OH
Exactly the same thinking applied when H+ added to buffered
solution of weak acid and conjugate base salt. A- (conjugate
base) has high affinity for added H+, and will form weak acid
HA; H+ + A- HA
H+ ions cannot accumulate (replaced by HA)
Net change of A- to HA, but if [A-] and [HA] are very large
compared to [H+], little pH change will occur (as expected, in
buffered solution).
[H+] = Ka[HA]
___
[A-]
Buffer Capacity
Amount of H+/OH- a buffer can absorb without a considerable
change in pH
Buffer 1:
Ka = [NH3][H+] / [NH4+]
[H+] = Ka. [NH4+] / [NH3] = 5.6 x 10-10 M; pH = ?
What happens to the pH if 0.1 moles of NaOH is added?
After the reaction with NaOH, calculate the [H+] and thus the
pH….
1.0 M NH3, and 1.0 M NH4+
Buffer Capacity
Buffer 2: 0.01 M NH3, and 0.01 M NH4+
Ka = [NH3][H+] / [NH4+]
[H+] = Ka. [NH4+] / [NH3] = 5.6 x 10-10 M; pH = ?
What happens to the pH if 0.1 moles of NaOH is
added?
After the reaction with NaOH, calculate the [H+] and
thus the pH….
Henderson-Hasselbalch equation
Useful equation; allows for the calculation of buffer pH if the
concentration of weak acid (HA) and conjugate base (A-)are
known.
A more convenient method of pH calculation – will give the same
answer as with our previous method of pH calculation.
Buffered solution contains 0.50 M HC2H3O2 (Ka = 1.8x10-5) and
0.50 M NaC2H3O2. Calculate pH of solution (we did this a few
slides ago, but let’s check that H-H equation gives us the same
answer).
Henderson-Hasselbalch equation
Titrations / pH curves
Experimental method to determine the concentration of an acid
(or base).
Think about this: If you were given 25 mL of a unknown
concentration of HCl, a solution of 0.100 M NaOH and a burette,
how could you determine the concentration of the HCl?
What would happen to the pH of the solution of HCl as the NaOH
was added to it?
Titrations / pH curves
Titration Calculations
Solubility Equilibria
Fluoride – used to combat tooth decay
One product of F- reaction at site of teeth is CaF2(s)
CaF2(s) Ca2+(aq) + 2F-(aq) (dissolving in water)
Consider equlibrium set up between these species:
CaF2(s) Ca2+(aq) + 2F-(aq)
Equilibrium Expression for this process?
Ksp = [Ca2+][F-]2
(Why is CaF2 not included in expression?)
Ksp = Solubility Product Constant (solubility product)
Solubility vs Ksp
Ksp: Equilibrium Constant (solubility product)
Solubility: Equilibrium Position
Copper (I) bromide has a measured aqueous solubility of 2.0 x
10-4 mol/L at 25 °C. Calculate its Ksp value.
Equilibrium reaction?
Equilibrium Expression? Ksp =
(equilibrium concentrations)
Initial Concentrations?