Drill: A 0.100 M solution of HZ ionizes 20.0 %. Calculate: K aHZ
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Drill: A 0.100 M solution of HZ ionizes 20.0 %. Calculate: K aHZ
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Drill: A 0.100 M solution of HZ ionizes 20.0 %. Calculate: K aHZ. Buffer Solutions. Buffer Solution. A solution that resists changes in pH. Buffer Solution. Made from the combination of a weak acid & its salt. Buffer Solution. Made from the combination of a weak base & its salt. - PowerPoint PPT Presentation
Transcript of Drill: A 0.100 M solution of HZ ionizes 20.0 %. Calculate: K aHZ
Drill: A 0.100 M solution of HZ ionizes 20.0 %.Calculate: KaHZ
Buffer Solutions
Buffer Solution
•A solution that resists changes in
pH
Buffer Solution•Made from the combination of a weak acid & its
salt
Buffer Solution•Made from the combination of a weak base & its
salt
Buffer Examples•Mix acetic acid &
sodium acetate
•Mix ammonia & ammonium chloride
Buffer Solution•A buffer solution
works best when the acid to salt ratio is
1 : 1
Buffer Solution•A buffer solution
works best when the base to salt ratio is
1 : 1
Buffer Solution•The buffering capacity of a solution works best when the pH is near the
pKa
pKa or pKb
•pKa = - log Ka
•pKb = - log Kb
BufferEquilibria
To solve buffer equilibrium
problems, use the same 5 steps
5 Steps of Equilibrium Problems
1) Set up & balance reaction
5 Steps of Equilibrium Problems
2) Assign Equilibrium amounts in terms of x
(ICE)
5 Steps of Equilibrium Problems
3) Write the equilibrium expression (K = ?)
5 Steps of Equilibrium Problems
4) Substitute Equilibrium amounts into the K
5 Steps of Equilibrium Problems
5) Solve for x
Buffer Problems•Calculate the pH of a
solution containing
0.10 M HAc in 0.10 M NaAc: Ka = 1.8 x 10-5
Buffer Problems•Calculate the pH of
0.10 M NH3 in
0.20 M NH4NO3:
•Kb = 1.8 x 10-5
Buffer ProblemsCalculate the pH of a
solution containing
0.10 M HBz in 0.20 M NaBz: Ka = 6.4 x 10-5
Drill:Calculate the pH of a
solution containing
0.30 M HZ in 0.10 M NaZ: Ka = 3.0 x 10-5
Buffer ProblemCalculate the pH of a
solution containing
0.50 M R-NH2 in 0.10 M R-NH3I: Kb = 4.0 x 10-5
Derivations from an
equilibrium constant
HA H+ + A-
[H+][A-]
[HA]Ka =
HA H+ + A-
[H+][A-]
[HA]Ka =
Cross multiply to isolate [H+]
HA H+ + A-
[Ka][HA]
[A-][H+]=
HA H+ + A-
[HA]
[A-][H+] = (Ka)
HA H+ + A-
[HA]
[A-][H+] = (Ka)
Take –log of each side
pH =
[HA]
[A-]pKa - log
Henderson-Hasselbach Eq
[A-]
[HA]pH = pKa + log
Henderson-Hasselbach Eq
[B+]
[B]pOH = pKb+ log
Buffer Problems•Calculate the salt to acid
ratio to make a buffer solution with pH = 5.0
•Ka for HBZ = 2.0 x 10-5
Derivations from an
equilibrium constant
HA H+ + A-
[H+][A-]
[HA]Ka =
[H+][A-]
[HA]Ka =
Divide both sides by [H+]
Ka [A-]
[H+] [HA]=
You Get the Salt to Acid Ratio
Drill:•Calculate the salt to acid
ratio to make a buffer solution with pH = 5.0
•Ka for HBZ = 2.0 x 10-5
Buffer ProblemsCalculate the salt to base
ratio to make a buffer solution with pH = 9.48
•Kb for MOH = 2.0 x 10-5
Equivalence PointPoint at which the # of moles of the two
titrants are equal
Titration Curves
0
2
4
6
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10
12
14
0.00 10.00 20.00 30.00 40.00 50.00
0
2
4
6
8
10
12
14
0.00 10.00 20.00 30.00 40.00 50.00
[HA]=[A-]
[HA]=[OH-]
Drill:Calculate the pH of a buffer solution containing 0.50 M HX in 0.25 M KX.
Ka = 2.5 x 10-5
0
2
4
6
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10
12
14
0 20 40 60 80 100
[H2A] = [HA-]
[H2A] = [OH-]
[HA-] = [A-2]
[OH-] = [A-2]
Calculate the HCO3- to
H2CO3 ratio in blood with pH = 7.40
•Ka1 for H2CO3 = 4.4 x 10-7
150 ml of 0.10 M NaOH is added to 100.0 ml of
0.10 M H2CO3. Calculate pH.
•Ka1 for H2CO3 = 4.4 x 10-7
•Ka2 for H2CO3 = 4.8 x 10-11
