Suppose you have H3O from added HCl 0.1 M 101 M

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Suppose you have H3O from added HCl 0.1 M
10-1 M
Then 10 -1OH- 10-14
In pure H2O, OH- H3O 10-7, but
Weak Acids and Bases
Add acetic acid to H2O
KKa
What are CH3COOH, H3O, CH3COO-?
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If we ignore H3O from H2O H2O ? H3O OH- ?
CH3COOH H2O ? H3O CH3COO-
(Stoichiometry)
? have H3O CH3COO-
H3O from H2O ? 10-7
Let Co initial CH3COOH? HOAc
Small K means equilibrium is far to left!
KKa
Guess, HOAc? Co , since KKaltltlt1 (? H3Oltltlt
Co)
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Co - H3O 0.9957 M ? Co 1 M (0.43 error)
Example Co 0.01 M ? H3O 4.3 ? 10-4
Co - H3O 10 ? 10-4 - 1.36 ? 10-4 8.64 ?
10-4 vs Co 10 ? 10-4 (13.6 error!!)
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The approximation that HOAc ? Co is good
only for small Ka, large Co.
Note In all cases H3O gtgt10-7 Thus, OK to
neglect H3O from H2O H2O ? H3O OH-
Could obtain an exact solution by solving a
quadratic
Such equations are trivial to solve with a modern
graphing calculator.
(KKa)
A powerful, approximate alternative method
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Method of Successive Approximations
Take OH-gtgt10-7 (ignore hydrolysis of
H2O)? OH-CH3NH3
(Stoichiometry)
Let initial concentration CH3NH2 Co 0.1
First Approximation OH- CH3NH3 x1
CH3NH2 Co - x1, Assume x1 ltlt Co ?
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x1 7.1 ? 10-3
Co - x1 (100 - 7.1) ? 10-3
CH3NH2 H2O ? CH3NH3 OH- (weak base) Co - x 1
? Co x 1 x 1
Second Approximation OH- CH3NH3 x2
(unknown)
(x2)2 (92.9 ? 10-3)(5.0 ? 10-4) 46.45 ? 10-6
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Value we chose was only off by 0.3 out of 93.2
(about 0.3)
Try further iterations (x3) and will find no
change in values to three significant figures.
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Hydrolysis
HOAc H2O ? H3O OAc -
Ka 1.85 ? 10-5
HOAcCH3COOH, OAc - CH3COO -
HOAc is an acid, therefore OAc- is a (conjugate)
base
Add NaOAc to H2O, which dissociates completely to
Na, OAc -
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KaOAc -H3O/HOAc is the ionization
constant for the reaction
HOAc H2O H3O OAc -
Note Kh is very small, indicating that little
OAc - combines With H2O to form HOAc
Since KhKw/Ka , the smaller Ka (weaker the acid)
the Larger Kh (or OAc - is more extensively
hydrolyzed)
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pKa -log10(Ka)
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Buffer Solutions
Consider acetic acid HOAc H2O ? H3O OAc
- HOAcCH3COOH, OAc - CH3COO -
HOAc HOAco (initial concentration HOAc)
e.g., add 0.70 mole of HOAc to make 1 liter of
solution
Now suppose add 0.60 moles NaOAc, which
dissociates completely to Na, OAc -
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First, look at HOAc H2O ? H3O OAc -
How keep Ka1.85x10-5 when throw in all of this
OAc - ?
This just uses up H3O that came from ionization
of HOAc, thereby forcing HOAc closer to HOAco
!!
Inverse teeter-totter effect
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Now, look at OAc - H2O ? HOAc OH -
Very little OH - made this way (lose little OAc -
and make little HOAc) since
So when put 0.6 moles OAc - in a liter of pure
water have OAc -OAc -o 0.6 M
How to keep Kh HOAc OH- / OAc - 5.4
? 10 -10 when HOAc is large?
Result is to force OAc - even closer to its
initial value OAc -o 0.60 M
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Addition of large amounts of a weak acid and its
conjugate base to the same solution nullifies
both the ionization of the acid and the
hydrolysis of the base!!
Ka OAc - H3O / HOAc 1.85 ? 10 -5
Kh HOAcOH- / OAc - 5.4 ? 10 -10
OAc- ? 0.6 M and HOAc ? 0.7 M
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H3O2.185x10-5 M This result is accurate to 4
significant figures (gives same result as
solving quadratic to this accuracy)!
A solution containing substantial amounts of a
weak acid and its conjugate base is known as a
buffer solution.
Thus, have H3O acid/salt Ka Depends
only on Ka and the ratio of acid to salt
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Example 1 liter of 0.25 M HCl, add 0.600 moles
of NaOAc(s) Assume no volume change occurs upon
addition of salt.
Find OAc -, HOAc, H3O, OH-
NaOAc ? Na OAc - H2O HCl ?
H3O Cl-
Recognizing a buffer when you see it. Not always
an easy task.
Buffer conditions!
OAc - .350 HOAc
.250