Acid/Base Properties of Salts
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Transcript of Acid/Base Properties of Salts
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Acid/Base Properties of Salts
Hiding in plain sight
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Recognizing Bases
Sometimes it seems that all acids and bases are labeled with H+ or OH-
Remember that a base is ANYTHING that can accept a proton
The salt formed by the dissociation of an acid is the “conjugate base” of the acid. This must mean that the salt is a base, whether it has an OH- or not.
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Pick a salt, any salt
How about sodium acetate, NaOAc? An excellent choice.
Sodium acetate is an ionic solid. Ionic solids dissociate in aqueous
solution Aqueous NaOAc will exist as anions
and cations.
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NaOAc in aqueous solution
NaOAc (aq) Na+(aq) + OAc-
(aq)
So what can we say about Na+ and OAc- in aqueous solution?
We need to think “backwards”. How did NaOAc get the Na+ in the first place? (Or, more accurately, what is one way it could have gotten it?)
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Salts are products of acid/base reactions
You might recall that the reaction of an acid and a base yields a salt and water:
NaOH + HOAc H2O + NaOAc
Keep in mind, this is the “net reaction”, it doesn’t give any detailed information on mechanisms.
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A more detailed reaction
Think of a titration:
NaOH (aq) + HOAc (aq) H2O (l) + NaOAc (aq)
But what are NaOH (aq) and HOAc (aq)?
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Acid (and Base) Dissociation Reactions
NaOH(aq) + H2O(l) H-OH (l) + OH-(aq) + Na+
(aq) Kb
Or: NaOH Na+(aq) + OH-
(aq)
OH- (aq) + H2O(l) H-OH (l) + OH-(aq) Kb
HOAc (aq) + H2O(l) H3O+(aq)
+ OAc-(aq) Ka
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Putting it all together…
NaOAc (aq) Na+(aq) + OAc-
(aq)
NaOH(aq) + H2O(l) OH-(aq) + Na+
(aq) + H-OH (l)
HOAc (aq) + H2O(l) H3O+(aq)
+ OAc-(aq)
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Putting it all together…
NaOAc (aq) Na+(aq) + OAc-
(aq)
NaOH(aq) + H2O(l) OH-(aq) + Na+
(aq) + H-OH (l)
Base acid conjugate conjugatebase acid
HOAc (aq) + H2O(l) H3O+(aq)
+ OAc-(aq)
Base acid conjugate conjugateacid base
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In short…
We have the conjugate acid (Na+) of a base (NaOH), and the conjugate base (OAc-) of an acid (HOAc).
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You can always tell…
Take the anion and add an H+, that’ll tell you what acid the anion “came from”.
Take the cation and either add an OH- or take away an H+ and that will tell you what base the cation came from.
Ignore anything strong – it won’t, it CAN’T go back!!!
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Kb becomes Ka as Ka becomes Kb
NaOAc (aq) Na+(aq) + OAc-
(aq)
Na+(aq) + 2 H2O (l) NaOH(aq) + H3O+
(aq)
(or, if you prefer)
Na+(aq) + H2O (l) NaOH(aq) + H+
(aq) Ka=Kw/Kb
OAc-(aq) + H2O(l) OH-
(aq) + HOAc (aq) Kb=Kw/Ka
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But one is strong and the other weak.
NaOH(aq) + H2O(l) H-OH (l) + OH-(aq) + Na+
(aq) Kb=
Na+(aq) + H2O (l) NaOH(aq) + H+
(aq) Ka=Kw/Kb = 0
HOAc (aq) + H2O(l) H3O+(aq)
+ OAc-(aq) Ka =1.8x10-5
OAc-(aq) + H2O(l) OH-
(aq) + HOAc (aq) Kb=Kw/Ka
Kb = 10-14/1.8x10-5=5.56x10-10
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Net Result
NaOAc gives rise to a single equilibrium reaction that must be considered:
OAc-(aq) + H2O(l) OH-
(aq) + HOAc (aq)
Kb=5.56x10-10
The salt is a base!!!!
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Sample problem
What is the pH of a 0.100 M solution of NaOAc?
What do we need?
Balanced equation
What is it?
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NaOAc (aq) Na+ (aq) + OAc- (aq)
Where does Na+ “come from”?NaOH
Where does the OAc- (aq) “come from”?
HOAc
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NaOAc (aq) Na+ (aq) + OAc- (aq)
NaOHStrong Base
HOAcWeak Acid
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OAc-(aq) + H2O (l) HOAc (aq) + OH-
(aq)
OAc-(aq) + H2O (l) HOAc (aq) + OH-
(aq)
I 0.100 M - 0 0C –x - +x +xE 0.100-x - x x
Now all we need is K!Kb(OAc-) = Kw/Ka(HOAc)
= 1x10-14/1.8x10-5 = 5.56x10-10
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OAc-(aq) + H2O (l) HOAc (aq) + OH-
(aq)
OAc-(aq) + H2O (l) HOAc (aq) + OH-
(aq)
I 0.100 M - 0 0
C –x - +x +x
E 0.100-x - x x
Kb = 5.56x10-10 = [OH-][HOAc]
[OAc-]
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OAc-(aq) + H2O (l) HOAc (aq) + OH-
(aq)
Kb = 5.56x10-10 = [OH-][HOAc]
[OAc-]5.56x10-10 = x*x 0.100-xAssume x<<0.15.56x10-11 = x2
7.45x10-6 = x, GOOD assumption
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OAc-(aq) + H2O (l) HOAc (aq) + OH-
(aq)
OAc-(aq) + H2O (l) HOAc (aq) + OH-
(aq)
I 0.100 M - 0 0C –7.45x10-6 - + 7.45x10-6 + 7.45x10-6
E 0.100 - 7.45x10-6 7.45x10-6
pOH = -log (7.45x10-6) = 5.12pH = 14 – 5.12 = 8.88
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More fun with salts…
What can we say about:
NaCl?
NaCl = Na+ + Cl-
It came from…
Cation + OH- = NaOHAnion + H+ = HCl
So….
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If you can’t find Ka …
…maybe it’s a strong acid.Or if you do find it, it is HUGE!
Strong acids:HCl Ka ~106
HBrHIHNO3
HClO4
H2SO4 (diprotic – more on this next week)
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If you can’t find a Kb…
…maybe it’s a strong base. (Or Kb is huge.)
List of strong bases:LiOHNaOH Kb~108
KOHSr(OH)2
Ca(OH)2
Ba(OH)2
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What kind of acid/base is it?
NaOH?STRONG base
HCl?STRONG acid
The salt is…NEUTRAL we “ignore both”
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More fun with salts…
What can we say about:KCl
Came from:KOH and HCl
Also neutral!
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More fun with salts…
What can we say about:Ca(OAc)2
Ca(OAc)2 → Ca2+ + 2 OAc-
It came from:
Ca(OH)2 and HOAc
Only the HOAc matters, it’s weak.
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More fun with salts…
What can we say about:NH4Cl
It came from?NH3 (or NH4OH) and HCl
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More fun with salts…
What can we say about:NH4OAc
It came from:NH3 and HOAc
BOTH WEAK!!!
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If they are both weak…
Look at the K. Biggest K wins.
Although for NH4OAc..
Ka(NH3) = Kw = 1.00x10-14 = 5.68x10-10
Kb(NH3) 1.76x10-5
Kb(HOAc) = Kw = 1.00x10-14 = 5.59x10-10
Ka(HOAc) 1.79x10-5
NH4OAc ends up being neutral!
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If they don’t cancel…???
We won’t worry about the exact calculation (you’d need to do both simultaneous equilibria), but you can tell whether it is acidic or basic:
NH4IO3 (ammonium iodate)
It comes from:NH3 and HIO3
Kb(NH3) = 1.76x10-5
Ka(HIO3) = 1.7x10-1
HIO3 is a better acid than NH3 is a base. Which means that NH4
+ is a better acid than IO3- is a base.
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If they don’t cancel…???
NH4IO3 (ammonium iodate)
It comes from:NH3 and HIO3
Kb(NH3) = 1.76x10-5
Ka(HIO3) = 1.7x10-1
HIO3 is a better acid than NH3 is a base. Which means that NH4
+ is a better acid than IO3- is a base.
A solution of NH4IO3 would be acidic.
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Another little problem
What is the pH of a 0.250 M solution of Ca(IO3)2?
Where do we start?
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Split the salt into ions
Ca(IO3)2 Ca2+ + 2 IO3-
Where did the ions come from?
Ca(OH)2 and HIO3
Which are…?
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Ca(OH)2 is strong
HIO3 is weak.
Ignore the Ca2+
Now, we need to…
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Write a balanced equation!
IO3-(aq) + H2O (l) HIO3(aq) + OH-(aq)
Once I have a balanced equation:ICE chartK equation
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ICE ICE BABY ICE ICE
IO3-(aq) + H2O (l) HIO3(aq) + OH-(aq)
-
-x - +x +x
-
I
C
E
What else?
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Another little problem
What is the pH of a 0.250 M solution of Ca(IO3)2?
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ICE ICE BABY ICE ICE
IO3-(aq) + H2O (l) HIO3(aq) + OH-(aq)
0.500 M - 0 0
-x - +x +x
0.500-x - x x
I
C
E
Now K
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K = [HIO3][OH-] [IO3
-]
This is the Kb of IO3
Kb(IO3) = Kw/Ka
Ka(HIO3) = 1.7x10-1
Kb = 1x10-14/1.7x10-1 = 5.88x10-14
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5.88x10-14 = [HIO3][OH-]
[IO3-]
5.88x10-14 = [x][x] [0.5-x]
How do we solve this?
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Assume x<<0.5
5.88x10-14 = [x][x] = [x][x] [0.5-x] 0.5
2.941x10-14 = x2
1.715x10-7 = x
Good assumption?YES, MA’AM!
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ICE ICE BABY ICE ICE
IO3-(aq) + H2O (l) HIO3(aq) + OH-(aq)
0.500 M - 0 0
- 1.715x10-7 - + 1.715x10-7
+ 1.715x10-7
0.500 - 1.715x10-7 1.715x10-7
I
C
E
How do we finish?
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Another little problem
What is the pH of a 0.250 M solution of Ca(IO3)2?
[OH-] = 1.715x10-7 pOH = - log (1.715x10-7) = 6.77
pH = 14 – pOH = 14 – 6.77 = 7.23
Make sense?You bet! Strongish acid (HIO3), very weak conjugate.
Almost neutral.