Neutralization
Recall
(from grade 10 and 11) that the reaction of an acid and a base results in the production of a salt
(ionic compound) and water.
HCl(aq)
+ NaOH(aq) →
NaCl(aq) + H2O(l)
Salts
If
the above reaction proceeds in the opposite direction, the salt may react with
water by splitting the ionic compound and
forming an acid or base. Salts will
produce either acidic, basic or neutral solutions when placed in water. The reaction of an ion with water to form an
acidic or basic solution is called hydrolysis.
Example:
What type of solution will sodium chloride form when placed in water?
Answer: The
solution to this question involves three parts:
First,
show the dissociation of the solid salt into its aqueous ions.
NaCl(s) → Na+(aq) + Cl-(aq)
Second,
show the predicted products of the reaction of the cation with water. Critically examine those products. In this case, the reaction would not occur because
NaOH is a strong base and it would dissociate completely in solution, thus it
could not form as an intact entity. So, the arrow employed points in the direction
of reactants. This tells us that there is no hydrolysis.
Na+(aq) + H2O(l) ← NaOH(aq)
+ H+(aq)
Third,
show the predicted products of the reaction of the anion with water. Critically examine those products. In this case, the reaction would not occur because
HCl is a strong acid and it would dissociate completely in solution, thus it
could not form as an intact entity. So, the arrow employed points in the direction
of reactants. This tells us that there is no hydrolysis.
Cl-(aq) + H2O(l) ←
HCl(aq) +OH-(aq)
Noting
that neither aqueous ion undergoes hydrolysis means that sodium chloride forms a neutral solution.
Here’s
another example, omitting all the wordy guidance. This is how I would expect you to answer a
question of this type:
Example:
What type of solution will potassium fluoride form when placed in water?
Answer:
KF(s) →
K+(aq) + F-(aq)
K+(aq) + H2O(l) ←
KOH(aq) + H+(aq) ∴ K+ does not hydrolyze
F-(aq) + H2O(l) ↔
HF(aq) + OH-(aq) ∴ F- does hydrolyze
∴ potassium fluoride forms a basic
solution
Okay,
maybe a few words. Because K+
does not react, it has no effect on the solution’s pH. Because the F- does react (undergoes
hydrolysis), it will affect the solution’s pH.
Since the F- is accepting a proton from water, it is acting as
a base (also, notice that OH- is being produced). So, in this case we have a basic
solution.
Salt Hydrolysis Summary
- salt of a strong acid and a strong base ↝ neutral (ex. NaCl, NaNO3)
- salt of a weak acid and a strong base ↝ basic (ex. KF, NaAc)
- salt of a strong acid and a weak base ↝ acidic (ex. NH4Cl)
- salt of a weak acid and a weak base ↝ it depends (ex. NH4Ac) ↝ in this case, both ions will undergo hydrolysis; you would need to find the appropriate K value for both ions and if Ka > Kb then the solution is acidic, if Ka < Kb then the solution is basic and if Ka = Kb then the solution is neutral
Predicting the Acidic or Basic Nature
of a Salt Solution
This
is an example of my favourite kind of question.
It has it all. Just like me. Hmm, maybe that's why it's my favourite...
It is just like the question we did in the last lesson. However, you no longer need to
trust that it is the anion that is reacting.
Now, you can prove to me which part of the salt undergoes hydrolysis to
affect the solution’s pH. So, start by
showing the three lines for salt hydrolysis (as taught above), then get into
the ICE table and math (as shown in the last lesson).
Example:
What is the pH of a 0.10 M NaF solution?
Answer:
NaF(s)
→ Na+(aq) + F-(aq)
Na+(aq) + H2O(l) ←
NaOH(aq) + H+(aq) 😒 hydrolysis doesn't occur
F-(aq) + H2O(l) ↔
HF(aq) +OH-(aq) 😁 hydrolysis does occur
I 0.10 ----- 0 0
C -x ----- +x +x
E
0.10 – x
----- x x
KaKb = Kw
(6.8 x 10-4)Kb = 1.00 x
10-14
Kb = 1.5 x 10-11
Kb = [HF][OH-]/[F-]
1.5 x 10-11 = x2/(0.10
– x)
x = 1.2 x
10-6
pOH
= -log [OH-]
pOH
= -log(1.2 x 10-6)
pOH
= 5.92
pH
+ pOH = 14.00
pH
+ 5.92 = 14.00
pH = 8.08
As an aside, students are sometimes flummoxed by the ammonium ion and how it behaves in water. I have created this video to clear up any confusion.
