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WEAK ACIDS AND BASES
[MH5; Chapter 13]
- Recall that a strong acid or base is one which completely ionizes in water.
HCR! H +^ + CR —
NaOH! Na +^ + OH—
- The above equations fit the Arrhenius definition of acids and bases; in water, acids produce H +^ and bases produce OH —^.
- The Brønstead-Lowry definition of acids and bases is better:
- An acid is a proton (H +^ ) donor.
- A base is a proton (H +^ ) acceptor.
- This means that in any acid-base reaction, a proton is transferred from the acid to the base.
- Consider the weak acid, HA: HA + H 2 O º H 3 O +^ + A—
- The acid has transferred a proton to the water (which in this case is acting as a base).
- The H 3 O +^ is what makes the solution acidic.
- The other species formed in this process is (^) A—^ ; we call this the conjugate base of the weak acid, HA.
- Now look at a weak base, B: B + H 2 O º BH +^ + OH —
- The base has taken a proton from the water (which in this case is behaving like an acid).
- The OH —^ is what makes the solution basic.
- The other species formed in this process is BH +^ ; this is the conjugate acid of the weak base B.
- Notice that water appears in both of these equations; it can either accept or donate a proton.
- Species such as this are termed amphiprotic or amphoteric.
- So, what is the difference between a strong acid (or base) and a weak acid (or base)?
- Strong acids and bases completely ionize in solution, whereas weak acids and bases do not.
- Because the ionization of a weak acid or base in water is incomplete, an equilibrium is established.
- It is, therefore, controlled by an equilibrium constant; K (^) a for acids and Kb for bases.
- For the weak acid, HA:
- For the weak base, B:
OH-^ H 2 O
_H+ + H+
H 3 O+
Calculations Involving Weak Acids
- Suppose that we have a solution of acetic acid, CH 3 COOH (Ka = 1.8 x 10 — 5^ ); let its initial concentration be represented by “ c ” molL —^. At equilibrium, “ x ” mol L —1^ have dissociated..........
CH 3 COOH º^ H+^ + CH 3 COO —
- This is a quadratic equation in “ x ”.
- The quadratic equation may be simplified by assuming that “ x ” is negligible compared to “ c ”.
- CH 3 COOH is a weak acid; so we may assume that the small amount dissociation makes little difference to the amount of undissociated acid left.
- Now let “c” = 0.10 molL —^ ..........
- If x is negligible compared to 0.10 molL —1^ :
So...
- In this case, x is about 1.3% of 0.10, a negligible error within the accuracy of the original data we were given.
- If we were to solve the equation exactly, x = 1.333 × 10—3^ M, so the approximate answer 1.342 × 10 —3^ M only differs from the exact answer by 0.009 × 10 —3^ , an ”error” of less than 1%.
General Guideline:
- This (^) assumption ; a ‘negligible’ amount of the acid being ionized - is valid if x # 5% of the initial concentration of acid......
- This will generally be the case if the concentration of the acid, c , divided by the Ka value is > 100.
EXAMPLE: [CH 3 COOH] = 0.10 M; Ka = 1.8 × 10 —
Test the assumption that c - x. c:
- What if we need to do the exact calculation?
- We have to solve the quadratic: x 2 + Ka x – Ka C = 0 (from expansion of K (^) a = x 2 /(C – x)
- What else can we calculate using K (^) a ??
- The Ka expression says:
- Depending on the information given, we could also calculate the initial concentration of the weak acid or the actual value of K (^) a...........
Strategy
- Always write the reaction for the weak acid given in the question.
- Write the equilibrium constant expression for the reaction.
- Identify what you know and what you are asked to find.
- Usually, it’s a fairly straightforward matter of substituting for the appropriate variables in the Equilibrium Constant Expression.
- Sometimes there are a few intermediate calculations to perform....
EXAMPLE: Nitrous acid, HNO 2 has a Ka value of 6.0 x 10 —^. a) Calculate the initial concentration of HNO 2 if a solution of this acid has a pH of 3.65.
b) Calculate the % ionization of HNO 2 in this solution.
EXAMPLE 1:
What is the % ionization and pH of a solution of a 0.0850 M solution of NH 3? [Kb = 1.8 x 10 —5^ ]
EXAMPLE 2:
Calculate the K (^) b for the weak base B, if a 0.00365 M solution of that base is 8.50 % ionized.
- How do conjugate species behave when they are put in water?
- Consider the generic weak acid, HA ; its conjugate base is A—.
- This species will act as a base in what is often called hydrolysis.
- What is Kb for A —^ , and how is it related to K (^) a for HA?
- Or...to get the same result, we could add the equations, and then (as we have seen earlier in these notes; p. 135) their K’s are multiplied:
- The relationship Ka x Kb = Kw is always true for a conjugate acid-base pair.
- The weaker the acid, the more basic is its conjugate base.
We can also use the log scale:
pKa + pKb = pKw = 14.
EXAMPLE 1:
HF, a weak acid: K (^) a = 7.24 x 10 —4^ ; pKa = 3. F —^ , (conjugate base of HF) pKb =
Kb =
EXAMPLE 2:
HCN, a very weak acid: K (^) a = 4.00 x 10 —10^ ; pKa = 9. CN —^ , (conjugate base of HCN) pKb =
Kb =
Salts
- You may have noticed that the conjugate base of a weak acid, or the conjugate acid of a weak base is always an ion.
- So where does this ion come from?
- It is always produced when the “parent” weak acid or base ionizes, but these conjugate species can be also be found in ionic compounds.
- And ions are formed when an ionic solid is dissolved in water........
- Remember those solubility rules ????
- This is where they come in handy; so you will know whether or not a solid will dissolve in water to produce ions!!
- First we consider salts that yield the conjugate base of a weak acid.
- Recall that the conjugate base will always behave like a base when in aqueous solution: B + H 2 O º BH +^ + OH —
EXAMPLE 1: Consider the salt CH 3 COONa. In water:
CH 3 COONa! CH 3 COO —^ + Na +
- The CH 3 COO —^ will now undergoes hydrolysis with water:
- The ion that is the conjugate species of a weak acid or base is the species that will undergo the (^) hydrolysis with water.
- Remember that a conjugate species differs from its “parent” species by only one H +^ !!
EXAMPLE 2:
Calculate the pH of a 0.100 M solution of KF. [Ka for HF = 7.24 x 10 —4^ ]
- Salts that produce the conjugate acids of weak bases will exhibit acidic behaviour in solution:
HA + H 2 O º H 3 O +^ + A —
EXAMPLE 1: What happens when NH^4 CR is placed in water?
NH 4 CR! NH 4 +^ + CR—
The NH 4 +^ is the conjugate acid of NH 3 , so in water, hydrolysis occurs.
EXAMPLE 2: Calculate the pH of a solution of 0.175 M NH 4 NO 3.
In water:
Then:
Equivalence Point of a Titration
- This is the point where the stoichiometric quantities of acid and base, defined by the equation, have been mixed together.
- It is really important to note that the solution is NOT always neutral (i.e. pH = 7) at the equivalence point !!
- This is why, earlier, we used the term equivalence point rather than neutralization point.
- The pH at the equivalence point is only truly neutral (pH = 7) for a titration of a strong acid with a strong base.
HC R + NaOH! NaC R + H 2 O
strong strong neutral
- So why is the solution of NaCR neutral?
- Because neither Na +^ (aq) nor CR—^ is the conjugate of a weak species.
- If both ions which form the salt (NaCR in this case) originally came from strong species, they may termed spectator ions and the solution will be neutral.
In contrast; look at the reaction of a weak acid with a strong base :
CH 3 COOH + NaOH! CH 3 COO —Na+^ + H 2 O weak strong basic solution
- The salt formed in the reaction is CH 3 COO —^ Na +^.
- The CH 3 COO —^ is a base ; the conjugate base of the weak acid CH 3 COOH.
- That means that this solution will be basic.
