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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...........
- In contrast a weak acid or base is only partially ionized in aqueous solution..................
- The ionization of a weak acid or base in water is an equilibrium situation.
- It is, therefore, governed by an equilibrium constant; K (^) a for acids and Kb for bases.
- We write these equilibrium constants as:
Notice that....
- [H 2 O] (constant at. 55.6 M) does not appear in K (^) a or Kb.
- The larger the K (^) a the stronger the acid; the larger the K (^) b the stronger the base.
EXAMPLE: CCR 3 COOH; Ka = 2.0 x 10–1^ is a stronger acid than CH 3 COOH; Ka = 1.8 x 10-
CH 3 NH 2 ; Kb = 6.4 x 10-4^ is a stronger base than NH 3 ; Kb = 1.8 x 10-
- H’s on C are not (generally) acidic in water - e.g. CH 3 COOH monoprotic (one H +) and one Ka
- We may define pKa analogous to pH and pK (^) w :
pKa = - log Ka pKb = - log Kb
EXAMPLES: pKa for CCR 3 COOH = 0.70 [ - log (2.0 x 10 -1^ )] pKa for CH 3 COOH = 4.74 [ - log (1.8 x 10 -5^ )]
pKb for CH^3 NH^2 = 3.19^ [ - log (6.4 x 10^ -4^ )] pKb for NH 3 = 4.74 [ - log (1.8 x 10 -5^ )]
- Because of the minus sign, the smaller the value of K (^) a or Kb the larger the value of pK (^) a, or pKb
OR
- The weaker the acid or base, the larger the pK (^) a or pKb ..........
- 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 acceptable 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:
- The quantity [x /c] x 100 is called the degree of dissociation, expressed on a % basis; may also be called percent dissociation or ionization.
Percent = Amount dissociated x 100% dissociation Original amount
=
- We can also use x to calculate pH; because in a weak acid equilibrium; x = [H +^ ]
[H +^ ] = 1.34 × 10 —3^ M, therefore pH = 2.
- The degree of dissociation increases as the solution is diluted, although [H+^ ] decreases.
EXAMPLE: For solutions of HF in water: Ka for HF = 6.7 x 10 —
Initial [HF] Equil. [H +^ ] % Dissoc. pH 1.0 2.6 x 10 —2^ 2.6 1. 0.1 7.9 x 10 —3^ 7.9 2. 0.01 2.3 x 10 —3^22 2. decreasing increasing increasing
EXAMPLE 1:
Nitrous acid, HNO 2 has a Ka value of 6.0 x 10 —^. Calculate the initial concentration of HNO 2 if a solution of this acid has a pH of 3.65.
EXAMPLE 2:
A 0.025 M solution of formic acid, HCOOH, has a pH of 2.75. a) Calculate the % ionization of this solution. b) Calculate the K (^) a for formic acid.
EXAMPLE 1:
What is the % ionization and pH of a solution of a 0.085 M solution of NH 3? [Kb = 1.8 x 10 —5^ ]
EXAMPLE 2:
A solution of methylamine, CH 3 NH 2 , has a pH of 10.45. Calculate the initial concentration of methylamine in this solution. [Kb of CH 3 NH 2 = 6.4 x 10 — 4^ ]
Reactions of Conjugate Species
- Every weak acid (HA), will produce its conjugate base (A—^ ) when it ionizes in water.
- The conjugate base of any acid is the species that is obtained from the acid by removal of one H +^ (or proton).
- Every weak base (B), will produce its conjugate acid (BH +^ ) when it ionizes in water.
- Similarly, the conjugate acid of any base is the species that is obtained from the base by addition of a proton (or H +^ ).
- It is essential to realize that in any conjugate acid/base pair, the acid always has one more H than the base!
- You must be able to recognize conjugate bases and acids, based on the original weak acid or base!
EXAMPLES:
- Consider the weak acid HNO 2 ; its conjugate base is NO 2 —^.
- Now look at the weak base NH 3 ; its conjugate acid is NH 4 +^.
- HNO 2 / NO^2 —^ and^ NH^3 / NH^4 +^ are^ conjugate acid-conjugate base pairs [differing by one H +^ ]
- How do these conjugate species behave?
- What happens if you react them with water?
- Consider the generic weak acid, HA ; its conjugate base is A—.
- This species will act as a base:
- This reaction 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. 137) their K’s are multiplied:
- Now look at the generic weak base B , whose conjugate acid is BH +.
- BH +^ will act like an acid.......
- We also call this a hydrolysis reaction.
- We can determine K (^) a for this conjugate acid in the same manner as we determined K (^) b for a conjugate base previously..................
- As was the case with the weak acids, the (^) weaker the base, the more acidic is its conjugate acid.
EXAMPLE 1: NH 3 is a weak base; K (^) b = 1.8 x 10—5^ ; pKb = 4. NH 4 +^ (conjugate acid of NH 3 ); pK (^) a = Ka (NH 4 +^ ) =
EXAMPLE 2:
Aniline, C 6 H 5 NH 2 , is a very weak base; Kb = 4.0 x 10—10^ ; pKb = 9. C 6 H 5 NH 3 +^ (conjugate acid of C 6 H 5 NH 2 ) ; pK (^) a = Ka =
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........
- 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:
A 0.0285M solution of the sodium salt, NaA of the weak monoprotic acid, HA, has a pH of 9.65. Calculate K (^) a for the acid, HA.
- 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:
The salt in question is NH 4 CR. In water:
The NH 4 +^ is the conjugate acid of NH 3 , so in water...
EXAMPLE 2: Calculate the pH of a solution of 0.175 M NH 4 NO 3.
In water:
Then.....
