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Acid-Base Strength:Ka, Kb, Kw
Mrs. Kay
Chemistry 12
Chapter 15 Pages: 583-584,587-597
Relative Strengths Of Binary AcidsH –XThe greater the tendency for the transfer of a proton from HX to H2O, the more the forward reaction is favored and the stronger the acid.
in a periodic group:• The weaker the bond, the stronger the acid.• The larger the resultant anion’s radius, the stronger is
the acid.• The strengths of binary acids increase from top to
bottom in a group of the periodic table.
Relative Strengths Of Binary Acids
H –Xin a periodic group:
Bond dissociation energy: the weaker the bond, the stronger the acid.Bond dissociation energy 569 > 431 > 368 > 297(kJ/mol) HF HCl HBr HI
Acid strength Ka 6.6x10-4 < ~106 < ~108 < ~109
Anion radius: the larger the anion’s radius, the stronger the acid.Anion radius (ppm) 136 < 181 < 195 < 216(kJ/mol) HF HCl HBr HI
Acid strength Ka 6.6x10-4 < ~106 < ~108 < ~109
The strength of binary acids increase from top to bottom in a group of the periodic table.
Relative Strengths Of Binary Acids
H –X
in a period:• The larger the electronegativity difference between H and X,
the more easily the proton is removed and the stronger is the acid.
EN 0.4 < 0.9 < 1.4 < 1.9
Acid strength CH4 NH3 H2O HF
• The strengths of binary acids increase from left to right across a period of the periodic table.
Representative Trends In Strengths of Binary Acids
The Acid dissociation constant, Ka
Ka
• A weak acid only ionizes to a small extent and
comes to a state of chemical equilibrium. • We can determine how much it ionizes by
calculating the equilibrium constant for this reaction, the ionization constant, Ka.
• The larger the Ka the more acid ions are found in solution and the stronger the acid because the more easily it donates a proton.
• The reverse is true for the smaller the Ka
HCOOH (aq) + H2O (l) < -- > H3O+ (aq) + HCOO- (aq)
• Ka = [H3O+][HCOO-]
[HCOOH]
• Notice how the Ka ignores the water since we are dealing with dilute solutions of acids, water is considered a constant, and when multiplied by both sides it is cancelled out.
Equilibrium In Solutions Of Weak Acids And Weak Bases
weak acid: HA + H2O H3O+ + A-
[H3O+][A-]
Ka =
[HA]
weak base: B + H2O HB+ + OH-
[HB+][OH-]
Kb =
[B]
You need to be able to write acid and base ionization equations!!!
Practice:
1. A solution of a weak acid, “HA”, is made up to be 0.15 M. Its pH was found to be 2.96. Calculate the value of Ka.
• Steps to follow:1. Write balanced equation
2. Calculate [H+] using 10-pH
3. Set up chart for equilibrium (ICE or i Δ f)
4. Solve using Ka expression
Answer
1. HA (aq) + H2O (l) < -- >H3O+ (aq) + A-(aq)
2. [H3O+]= 10 -2.96 = 0.0011 M3. Set up table:
HA H2O < -- > H3O+ A-
0.15 0 0
-.0011 +0.0011 +0.0011
0.139 0.0011 0.0011
i
Δ
f
4. Ka = [H3O+][ A-]
[HA]
= [0.0011][0.0011]
[0.139]
= 8.7 x 10 -6
Percent Dissociation
The fraction of acid molecules that dissociate compared with the initial concentration of the acid.
Percent Dissociation = [H3O+] x 100%
[HA i]
For the previous question:
Percent Dissociation = [0.0011] x 100% =0.73 %
[0.15]
Practice:
• The ionization constant, Ka, for a hypothetical weak acid, HA, at 25°C is 2.2 x 10-4.
a) Calculate the [H3O+] of a 0.20 M solution of HA.
b) Calculate the percent ionization of HA.c) Calculate the [A-].d) What initial concentration of HA is needed
to produce a [H3O+] of 5.0 x 10-3 M?
Answer
a) HA(aq) + H2O(l) < -- >H3O+(aq) + A-(aq)
Ka = [H3O+][A-] = 2.2 x 10-4
[HA]2.2 x 10-4 = (x)(x)
0.20 Mx2 = (2.2 x 10-4) (0.20 M)x = 0.0066 M
The [H3O+] is 0.0066 M.
Because it is a 1:1 ratio they are both the same concentration (x)
b) % ionization = 0.0066 M x 100% = 3.3%
0.20 M
c) From the stoichiometry of the reaction,
[H3O+] = [A-]
Therefore, [A-] = 0.0066 M
Found d) from table set up
d) 2.2 x 10-4 = (5.0 x 10-3 M)(5.0 x 10-3 M)x – 5.0 x 10-3 M
• 2.2 x 10-4(x – 5.0 x 10-3M) = 2.5 x 10-5
• 2.2 x 10-4x – 1.1 x 10-6 = 2.5 x 10-5
• x = (2.5 x 10-5 + 1.1 x 10-6) / 2.2 x 10-4
• x= 0.12 M• The initial concentration of HA required is
0.12 M.
Acid And Base Ionization Constants
weak acid: CH3COOH + H2O H3O+ + CH3COO-
[H3O+][CH3COO-]
Acid ionization constant: Ka =
[CH3COOH]
weak base: NH3 + H2O NH4+ + OH-
[NH4+][OH-]
Base ionization constant: Kb =
[NH3]
Acid and base ionization constants are the measure of the strengths of acids and bases.
Kb
• When using weak bases, the same rules apply as with weak acids, except you are solving for pOH and using [OH-]
Another relationship
• Useful to know:
Ka x Kb = Kw = 1.0 x 10-14
Buffer Solutions
• A buffer solution is a solution that changes pH only slightly when small amounts of a strong acid or a strong base are added.
• A buffer contains
a weak acid with its salt (conjugate base) or
a weak base with its salt (conjugate acid)
CH3COOH/CH3COONa
NH3/NH4Cl
Depicting Buffer Action
How A Buffer Solution Works
• The acid component of the buffer can neutralize small added amounts of OH-, and the basic component can neutralize small added amounts of H3O+.
• Pure water does not buffer at all.
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