Name:_______________________________________________ Date:_______________ Period:______
Ka, Kb, Kw (1)
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Acids like HBr, HNO3 and HClO3 are considered strong acids because they essentially dissociate 100% in water. That is
to say that the concentration of hydrogen ions is equal to the concentration of acid. That is not the case with weak acids.
Most weak acids do not dissociate very much and one way to compare dissociation ability is to look at the acid
dissociation constant (Ka).
Model 1: Acid Dissociation Reaction (Ka)
The general reaction for an acid dissociating in water is as follows:
HA (aq) ↔ H+ (aq) + A- (aq)
OR
HA (aq) + H2O (l) ↔ H3O+ (aq) + A- (aq)
1. Answer the following questions based on Model 1:
a. What is the general formula for an acid?
HA
b. What is the general formula for the conjugate base (the anion)?
A-
c. Write the equilibrium expression for both reactions in Model 1. (Use Ka as the constant)
𝐾𝑎 =
[𝐻 + ][𝐴− ] [𝐻3 𝑂+ ][𝐴− ]
=
[𝐻𝐴]
[𝐻𝐴]
d. Why can the acidic hydrogen be written as H+ OR H3O+?
H+ does not stay as H+ in water because it will react with water to make H3O+
e. Write the reaction for the dissociation of hydrofluoric acid.
HF(aq) H+(aq) + F-(aq)
f.
Write the equilibrium expression for the dissociation of hydrofluoric acid.
𝐾𝑎 =
[𝐻 + ][𝐹 − ]
[𝐻𝐹]
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Model 2: Acid Dissociation Constants
Acid
Hydroiodic Acid
Acetic Acid
Carbonic Acid
Ascorbic Acid
Hydrofluoric Acid
Citric Acid
Formula
HI
HCH3COO
H2CO3
H2C6H6O6
HF
H3C6H5O7
Ka (25°)
>1
1.76 x 10-5
4.3 x 10-7
7.9 x 10-5
7.2 x 10-4
8.4 x 10-4
pKa (25°)
<1
4.75
6.37
4.10
3.14
3.08
2. Answer the following questions based on the data table in Model 2.
a. Are the Ka values of weak acids greater than or less than 1? What does this tell you about the dissociation of
weak acids in water?
Ka for weak acids will be less than 1 because weak acids do not dissociate completely so reactants are favored.
b. Why is the Ka value for HI just listed as greater than 1? What does this tell you about HI?
Since Ka > 1 we know this equilibrium reaction is product favored which mean the equilibrium will lie to the right. This
tells us that HI is a strong acid. It will dissociate completely
c. Based on the Ka values, list the acids in the table from strongest to weakest.
HI, Citric Acid, HF, Ascorbic acid, Acetic acid, Carbonic Acid
d. How is pKa calculated? Write an equation.
pKa = -log Ka
e. Explain why pKa is used instead of Ka sometimes.
For weak acids the values of Ka are extremely small so pKa makes it easier to compare Ka values
f.
What is the relationship between pKa and acid strength?
Since a larger Ka means a stronger acid a lower pKa value means a strong acid
Time for the math!!
Since Ka is just like any other equilibrium constant, the same ICE chart can be used. However, you must know the
relationship between hydrogen ion concentration and pH. In addition, there are assumptions that can be made to make the
algebra easier. Use the next couple of problems to guide you through the process of using Ka. Make sure you keep track of
your significant figures!
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3. The pH of a 0.10 M solution of formic acid, HCOOH, at 25°C is 2.38. Calculate Ka for formic acid at this
temperature.
a. Write the reaction and equilibrium expression for the dissociation of HCOOH in water.
HCOOH(aq)  H+ (aq) + HCOO-(aq)
𝐾𝑎 =
[𝐻 + ][𝐻𝐶𝑂𝑂 − ]
[𝐻𝐶𝑂𝑂𝐻]
b. Calculate the equilibrium hydrogen ion concentration from the pH.
pH=2.38 [H+]=10-2.38 = 4.2 x 10-3M
c. Fill out the information you know in the ice chart below (assume the initial concentration of hydrogen and
formate are zero) (Remember your addition and subtraction sig fig rules):
HCOOH(aq)
H+ (aq)
HCOO-(aq)
I
0.10 M
0
0
C
-4.2 x 10-3 M
+4.2 x 10-3 M
+4.2 x 10-3 M
E
0.10 M
4.2 x 10-3 M
4.2 x 10-3 M
d. Plug the equilibrium concentrations into the equilibrium expression and evaluate for Ka.
𝐾𝑎 =
[4.2 × 10−3 ][4.2 × 10−3 ]
= 1.8 × 10−4
[0.10]
e. How do the initial and equilibrium concentrations of formic acid compare?
When using significant figures they are the same
f.
The point: For an acid dissociation, you can assume that the dissociation of an acid is SO SMALL that
when subtracting it from large numbers, like the initial concentration of acid, we can just negate that
change in value.
4. Calculate the pH of a 0.20 M solution of HCN. The acid dissociation constant is 4.9 x 10-10.
a. Write the reaction and equilibrium expression for the dissociation of HCN.
HCN(aq)  H+(aq) + CN-(aq)
𝐾𝑎 =
[𝐻 + ][𝐶𝑁 − ]
[𝐻𝐶𝑁]
= 4.9 × 10−10
b. Fill out the following ICE chart:
HCN(aq)
H+(aq)
CN-(aq)
I
0.20M
0
0
C
-x
+x
+x
E
0.20 - x
x
x
c. Plug the equilibrium concentrations into the equilibrium expression. Evaluate the expression, omitting any x’s
that are being added to or subtracted to or from large values. Solve for pH.
For 0.20-x we can ignore the x since it is significantly smaller than 0.20
[𝑥][𝑥]
[0.20]
= 4.9 × 10−10
Solve for x=9.9x10-6=[H+] so pH=log(9.9x10-6) = 5.00
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Model 3: Base Dissociation Reaction (Kb)
The general reaction for a base dissociating in water is as follows:
B (aq) + H2O (l) ↔ HB+ (aq) + OH- (aq)
5. Use Model 3 to answer the questions that follow.
a. What is the general formula for the base?
B
b. What is the formula for the conjugate acid?
HB+
c. How does the base produce hydroxide ions when dissolved in water?
It attracts a H+ from the water so the water donates a proton (H+) to the B
d. Write a generic equilibrium expression for base dissociation (Kb).
[𝐻𝐵+ ][𝑂𝐻 − ]
𝐾𝑏 =
[𝐵]
e. Write the reaction for ammonia (NH3) in water.
NH3(aq) + H2O(l)  NH4+(aq) + OH-(aq)
f.
Write the equilibrium expression for the dissociation NH3.
[𝑁𝐻4+ ][𝑂𝐻 − ]
𝐾𝑏 =
[𝑁𝐻3 ]
Model 4: Base Dissociation Constant (Kb)
Acid
Formula
Ka (25°)
Conjugate
Base
Kb (25°)
Product of
Kb x K a
Hydroiodic Acid
HI
>1
I-
Acetic Acid
HCH3COO
1.76 x 10-5
CH3COO-
Insignificantly
Small
5.68 x 10-10
1.00 x 10-14
Carbonic Acid
H2CO3
4.3 x 10-7
HCO3-
2.33 x 10-8
1.00 x 10-14
Ascorbic Acid
H2C6H6O6
7.9 x 10-5
HC6H6O6-
1.27 x 10-10
1.00 x 10-14
Hydrofluoric Acid
HF
7.2 x 10-4
F-
1.39 x 10-11
1.00 x 10-14
Citric Acid
H3C6H5O7
8.4 x 10-4
H2C6H5O7-
1.19 x 10-11
1.00 x 10-14
6. Use the data in Model 4 to answer the following questions.
a. Based on the Kb values, list the bases in the table from strongest to weakest.
Carbonic acid, Acetic acid, Ascorbic Acid, HF, Citric Acid, HI
b. What is the relationship between the strength of an acid and the strength of its conjugate base?
Strong Acid has a weak conjugate base
c. Multiply Ka and Kb together to fill in the last column in the table using appropriate sig figs.
d. What is the mathematical relationship between the Ka of an acid and the Kb of its conjugate base? Which
equilibrium constant is equal to that value?
Ka x Kb =1.00 x 10-14 = Kw
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7. Use the following steps to prove that Ka x Kb = Kw.
a. Write the equilibrium expression for the dissociation of hydrofluoric acid.
HF(aq) H+(aq) + F-(aq)
[𝐻 + ][𝐹 − ]
𝐾𝑎 =
[𝐻𝐹]
b. Write the equilibrium expression for the reaction of F- with water.
F-(aq) + H2O  HF(aq) + OH-(aq)
𝐾𝑏 =
[𝐻𝐹][𝑂𝐻 − ]
[𝐹 − ]
c. Multiply the two expressions together.
𝐾𝑎 × 𝐾𝑏 =
[𝐻 + ][𝐹 − ] [𝐻𝐹][𝑂𝐻 − ]
×
= [𝐻 + ][𝑂𝐻 − ] = 𝐾𝑤
[𝐻𝐹]
[𝐹 − ]
Model 5: Polyprotic acid
8. Use the chart in Model 5 to answer the following questions.
a. What is a polyprotic acid?
An Acid that has more than 1 proton (H+) to donate
b. What is the relationship between the acidity of the first hydrogen and the acidity of subsequent hydrogens in a
polyprotic acid? Explain why this is the case.
The 1st hydrogen is a stronger has higher acidity then the 2nd hydrogen (lower acidity). This is because once you
remove a H+ the ion becomes negative so it holds onto the next proton (H+) stronger.
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9. Since the acidity of the second hydrogen is so low, we can just ignore the amount of hydrogen that is produced by
removal of the second or third hydrogens of a polyprotic acid.
a. Using the Ka values presented in Model 5, solve the following problem:
The solubility of CO2 in pure water at 25°C and 0.1 atm pressure is 0.0037 M. The common practice is to
assume that all of the dissolved CO2 is in the form of carbonic acid, which is produced when carbon
dioxide reacts with water. What is the pH of a 0.0037 M solution of carbonic acid?
CO2(aq) + H2O(l)  H2CO3(aq)
i)
Write the first hydrogen ionization and Ka1 of Carbonic acid.
H2CO3(aq)  H+(aq) + HCO3-(aq)
ii)
Kaq = 4.3 x 10-7
Complete the ICE the table and calculate the pH.
H2CO3
H+(aq)
HCO3-
I
0.0037
0
0
C
-x
+x
+x
E
0.0037 (ignore x)
x
x
𝑥2
𝐾𝑎 = 0.0037 = 4.3 × 10−7 solve for x = 4.0 x 10-5 = [H+]
pH=-log(4.0 x 10-5) = 4.40
b. If you are looking for a real challenge, try the following problem:
Calculate the concentration of [CO3-2] in the solution from part a.
HCO3-(aq)  H+(aq) + CO3-2(aq) Ka2=5.6 x 10-11
HCO3-
H+(aq)
CO3-2
I
4.0 x 10-5
4.0 x 10-5
0
C
-y
+y
+y
E
4.0 x 10-5 – y
(ignore y)
4.0 x 10-5 + y
(ignore y)
y
𝐾𝑎2 =
[𝐻 + ][𝐶𝑂3−2 ] (4.0 × 10−5 )𝑦
=
= 5.6 × 10−11
[𝐻𝐶𝑂3− ]
(4.0 × 10−5 )
y=5.6 x 10-11 M= [CO3-2]
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