Problem Set #2: Strong vs Weak
KEY
1. What is the difference between strong acids and weak acids? Use H2SO4 and H3BO3 as
examples.
A strong acid completely dissolves in water. For example, H2SO4 dissolves in water
producing 2H+ ions as shown below:
H2SO4(aq) + H2O(l)  2H+(aq) + 1 SO4-2(aq)
** No H2SO4(aq) remains after dissociation occurs**
A weak acid does not completely dissolve in water. A weak acid only partially dissolves in
water, creating an equilibria as shown below:
H3BO3(aq) + H2O(l)  H2BO3-1(aq) + H+1(aq)
OR EQUIVALENTLY…..
H3BO3(aq) + H2O(l)  H2BO3-1(aq) + H3O+1(aq)
2. What is the difference between strong bases and weak bases? Use NaOH and N2H4 as
examples.
A strong base completely reacts in water. For example, NaOH(aq) dissolves in water
producing 1OH-1 ion as shown below:
NaOH(aq) + H2O(l)  Na+(aq) + 1 OH-1(aq)
** No NaOH(aq) remains after dissociation occurs**
A weak base does not completely react in water. A weak base only partially reacts in water,
creating an equilibria as shown below:
N2H4(aq) + H2O(l)  N2H5+1(aq) + OH-1(aq)
3. Recall that for weak acids, the larger the Ka value – the stronger the acid. And for weak
bases, the larger the Kb value – the stronger the base.
(a) Write an equation for the chemical equilibrium that occurs when acetic acid is mixed
with water. Find the Ka value for this reaction in the table **WEAK ACIDS AND
WEAK BASES*
Acetic acid is given in the table as HC2H3O2, so the chemical equation for the equilibrium
is as shown below:
HC2H3O2(aq) + H2O(l)  C2H3O2-1(aq) + H+(aq) Ka=1.74x10-5
(b) Relative to other weak acids in the table, is acetic acid a relatively strong acid or
relatively weak acid? Explain
The large the Ka value the stronger the acid. Therefore, acetic acid is stronger than citric
acid and carbonic acid BUT weaker than chromic acid and sulfuric acid.
(c) The conjugate base for this reaction is C2H3O2-1(aq).
(d) Write and equation for the chemical equilibrium that occurs when the conjugate base
from part (c) mixes with water. Calculate the Kb for this base using the formula
Kw=Ka(Kb).
C2H3O2-1(aq) + H2O(l)  HC2H3O2(aq) + OH-1(aq)
Kw =Ka(Kb)
1.0x10-14 = 1.74x10-5(Kb)
Kb = 5.75x10-10
4. Recall that for weak acids, the larger the Ka value – the stronger the acid. And for weak
bases, the larger the Kb value – the stronger the base.
(a) Write an equation for the chemical equilibrium that occurs when CH3NH2 is mixed
with water. Find the Kb value for this reaction in the table called **WEAK ACIDS
AND WEAK BASES**.
CH3NH2 (aq) + H2O(l)  CH3NH3+1(aq) + OH-1(aq) Kb=4.27x10-4
(b) Relative to other weak bases in the table, is CH3NH2 a relatively strong base or
relatively weak base? Explain
The large the Kb value the stronger the base.
(c) The conjugate acid for this reaction is CH3NH3+1(aq).
(d) Write and equation for the chemical equilibrium that occurs when the conjugate acid
from part (c) mixes with water. Calculate the Ka for this base using the formula
Kw=Ka(Kb).
CH3NH3+1(aq) + H2O(l)  CH3NH2 (aq) + H3O+1(aq)
Kw =Ka(Kb)
1.0x10-14 = (Ka)4.27x10-4
Kb = 2.34x10-11
5.
(a)
(b)
(c)
(d)
Calculate:
pKa if Ka=1.86x10-7 pKa = -log(1.86x10-7 ) = 6.73
pKb if Kb = 4.56x10-11 pKb = -log(4.56x10-11 ) = 10.34
Ka if pKa = 4.75
Ka = 1.78x10-5
Kb if pKb = 10.93
Kb = 1.17x10-11
6. If HC6H6O6-1 + H2O(l)  C6H6O6 -2 + H3O+1:
(a) Write a mathematical expression for Ka.
Ka = [C6H6O6 -2][H3O+1] divided by [HC6H6O6-1]
(b) Calculate the [H+1] if pKa = 11.79 for this acid, and [HC6H6O6-1] = 0.56mol/L and
[C6H6O6-2] = 0.034mol/L.
Ka = [C6H6O6 -2][H3O+1] divided by [HC6H6O6-1]
1.62x10-12 = (0.034)(x) divided by 0.56
x = [H+] =2.67x10-11
+1
7. If N2H4 + H2O(l)  N2H5 + OH-1 and Kb =1.26x10-6
(a) Write a mathematical expression for Kb.
Kb = [N2H5+1][OH-1] divided by [N2H4]
(b) Calculate the [OH-1] if [N2H5+1] = 0.054mol/L and [N2H4] = 0.034mol/L.
Kb = [N2H5+1][OH-1] divided by [N2H4]
1.26x10-6 = (0.054)(x) divided by 0.034
x = [OH-1] =7.93x10-7
8. Write all three chemical equations for the dissociation of the polyprotic acid H 3PO4.
Discuss the relative strength of each acid: H3PO4, H2PO4-1, HPO4-2, and PO4-3.
H3PO4(aq) + H2O(l)  H2PO4-1(aq) + H+1(aq) Ka = 6.92x10-3
H2PO4-1(aq) + H2O(l)  HPO4-2(aq) + H+1(aq) Ka = 6.17x10-8
HPO4-2(aq) + H2O(l)  PO4-3(aq) + H+1(aq) Ka = 2.09x10-12
Strongest acid  Weakest acid
H3PO4(aq)  H2PO4-1(aq)  HPO4-2(aq)  PO4-3(aq)