This preview shows pages 1–3. Sign up to view the full content.
Solutions for problems assigned with Lecture 2
Garrett & Grisham, pg. 45
1.
Calculate the pH of the following.
a.
5 x 10
4
M HCl
Answer:
HC1 is a strong acid and fully dissociates into [H
+
] and [C1

].
Thus,
[H
+
] = [C1

] = [HC1]
total added
pH
=−
log
10
[H
+
]
=−
log
10
[ HCl
total
]
=−
log
10
(5
×
10
−
4
)
=
3.3
b.
7 x 10
5
M NaOH
Answer :
For strong bases like NaOH and KOH,
pH
=
14
+
log
10
[ Base]
= 14 + log
10
(7
×
10
−
5
)
=
9.85
c.
2
μ
M HC1
Answer:
pH
=−
log
10
[H
+
]
=−
log
10
[ HCl
total
]
=−
log
10
(2
×
10
−
6
)
=
5.70
d.
3 x 10
2
M KOH
Answer:
pH
=
14
+
log
10
(3
×
10
−
2
)
=
12.5
e. 0.04 mM HC1
Answer:
pH
=−
log
10
[H
+
]
=−
log
10
[ HCl
total
]
=−
log
10
(0.04
×
10
−
3
)
=
4.4
f.
6 x 10
9
M HC1= 0.06 x 10
7
M HCl
Answer :
Beware!
Naively one might fall into the trap of simply treating this like another strong
acid problem and solving it like so:
pH
=−
log
10
[H
+
]
=−
log
10
[HCl
total
]
=−
log
10
(6
×
10
−
9
)
=
8.22
However, something is odd.
This answer suggests that addition of a small amount of a strong
acid
to water will give rise to a basic pH!
What we have ignored is the fact that water itself will
contribute H
+
into solution so we must consider the ionization of water as well.
There are two
approaches we can take in solving this problem.
As a close approximation we can assume that:
[H
+
]
=
10
−
7
+
[ HCl ] or,
[H
+
]
=
10
−
7
+
6
×
10
−
9
=
10
−
7
+
0.06
×
10
−
7
=
1.06
×
10
−
7
pH
=−
log
10
(1.06
×
10
−
7
)
=
6.97
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThe exact solution uses the ion product of water.
[H
+
][OH
−
]
=
K
W
=
10
−
14
(the ion product of water)
(1)
In solution HC1 fully dissociates into H
+
+ C1

.
For any solution the sum of negative and
ing with monovalent ion we can write:
positive charges must be equal.
Since we are deal
[H
]
+
[OH
−
]
(2)
+
]
=
[Cl
−
Now because HC1 is fully dissociated:
[Cl
−
]
=
[ HCl
total
]
=
6
×
10
−
9
(3)

Substituting this into equation (2), solving equation (2) for [OH ] and substituting in equation (1)
we have a quadratic equation in H
+
:
[H
+
]
2
−
6
×
10
−
9
[H
+
]
−
10
−
14
=
0
4ac
whose general solution is given by
[H
+
]
=
−
b
±
b
2
−
2a
=
6
This is the end of the preview. Sign up
to
access the rest of the document.
 Spring '09
 KENNETHHILT

Click to edit the document details