The presence of bicarbonate ion in river water gives the water a capacity to res
ID: 637063 • Letter: T
Question
The presence of bicarbonate ion in river water gives the water a capacity to resist
pH change when, for example, acid rain falls into it.
Consider river water in which [H
2
CO
3
] = 1.5 X 10
-5
M
and [HCO
3
-
] = 1.0 X 10
-4
M
.
H
2
CO
3
(
aq
) + H
2
O (
l
) HCO
3
-
(
aq
) + H
3
O
+
(
aq
) pKa = 4.3 X 10
-7
A. What is the pH of the river water?
B. What is the pH of the river water after adding 10.0 mL of 1.0 X 10
-3
M
HNO
3
, a strong acid,
to 1.0 L of river water?
C. What is the pH of 1.0 L of distilled water after adding 10.0 mL of 1.0 X 10
-3
M
HNO
3
?
Explanation / Answer
Given: [H2CO3] = 1.5 x 10-5 M & [HCO3-] = 1.0 x 10-4 M; ka = 4.3 x 10-7
To find:
A. What is the pH of the river water?
Solution: pH = pKa + log{[HCO3-]/ [H2CO3]}
pH = -logKa + log{[HCO3-]/ [H2CO3]}
pH = -log(4.3 x 10-7) + log{1.0 x 10-4 M/ 1.5 x 10-5 M}
pH = 7 - log4.3 + log{1.0 / 0.15}
pH = 7 - 0.633 + 0.824 = 7.19
B. What is the pH of the river water after adding 10.0 mL of 1.0 X 10-3 MHNO3?
Upon adding HNO3, [HCO3-] & [H2CO3] would change as the strong acid HNO3 would react with HCO3- leading to a decrease in [HCO3-] and hence increasing [H2CO3].
[HNO3] added = moles HNO3 /total volume
moles HNO3 = molarity* volume = 1.0 X 10-3 moles/L * 10.0 x 10-3 L = 1.0 X 10-5 moles HNO3
[HNO3] added = 1.0 x 10-5 moles HNO3 / 1.10 L = 9.09 x 10-6 M
The reaction and ICE table would be:
HCO3- (aq) + HNO3 (aq) ---------> H2CO3 (aq) + NO3-(aq)
I 1.0 x 10-4 M 9.09 x 10-6 M 1.5 x 10-5 M
C - 9.09 x 10-6 M -9.09 x 10-6 M + 9.09 x 10-6 M
E 9.09 x 10-5 0 2.41 x 10-5 M
Therefore the new pH = -log(4.3 x 10-7) + log{9.09 x 10-5 M/ 2.41 x 10-5 M}
pH = -log(4.3 x 10-7) + log{9.09 x 10-5 M/ 2.41 x 10-5 M}
pH = 7 - log4.3 + log{9.09 / 2.41}
pH = 7 - log4.3 + log{9.09 / 2.41}
pH = 6.94
C. What is the pH of 1.0L of distilled water after adding 10.0 mL of 1.0 X 10-3 MHNO3?
pH of distilled water = 7 => [H+] = 1.0 x 10-7 M
Upon addition of HNO3 the [H+] will increase.
HNO3 (aq) ------> H+ (aq) + NO3- (aq)
[HNO3] = 1.0 X 10-3 M, vlume of HNO3 = 10.0 mL = 10.0 x 10-3 L
moles of HNO3 = moles of H+ = 1.0 X 10-3 moles/L * 10.0 x 10-3 L = 10.0 x 10-6 moles
[H+ ]added = moles H+ added/ total volume = 10.0 x 10-6 moles/1.10L = 9.09 x 10-6 M
[H+] after addition of HNO3 = 1.0 x 10-7 M + 9.09 x 10-6 M = 9.19 x 10-6 M
pH after addition of HNO3 = -log[9.19 x 10-6] = 5.04
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