Page 48 - Soil and water contamination, 2nd edition

Basic environmental chemistry 35

Example 2.10 Complexation
-6
-1
In a groundwater sample the total dissolved copper (Cu) concentration is 10 mol l
-3
-1
2-
and the concentration of sulphate (SO ) is 1.5 mmol l . The ion strength is 7.2 10 .
4
-
Calculate the activity of the CuSO ion at 25 °C given the logK = –2.31 for the reaction
4
2+
-
CuSO ↔ Cu + SO 2-
4 4
Disregard the formation of other complexes of copper and other cation complexes with
-
sulphate and assume the activity coefficient for CuSO to be unity.
4
Solution
2+
First, use Equation (2.4) to calculate the activity coefficient s for Cu (a Cu 2+ = 6.0) and
2-
SO (a 2- = 4.0):
4 SO 4
log 2 . 0 148 2 . 0 711
Cu Cu
log 2 . 0 155 2 . 0 699
SO 4 SO 4
-
(It is not necessary to calculate the activity coefficient CuSO , since we assume that the
4
-
activity coefficient for CuSO equals 1).
4
-
The mass action expression for the dissociation of CuSO is
4
[Cu 2 ][SO 4 2 ] Cu 2 {Cu 2 } SO 2 {SO 4 2 } . 2 31 3
K 4 10 . 4 90 10
[CuSO 4 ] {CuSO 4 }
-
2+
We may assume that the total amount of Cu is distributed between Cu and CuSO .
4
-
2+
-1
This means that the sum of the Cu and CuSO concentrations in mol l equals
4
-1
-6
10 mol l . Thus,
2+
-
-6
{Cu } = 10 – {CuSO }
4
and
6 {CuSO } 10 {SO 2 }
Cu 2 4 SO 4 2 4
{CuSO 4 }
. 0 711 ( 10 6 {CuSO }) . 0 699 . 0 0015
4 . 4 90 10 3
{CuSO }
4
. 7 45 10 4 ( 10 6 {CuSO }) . 4 90 10 3 {CuSO }
4 4
. 7 45 10 10 . 7 45 10 4 {CuSO 4 } . 4 90 10 3 {CuSO 4 }
. 4 15 10 3 {CuSO } . 7 45 10 10
4
7
{CuSO } . 1 79 10 mol l -1
4
-
This means that almost 18 percent of the total dissolved Cu occurs in the CuSO form.
4
10/1/2013 6:44:15 PM
Soil and Water.indd 47 10/1/2013 6:44:15 PM
Soil and Water.indd 47