The following equilibria control the relative concentrations of molybdate and thiomolybdate ions in dilute aqueous solution. A solution initially containing 2.0×10
-7 M MoS
42- hydrolyzes in a closed system. The H
2S product accumulates until equilibrium is reached.
MoS
42- + H2O
MoOS
32- + H2S
MoOS
32- + H2O
MoO
2S
22- + H2S
MoO
2S
22- + H2O
MoO
3S
2- + H2S
MoO
3S
2- + H2O
MoO
42- + H2S
All ions and hydrogen sulphide are aqueous. Write the equilibrium equations that determine the system, noting the possibility that H
2S can ionize to HS
- and S
2- under certain pH conditions. (K values for all of these equilibria are available in tables)
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The equations that confuse me are the mass balances here. Molybdenum mass balance is an obvious summation over all species containing molybdenum, sulphur I don't get: I don't know if I should include the H
2S, HS
-, S
2- in the mass balance that starts from 8*10
-7, or not - because H
2S is produced as a side-product of the
same reaction as produces the molybdate conjugates.