Mass balance for the solubility of magnesium
hydroxide (see page 153)
• Approach 1: using magnesium as our reference
Step 1: because the starting material is Mg(OH)2,the amount of magnesium
ions (denoted by X) we have in the solution should be half of the hydroxide
released by Mg(OH)2 (denoted by Y). i.e. X = ½ Y
Note that the total amount of hydroxide (denoted by Z) in solution include
the hydroxide (Y) from Mg(OH)2 and from the hydroxide water that is equal
to the concentration of [H+] from water, i.e. Z = Y + [H+]; thus
Y = Z – [H+]
Step 2: X exists in water in the forms: [Mg2+], [MgOH+], therefore
X = [Mg2+] + [MgOH+],
hydroxide Z exists in the forms [OH-] and [MgOH+], therefore
Z = [OH-] + [MgOH+], thus Y = [OH-] + [MgOH+] - [H+];
Step 3: given that X = ½ Y
[Mg2+] + [MgOH+] = ½ ([OH-] + [MgOH+] - [H+]);
2( [Mg2+] + [MgOH+]) = [OH-] + [MgOH+] - [H+];
2( [Mg2+] + [MgOH+]) + [H+] = [OH-] + [MgOH+]; (this is the
same as textbook)
• Approach 2: Using hydroxide as our
reference
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Hydroxide (denoted by X) in solution comes from two sources: (1) Y amount
from Mg(OH)2, which is equal to TWICE amount of magnesium ions
(denoted by M here) Y = 2 M; and source (2) Z amount from water, that
equals the amount of [H+] from water, i.e. Z = [H+]
We now know X = Y + Z, i.e. X = Y + [H+], i.e. X = 2 M + [H+];
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M in solution exists in the forms [Mg2+], [MgOH+],
M = [Mg2+] + [MgOH+],
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Therefore X = 2([Mg2+] + [MgOH+]) + [H+];
Hydroxide X in solution exits in the forms of [OH-] and [MgOH+],
X = [OH-] + [MgOH+],
Combined the above steps: [OH-] + [MgOH+] = 2([Mg2+] + [MgOH+]) + [H+];
First: the calcium oxalate should not bear no charge, as pointed out by a
student in class. If so, the charge balance equation will be
2[Ca2+] = 2[C2O42-] + [HC2O4-] + [OH-]
Assuming the original question is correct, the charge balance equation
would be 2[Ca2+] = 2[C2O42-] + [HC2O4-] + [OH-] + 2[CaC2O42-]
If we also consider the autoprotolysis of H2O: H2O ↔ H+ + OHThe charge balance equation will become
2[Ca2+] + [ H+] = 2[C2O42-] + [HC2O4-] + [OH-]
(R1)
Mass balance equation for 9-11
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Using calcium as our reference
The amount of calcium (denoted by X) should be equal to the amount of
oxalate (denoted by Y), given that they come from the same source with a
1:1 stoichoimetric relationship.
X in the system exists as [Ca2+], i.e. X = [Ca2+]
Y in the system exists as [C2O42-], [HC2O4-], and [H2C2O4],
i.e. Y = [C2O42-] + [HC2O4-] + [H2C2O4],
Since X = Y, we have [Ca2+] = [C2O42-] + [HC2O4-] + [H2C2O4] (Final answer)
• Using oxalate as our reference yeilds the same result as above
• How about using hydroxide ion as the reference of mass balance?
• Hydroxide ion (denoted by X) in solution comes from water, which leads
to the formation of [HC2O4-] + [H2C2O4],
therefore [OH-] = [HC2O4-] + [H2C2O4],