Graduate An aqueos solution of ##M^+X^-## from Kubo's problems book

[MathematicalPhysicist]

My following question is from Kubo's textbook on Thermodynamics advanced problems.

I'll quote both the question and its solution in the book, and after that I'll ask my questions regarding the solution offered in the book.

Question 40:

An aqueous solution of $M^+X^-$ of concentration $x_1^0$ is poured into room I, and another solution $M^+R^-$ of concentration $x_2^0$ into the other room II, which is separated from the room I by a semi-permeable membrane permeable for $M^+$ ions and $X^-$ ions but not for $R^-$ ions.
What are the concentrations of $M^+X^-$ in room I and II, $x_1'$ and $x_1''$, when the equilibrium (Donnan's membrane equilibrium) is reached by diffusion of $M^+$ and $X^-$ ions through the membrane?
For the sake of simplicity, assume that the solutions are very dilute, and may be regarded as ideal dilute solutions. [Hint: the solution must always be electrically neutral.]

The solution offered in the book:

According to the condition of electrical neutrality,
$$(1a)\ \ \ \ \ \ x'_{M^+}=x'_{X^-}$$
$$(1b) \ \ \ \ \ x''_{M^+}=x''_{X^-}+x_2^0 , $$

although $M^+$ and $X^-$ are exchanged between I and II.
Also the conservation laws for $M^+$ and $X^-$ require the equations:
$$(2) \ \ \ \ \ x'_{M^+}+x''_{M^+}=x_1^0 , \ \ x'_{X^-}+x''_{X^-}=x_1^0$$
On the other hand, the equilibrium condition is:
$$(3)\ \ \ \ \ \bar{G}_{M^+X^-(I)}=\bar{G}_{M^+X^-(II)}.$$
If an ideal dilute solution is assumed, the latter condition becomes:
$$(4)\ \ \ \ \ RT\log(x'_{M^+}x'_{X^-})=RT\log(x''_{M^+}x''_{X^-}),$$
or $$x'_{M^+}x'_{X^-}=x''_{M^+}x''_{X^-},$$
where equations (4.33) have been used ($\phi^0_{M^+}$ and $\phi_{X^-}^0$ are the same for I and II).
Substituting equations (1a), (1b) and (2), we obtain $(x_{X^-})^2=(x_1^0-x_{X^-}'')^2=(x_{X^-}''+x_2^0)x_{X^-}''$, and hence:
$$x_{X^-}''=\frac{(x_1^0)^2}{2x_1^0+x_2^0}$$
In the same way,
$$x_{X^-}' = \frac{x_1^0(x_1^0+x_2^0)}{2x_1^0+x_2^0}$$
or
$$x'_{X^-}/x''_{X^-}=1+x_2^0/x_1^0$$
This means that $M^+ X^-$ itself is distributed between I and II in the ratio $x_1^0+x_2^0:x_1^0$

Now, the derivations are ok but I seem to be getting that: $x_2^0$.
From $(1a)-(1b),(2)$ I get the following:
$$x_1^0-x_{X^-}''=x_{X^-}'=x'_{M^+}=x^0_1-x''_{M^+}=x_1^0-x''_{X^-}-x_2^0$$
from which follows that $x_2^0=0$.

Have I done something wrong here?I don't think so...
I am confused can anyone explain this to me?

Thanks.

 

[mjc123]

There is a mistake in (2). x'_X- + x''_X- = x₁⁰, but x'_M+ + x''_M+ = x₁⁰ + x₂⁰
All this of course assumes that the volumes of the two chambers (we wouldn't usually say "rooms", are you translating?) are equal, which is not stated.

 

[MathematicalPhysicist]

I am not translating, Iv'e copied from I assume a translated copy of Kubo's textbook.