= Ideal binary Gibbs-Duhem cancellation
{c}
{title2=$x\,d\mu_A+(1-x)\,d\mu_B=0$}
An <ideal solution> has molar <chemical potentials>
$$
\mu_A=\mu_A^*(T,p)+RT\ln x,
\qquad
\mu_B=\mu_B^*(T,p)+RT\ln(1-x),
$$
where $R$ is the <molar gas constant> and the stars denote pure-component reference values at the same <temperature> and <pressure>.<Thermodynamics and Chemistry by Howard DeVoe>[Thermodynamics and Chemistry, §9.4.2, printed pp. 249–250, equations 9.4.5 and 9.4.8; definition of an ideal mixture]{ref}
Hold <temperature> and <pressure> fixed. The <derivatives> with respect to $x$ are $RT/x$ and $-RT/(1-x)$, so
$$
x\frac{d\mu_A}{dx}=RT,
\qquad
(1-x)\frac{d\mu_B}{dx}=-RT.
$$
Each <chemical potential> changes, yet their composition-weighted slopes cancel exactly. At $x=1/4$, the two slopes are $4RT$ and $-4RT/3$; multiplying by $1/4$ and $3/4$ gives $RT$ and $-RT$. Near either pure-component endpoint the dilute component's slope diverges, but its weighted contribution remains finite. The endpoint <chemical potential> itself requires a limiting interpretation because its <logarithm> diverges.
\Image[/ideal-binary-chemical-potential-cancellation.png]
{title=Ideal binary chemical potentials change while their weighted slopes cancel}
{description=Original ideal-solution calculations at fixed temperature and pressure. The left panel subtracts each pure-component chemical potential and divides by RT. The right panel shows the two weighted derivatives divided by RT; they sum to zero throughout the open composition interval.}
The same constraint has a geometric explanation. Subtract the pure-component reference contribution from the molar <Gibbs free energy> and divide by $RT$ to obtain
$$
f(x)=x\ln x+(1-x)\ln(1-x).
$$
Its <derivative> is $f'(x)=\ln[x/(1-x)]$. At fixed total <amount of substance>, exchanging a little B for A changes the <Gibbs free energy> with slope $\mu_A-\mu_B$. Consequently the tangent to $f$ at $x$ has intercepts
$$
f(x)-x f'(x)=\ln(1-x),
\qquad
f(x)+(1-x)f'(x)=\ln x
$$
at horizontal coordinates $0$ and $1$, respectively. The two <chemical potentials> come from the same curve: choosing two unrelated functions would generally destroy this tangent construction.
\Image[/ideal-binary-gibbs-energy-tangent.png]
{title=One ideal mixing Gibbs-energy curve determines both chemical potentials}
{description=Original ideal-solution geometry with each pure-component reference subtracted. The tangent at x=0.30 reaches the B chemical-potential shift at x=0 and the A chemical-potential shift at x=1. The plotted curve and tangent are theoretical calculations, not experimental data.}
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