= Solution
Assume the structural <surface charge density> stays fixed while a divalent chloride is added, without changing $T$ or $\epsilon$. Let $v=e\phi_0/(k_BT)$ and $\delta c$ be the increment of the bulk divalent <concentration>. Electroneutral addition also increases bulk chloride by $2\delta c$. Varying the squared <Grahame equation> at fixed $\sigma$ gives
$$
0=\delta c(e^{-2v}+2e^v-3)-\frac e{k_BT}\left[c_{1+}(0)+2c_{2+}(0)-c_{1-}(0)-2c_{2-}(0)\right]\delta\phi_0.
$$
At a strongly negative wall, coions are small. Dropping their contribution to the denominator and $2e^v$ in the numerator gives the requested leading approximation
$$
\boxed{\delta\phi_0\simeq\frac{k_BT}{e}\frac{\delta c(e^{-2v}-3)}{c_{1+}(0)+2c_{2+}(0)}.}
$$
The $-3$ includes the change in bulk osmotic concentration of one added divalent counterion and its two monovalent coions. The numerator's increment is a \b[bulk] <concentration>; the denominator contains \b[surface] <concentrations>.<Fixed-charge response to added divalent chloride>[Full differential response before the counterion approximation]{ref} Extra counterions screen the wall and make its negative <electric potential> less negative. At fixed bulk <calcium>, $c_{\rm Ca}(0)=c_{\rm Ca}^\infty e^{-2\phi_0/V_T}$ therefore decreases. Under the stipulated proportionality, synaptic activity decreases as well; this argument isolates electrostatic accumulation rather than assuming a separate specific magnesium effect on the <ion channels>.
For $\delta c=5\,\mathrm{mM}$, the printed linear approximation gives $\delta\phi_0=9.53434\,\mathrm{mV}$. Linearizing the activity too gives $\delta A/A=-2\delta\phi_0/V_T=-0.7422$, or approximately a $74\%$ reduction. Exponentiating this approximate voltage instead gives $A'/A=0.4761$. \b[Neither is a controlled finite-increment prediction:] the added divalent <concentration> is 2.5 times its initial value.
A consistent finite-change calculation is possible within the same <Poisson-Boltzmann equation>. Define, using $\mathrm{mM}$ throughout,
$$
S_m(u)=160(e^u-1)+(2+m)(e^{2u}-1)+(114+2m)(e^{-u}-1)+25(e^{-2u}-1).
$$
Fixed $\sigma$ requires $S_5(u')=S_0(2.84129)$. The physical positive root is $u'=2.58241$, giving
$$
\boxed{\phi'_0=-66.3488\,\mathrm{mV},\quad
\Delta\phi_0=6.65122\,\mathrm{mV},\quad
\frac{A'}A=e^{2(u'-u)}=0.595857.}
$$
Thus the self-consistent finite-change estimate is about a \b[40.4% reduction]. Both estimates rely on fixed surface charge, ideal ionic activities, and unchanged bulk <calcium>; the limitations of the large local <concentrations> remain.
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