A spin configuration specifies the spin variable at every site of a spin model. For $N$ Ising spins, it is a member of $\{-1,1{\}}^N$.

How can an invisible dissolved inventory be measured? The unsigned laboratory handout Solubility Product Constant, Ksp, for Calcium Hydroxide by Acid-Base Titration, Procedure, starts with saturated calcium hydroxide, supplied by the teacher or prepared from calcium and water. Its apparatus includes filter paper, a funnel, stoppered flasks, a volumetric pipette, a volumetric flask and a burette. Two successive rounds of filtration remove particles. The filtrate is stoppered to reduce carbon dioxide exposure. A $10.00\mathrm{mL}$ aliquot is diluted with deionized water and titrated with standardized hydrochloric acid; bromothymol blue gives a pale-green titration endpoint. The readout is the difference between final and initial burette readings; four consistent trials are required.^[ref]

Unlike AB, calcium hydroxide has one-to-two dissolution stoichiometry:For molar solubility $s$ in the ideal model with no background calcium or hydroxide, no additional chemical reactions, and negligible self-ionization of water, $[{\mathrm{Ca}}^{2+}]=s$ and $[{\mathrm{OH}}^{-}]=2s$. Hence $K_{\mathrm{sp},c}=4s^3$, with units ${\mathrm{M}}^3$; it is not the AB constant used in the figures. The neutralization chemical equation is ${\mathrm{H}}^{+}+{\mathrm{OH}}^{-}⟶{\mathrm{H}}_2\mathrm{O}$. Let $C_H$ be the known titrant molar concentration, $v$ its delivered volume at the equivalence point, and $V_a$ the original aliquot volume. Then the inventory calculation isDilution of the aliquot changes its concentration during the acid-base titration, but preserves its initial amount of substance of hydroxide. Therefore $V_a$, rather than the diluted flask volume, belongs in the denominator.

The handout identifies two biases: carried-over calcium hydroxide solid dissolves during acid-base titration, consuming extra titrant; carbon dioxide consumes hydroxide and forms calcium carbonate. Filtration and stoppering address these effects, while repeat trials test consistency. Its review questions also ask about dissolved ion pairs, which filtration cannot remove.^[ref]

Further controls would compare a water blank, independently verify titrant standardization, and record the saturation temperature; these are proposed checks, not reported measurements. A titration endpoint estimates the equivalence point and may introduce a systematic offset. Even perfect inventory measurement does not determine activity coefficients. Recovering the dimensionless thermodynamic solubility product additionally requiresThis explains both the power and the limit of the experiment: acid-base titration measures a dissolved inventory, while the link from that inventory to a thermodynamic equilibrium constant still depends on a chemical model.

The common-ion effect is the change in a chemical equilibrium caused by adding an ion already participating in it. In the fixed-activity coefficient AB model, adding the common B anion lowers the molar solubility. The shared ion participates regardless of which salt supplied it.^[ref]

The preceding solubility product and material balance give a short proof without a qualitative slogan. Taking a derivative of $s_{\ast }^2+c_0{s}_{\ast }=K_{\mathrm{sp},c}$ at fixed $K_{\mathrm{sp},c}$ givesThus the molar solubility decreases continuously as $c_0$ increases. At $c_0=0$, the derivative is $-1/2$; for a very large common-ion background, the rationalized expression approaches $K_{\mathrm{sp},c}/c_0$. The solubility product is fixed within this model even while the molar solubility changes.

Neglecting $s_{\ast }$ beside $c_0$ replaces $s_{\ast }(c_0+s_{\ast })=K_{\mathrm{sp},c}$ by $c_0{s}_{\mathrm{app}}=K_{\mathrm{sp},c}$. This approximation always overestimates the exact molar solubility, since it neglects the B anions supplied by AB itself. For $c_0>0$, its relative error has an exact simple form:At $c_0=0.030\mathrm{M}$, $s_{\mathrm{app}}=0.003333333\mathrm{M}$, approximately $10.1\%$ too high. At $c_0=0.100\mathrm{M}$, $s_{\ast }=0.000990195\mathrm{M}$ and $s_{\mathrm{app}}=0.001000000\mathrm{M}$, approximately $0.990\%$ too high. A small solubility product by itself does not justify the approximation: the common-ion background must be large compared with the resulting molar solubility.

For a desired fractional relative error at most $\epsilon >0$, substituting $s_{\ast }/c_0\le \epsilon$ into the exact balance gives the equivalent criterionFor this model, a $5\%$ tolerance requires $c_0\ge 0.0436436\mathrm{M}$. This bound concerns only the algebraic approximation. In a real chemical solution, changes in ionic strength and activity coefficients, or competing chemical reactions, can invalidate a constant $K_{\mathrm{sp},c}$ even when this bound is met.

A solubility product constrains a product of thermodynamic activities; the available inventories determine which combinations of concentrations are possible. Their intersection determines the molar solubility. Consider an original, abstract AB solid, with the preceding one-to-one dissolution chemical equation. Stipulate $c^{\circ }=1\mathrm{M}$, ${\gamma }_{\mathrm{A}}={\gamma }_{\mathrm{B}}=1$ and $K_{\mathrm{sp},c}=10^{-4}{\mathrm{M}}^2$ at fixed temperature and pressure. Thus $K_{\mathrm{sp}}=10^{-4}$ is dimensionless. These are chosen model parameters, not data for an identified material.

Initially, a completely dissociated soluble XB salt supplies B anions at molar concentration $c_0\ge 0$ and X cations at the same molar concentration; A is absent. All three ions have unit electric charge magnitude. X remains a spectator ion, and only AB can form a solid. Assume constant volume, no additional chemical reactions, and negligible contributions from self-ionization of water. Dissolving AB raises both reacting-ion concentrations by the same $s$:Charge neutrality is satisfied: $c_{\mathrm{A}}+c_{\mathrm{X}}-c_{\mathrm{B}}=s+c_0-(s+c_0)=0$. Adding B alone without its accompanying cation would omit the initial electric charge balance.

In the plane of the two reacting-ion concentrations, this inventory constraint is the line $c_{\mathrm{B}}=c_{\mathrm{A}}+c_0$. Saturation supplies the hyperbola $c_{\mathrm{B}}=K_{\mathrm{sp},c}/c_{\mathrm{A}}$ for $c_{\mathrm{A}}>0$. Their positive intersection solves a quadratic equation:The quadratic formula supplies two roots of a polynomial; only the positive one is a physical concentration. Uniqueness also follows because $s(s+c_0)$ strictly increases for $s\ge 0$, starting at zero and growing without bound. The second expression, obtained by multiplying by the conjugate, avoids subtraction of nearly equal numbers at large $c_0$.

For pure water, $c_0=0$ gives $s_{\ast }=\sqrt{K_{\mathrm{sp},c}}=0.010\mathrm{M}$. For $c_0=0.030\mathrm{M}$,Their product is $10^{-4}{\mathrm{M}}^2$, although neither concentration equals the pure-water value. This is the useful geometrical distinction: an inventory line and an equilibrium constant specify different constraints.

Check the solid inventory before imposing saturation. If the initially added AB has amount of substance $n_0$ in volume $V$, the saturated branch requires $n_0>V{s}_{\ast }$ for excess solid to remain. For example, choose $n_0=0.0020\mathrm{mol}$ and $V=0.100\mathrm{L}$. In pure water, $0.0010\mathrm{mol}$ dissolves and the same amount of substance remains as solid. At $c_0=0.030\mathrm{M}$, approximately $0.0003027756\mathrm{mol}$ dissolves, leaving $0.0016972244\mathrm{mol}$ as solid. In this model, an insufficient supply instead givesIf $n_0/V<s_{\ast }$, all AB dissolves and the ion product is below the solubility product. At $n_0/V=s_{\ast }$, the last solid disappears exactly at saturation.

The molar solubility $s$ is the amount of substance of a specified solid dissolved per volume of saturated chemical solution, under stated conditions. It counts dissolved formula units, not the sum of all dissolved ion concentrations. For AB with no initial A or B and no competing chemical reactions, one-to-one stoichiometry gives $c_{\mathrm{A}}=c_{\mathrm{B}}=s$. For an AB${}_2$ solid, the corresponding concentrations are $s$ and $2s$.^[ref]

A Foucault pendulum is a long simple pendulum whose suspension allows the bob to swing in both independent horizontal directions. Its slowly changing swing direction relative to the building reveals Earth's rotation. A hinge restricting the bob to one vertical plane would prevent this measurement: the support would impose the very direction we want the pendulum to select freely. Léon Foucault demonstrated the effect in 1851.^[ref]

The bob's back-and-forth oscillation and the swing direction's precession are different motions. The bob traverses its arc in seconds, while successive arcs change direction over hours. In a conical pendulum the bob itself circulates around the vertical; a nearly planar Foucault pendulum instead reverses along an almost straight horizontal trace, whose orientation changes slowly. The measured quantity is that orientation against a fixed floor or wall reference, rather than the bob's instantaneous azimuth.

Inside a spherical body of uniform density, the shell theorem gives gravitational acceleration $g(r)=g(R)r/R$. A simple pendulum at depth $h$ therefore has period $P_h=P_0/\sqrt{1-h/R}$. If its clock is calibrated at the surface, its indicated elapsed time is $t\sqrt{1-h/R}$ and it loses $t[1-\sqrt{1-h/R}]\simeq th/(2R)$. This is a mechanical clock-rate change caused by weaker local gravity, not gravitational time dilation. The uniform-density assumption is essential; Earth's actual density profile gives a different interior field.

A porous form of silicon dioxide used as an adsorption medium in thin-layer chromatography.

A signal peptidase removes a cleavable signal peptide; a suitable cleavage site and access are required.^[ref]

For the stipulated $180$-amino acid residue precursor under ER signal sequence, removal of the first $20$ amino acid residues leaves $160$. Now stipulate that positions $101$–$120$ of that precursor are instead retained as a stop-transfer sequence. Their positions in the mature chain become $81$–$100$, because the numbering origin changes when the first $20$ positions are removed. The mature chain still has $160$ amino acid residues, but its membrane-spanning segment is retained. A second hypothetical chain with an uncleaved internal signal anchor loses no positions at all. These are conditional bookkeeping examples, not predictions from invented biological sequences.

The period associated with one full cycle of sidereal time.