= Solubility product measurement by titration
How can an invisible dissolved inventory be measured? The unsigned laboratory handout https://www.webassign.net/question_assets/ucscgencheml1/updated_labs2020/lab-10-procedure-2020.pdf[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 (chemistry)> 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.<Solubility Product Constant, Ksp, for Calcium Hydroxide by Acid-Base Titration, University of California, Santa Cruz>[pp. 3–7, Procedure: gravity filtration steps 2–7, HCl preparation steps 8–14, and titrations steps 15–19]{ref}
Unlike AB, <calcium hydroxide> has one-to-two dissolution <stoichiometry>:
$$
\mathrm{Ca(OH)_2}(s)\rightleftharpoons\mathrm{Ca}^{2+}(aq)+2\mathrm{OH}^-(aq),\qquad
K_{\mathrm{sp},c}=[\mathrm{Ca}^{2+}][\mathrm{OH}^-]^2.
$$
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 $\mathrm H^++\mathrm{OH}^-\longrightarrow\mathrm H_2\mathrm O$. Let $C_H$ be the known <titrant> <molar concentration>, $v$ its delivered <volume (thermodynamics)> at the <equivalence point>, and $V_a$ the original <aliquot> <volume (thermodynamics)>. Then the inventory calculation is
$$
h=[\mathrm{OH}^-]=\frac{C_Hv}{V_a},\qquad
s=\frac h2,\qquad K_{\mathrm{sp},c}=\frac{h^3}{2}.
$$
Dilution 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 (thermodynamics)>, 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.<Solubility Product Constant, Ksp, for Calcium Hydroxide by Acid-Base Titration, University of California, Santa Cruz>[pp. 2–4, filtration and CO2 discussion; p. 9, Discussion Questions and Review Topics, ion pairs and dissolved CO2]{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 requires
$$
K_{\mathrm{sp}}=\gamma_{\mathrm{Ca}}\gamma_{\mathrm{OH}}^2\frac{K_{\mathrm{sp},c}}{(c^\circ)^3}.
$$
This 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.
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