(a) If the solubility of Ag2 Cr2 O4 is 6.5 ×10^-5 mol dm^-3, determine the solubility product fo

step 1 Okay for this question we have balanced chemical equations are AGCL giving out AG positive plus

(a) If the solubility of Ag2 Cr2 O4 is 6.5 ×10^-5 mol dm^-3,...

(a) If the solubility of $\mathrm{Ag}_{2} \mathrm{Cr}_{2} \mathrm{O}_{4}$ is $6.5 \times 10^{-5} \mathrm{mol} \mathrm{dm}^{-3},$ determine the solubility product for this salt. (b) $K_{\mathrm{sp}}$ for $\mathrm{AgCl}$ is $1.8 \times 10^{-10} .$ Why can silver chromate (which is red) be used to indicate the end point in titrations involving the precipitation of $\mathrm{AgCl} ?$

(a) If the solubility of $\mathrm{Ag}_{2} \mathrm{Cr}_{2} \mathrm{O}_{4}$ is $6.5 \times 10^{-5} \mathrm{mol} \mathrm{dm}^{-3},$ determine the solubility
product for this salt.
(b) $K_{\mathrm{sp}}$ for $\mathrm{AgCl}$ is $1.8 \times 10^{-10} .$ Why can silver chromate (which is red) be used to indicate the end point in titrations involving the precipitation of $\mathrm{AgCl} ?$

Key Concepts

Precipitation Titrations and Endpoint Detection

Precipitation titrations involve the gradual addition of one reagent to a solution containing an ion that forms an insoluble salt, leading to the formation of a precipitate. An endpoint indicator, such as a compound that forms a colored precipitate with a metal ion, is used to signal the completion of the reaction. The distinct color change helps in accurately determining the point at which all of the analyte has been precipitated, improving the reliability of the titration.

Solubility Product (Ksp)

The solubility product is an equilibrium constant that applies to sparingly soluble salts in water. It is defined as the product of the concentrations of the ions, each raised to the power of their stoichiometric coefficients, when the salt is in equilibrium with its ions in solution. The Ksp provides a quantitative measure of the extent to which a salt will dissolve under given conditions.

Relationship Between Solubility and Ksp

For a sparingly soluble salt that dissociates into its constituent ions in solution, the molar solubility can be related to the solubility product. By expressing the concentrations of the ions in terms of the molar solubility and using the stoichiometry from the dissolution equation, one can calculate the value of the Ksp. This relationship is fundamental in predicting the conditions under which a salt will precipitate.

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