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This question is providing us with the concentration of a lead (II) chromate solution at its saturation point and asks for its solubility product. Thus, after the corresponding calculation, the result turns out to be 1.80x10⁻¹⁴.
In chemistry, solubility tells us how much solid is dissolved in a solvent before exhibiting a precipitate (undissolved solid). This can be calculated with the concept of solubility product based on the equilibrium the solid creates with its ions, which for this case looks like;
[tex]PbCrO_4(s)\rightleftharpoons Pb^{2+}(aq)+CrO_4^{2-}(aq)[/tex]
Whose equilibrium expression is:
[tex]Ksp=[Pb^{2+}][CrO_4^{2-}][/tex]
Hence, since the solid and the ions are all in 1:1:1 mole ratios, we set the concentration of the ions equal to that of solid at equilibrium (saturation), in order to obtain the Ksp:
[tex]Ksp=(1.34x10^{-7})(1.34x10^{-7})\\\\Ksp=1.80x10^{-14}[/tex]
Learn more about chemical equilibrium: https://brainly.com/question/26453983