This week we will measure the equilibrium molar solubility of the ionic compound

Question

This week we will measure the equilibrium molar solubility of the ionic compound silver chromate, Ag_2 CrO_4, in three different solutions and use the results to: Determine the equilibrium constant associated with its solubility; the solubility product, K_sp: Ag_2 CrO_4 (s) rightwardsarrowoverleftwardsarrow 2 Ag^+_(aq) + CrO^2-_4(aq) K_sp = [Ag^+]^2 [CrO^2-_4] Show how the value of the solubility product is independent of the presence of initial Common ions. It turns out that the value of K_sp depends slightly on the presence of "uncommon ions", that is, spectator ions that might be present in the solutions. So, adding common ions with their associated spectator ions cause rather complicated changes in molar solubility and K_sp. To avoid these complicated fluctuations in our measured K_sp values from different concentrations of spectator ions, all solutions used in our measurements contain a large excess of NaNO_3 spectator ions. The chromate ion, CrO^2-_4, is yellow in aqueous solution and absorbs light of wavelength 375 nm. This absorption will enable the measurement of the concentration of chromate ion in the equilibrium solutions. The first step in our procedure is the calibration of a spectrophotometer like we did in the FeSCN^2+ experiment.

Explanation / Answer

1,

The solubility product is related to solubility,which can be explained as follows-

Ag2CrO42Ag+ +CrO42-

S                     2S          S

Ksp=[Ag+]2 [CrO42-]

=[2S ]2x[ S]

=4S2 x S

=4 S3

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