Shweta Thakur posted an Question
October 16, 2020 • 20:14 pm 30 points
  • IIT JAM
  • Chemistry (CY)

Plz explain this e is insoluble or sparingly soluble, to be more accurate.

Home Tools Puri Sharma Patha 384 704 4. Prefereatial Precipitation of an msoluble Salt. Silver chloride is insoluble or sparingly soluble, to be more accurate. So is silver iodide. The question. arises as to what would happen if potassium iodide solution is added to silver chloridc. Would the reaction KI KCI AgCI Sparingly soluble Agl Sparingly soluble take place towards the right, i.e., would the precipitate of silver chloride change into the precipitate of silver iodide ? For answer, it is necessary to ook to their respective solubility products. Solubility product of silver chloride Kp(AgC) [Ag ICI]= 1-56x10-10 (Table 7) and that of silver iodide, gpAgnAgI 0-94x1016 (Table 7) C 156x10 166 x 10 0.94 x 10-16-66 x 10 This means that at equilibrium, the concentration of Cl ions in solution is more than a million times greater than that of ions n other words, practically nothing of the 1onS can remain in solution at equilibrium. As the ions can only be removed as Agl, it means that the Feaction proceeds virtually to completion towards the right. As a rule, the compound with the lower solubility product gets precipitated in preference. Silver iodide has lower solubility product than silver chloride. Therefore, the former gets precipitated in preference to the latter 11.69 x8.26 in

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    Priyanshu kumar Best Answer

    Just for comparison... In general,the compound with lower solubility product gets precipitated first..Here AgI has lower Ksp than Agcl ..so it will precipitate first

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    why sir??

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    see the comparison cl- ions is millions time more available than I- ions ...that means nearly all iodide ions has been removed...so how it will remove .it will remove as ppt of AgI

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    so in general we say low ksp value gets ppt first

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    sir

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