Respuesta :

Answer: The answer is C. 0.1MCaCl2

Explanation:

We have AgCl whose least solubility is to be determined at 25degree Celsius

so upon looking at the options given we observe C 0.1MCaCl2 and D 0.1MHCl both have chlorine ions present and we know that here common ion effect will happen that causes decreases the solubility of the AgCl here

but when Calcium chloride dissociates it gives 2 moles of chlorine ions

[tex]CaCl2[/tex]→[tex]Ca^+2 + 2Cl^-[/tex]

so chlorine ion will be present in more amount [tex]2*0.1\\[/tex]=[tex]0.2[/tex] as compared to the chlorine produced by HCl

[tex]HCl[/tex]→[tex]H^+ +Cl^-[/tex] only 1 mole of chlorine is produced.

Hence AgCl will be least soluble in 0.1M CaCl2

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