Which one of the following arrangements represents the correct order of solubilities of sparingly soluble salts \(\mathrm{Hg_2Cl_2}\), \(\mathrm{Cr_2(SO_4)_3}\), \(\mathrm{BaSO_4}\) and \(\mathrm{CrCl_3}\) respectively?
Show Hint
- \(\mathrm{BaSO_4} \gt \mathrm{Hg_2Cl_2} \gt \mathrm{Cr_2(SO_4)_3} \gt \mathrm{CrCl_3}\)
- \(\mathrm{BaSO_4} \gt \mathrm{Hg_2Cl_2} \gt \mathrm{CrCl_3} \gt \mathrm{Cr_2(SO_4)_3}\)
- \(\mathrm{BaSO_4} \gt \mathrm{CrCl_3} \gt \mathrm{Hg_2Cl_2} \gt \mathrm{Cr_2(SO_4)_3}\)
- \(\mathrm{Hg_2Cl_2} \gt \mathrm{BaSO_4} \gt \mathrm{CrCl_3} \gt \mathrm{Cr_2(SO_4)_3}\)
The Correct Option is B
Solution and Explanation
\(\mathrm{BaSO_4}\): \(K_{sp} pprox 1 imes10^{-10}\), so \(s = \sqrt{K_{sp}} pprox 1 imes10^{-5}\) M.
\(\mathrm{Hg_2Cl_2}\): \(K_{sp} pprox 1.4 imes10^{-18}\), so \(s = (K_{sp}/4)^{1/3} pprox 7 imes10^{-7}\) M.
So \(\mathrm{BaSO_4} \gt \mathrm{Hg_2Cl_2}\), which rules out option (4).
The anhydrous chromium(III) salts are even less soluble in water because the \(\mathrm{Cr^{3+}}\) ion is kinetically inert (they dissolve very slowly), and the standard order is \(\mathrm{CrCl_3} \gt \mathrm{Cr_2(SO_4)_3}\) (the chloride is the more soluble of the two).
Hence the order is \(\mathrm{BaSO_4} \gt \mathrm{Hg_2Cl_2} \gt \mathrm{CrCl_3} \gt \mathrm{Cr_2(SO_4)_3}\).
Answer: (2) \(\mathrm{BaSO_4} \gt \mathrm{Hg_2Cl_2} \gt \mathrm{CrCl_3} \gt \mathrm{Cr_2(SO_4)_3}\)
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