A) \[6.4\times {{10}^{-5}}\,\text{mol}\,\,\text{litr}{{\text{e}}^{-1}}\]
B) \[6.4\times {{10}^{-3}}\text{ mol litr}{{\text{e}}^{-1}}\]
C) \[8\times {{10}^{-6}}\text{ mol litr}{{\text{e}}^{-1}}\]
D) \[8\times {{10}^{-3}}\text{ mol litr}{{\text{e}}^{-1}}\]
Correct Answer: D
Solution :
Given that, \[{{K}_{sp}}\] for \[HgS{{O}_{4}}=6.4\times {{10}^{-5}}\] \[HgS{{O}_{4}}\] is a binary salt, hence, for it, \[{{K}_{sp}}={{S}^{2}}\] \[\therefore \] \[S=\sqrt{{{K}_{sp}}}\] \[=\sqrt{6.4\times {{10}^{-5}}}=8\times {{10}^{-3}}\,mol\text{/}litre\]
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