Question

If $ {{K}_{sp}} $ for $ HgS{{O}_{4}} $ is $ 6.4\times {{10}^{-5}}, $ then solubility of its salt is:

• 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}} $

Answer 1

Answer: (D)

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 $