1.

Following cell has EMF 0.7995V. \[Pt|{{H}_{2}}(1\,atm)|HN{{O}_{3}}(1M)||AgN{{O}_{3}}(1M)|Ag\] If we add enough \[KCl\] to the Ag cell so that the final \[C{{l}^{-}}\] is 1M. Now the measured emf of the cell is 0.222V. The \[{{K}_{sp}}\] of \[AgCl\] would be -

A. \[1\times {{10}^{-9.8}}\]
B. \[1\times {{10}^{-19.6}}\]
C. \[2\times {{10}^{-10}}\]          
D. \[2.64\times {{10}^{-14}}\]
Answer» B. \[1\times {{10}^{-19.6}}\]


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