question_answer 1)

Equal volume of 0.002 M solutions of sodium iodate and cupric chlorate are mixed together. Will it lead to precipitation of copper iodate? (For cupric iodate \[{{K}_{sp}}=7.4\times {{10}^{-8}})\]

Answer:

Given, \[[NaI{{O}_{3}}]=0.002M\] \[[Cu{{(Cl{{O}_{3}})}_{2}}]=0.002M\] \[\therefore \] \[[IO_{3}^{-}]=0.002M\,\,\,\,\,;[Cu_{2}^{+}]=0.002M\] when equal volumes of these solutions are mixed then new concentrations of \[IO_{3}^{-}\] ion \[C{{u}^{2+}}\] ions will be: \[[IO_{3}^{-}]=0.001M\,\,\,;\,\,[C{{u}^{2+}}]=0.001M\] \[{{K}_{ip}}=[C{{u}^{2+}}][IO_{3}^{-}]=0.001\times {{(0.001)}^{2}}={{10}^{-9}}\] Since, \[{{K}_{ip}}