Equal volumes of 0.002 , M solutions of sodiu | vedclass

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(B) When equal volumes of sodium iodate and cupric chlorate solutions are mixed,the concentration of each species is halved,becoming $0.001 \, M$.$NaI

Vedclass · Accepted Answer

No,precipitation will not occur.

Vedclass · Answer

(B) When equal volumes of sodium iodate and cupric chlorate solutions are mixed,the concentration of each species is halved,becoming $0.001 \, M$.$NaIO_3 ightarrow Na^{+} + IO_3^{-}$$[IO_3^{-}] = 0.001 \, M$$Cu(ClO_3)_2 ightarrow Cu^{2+} + 2ClO_3^{-}$$[Cu^{2+}] = 0.001 \, M$The ionic product $(Q_{sp})$ for $Cu(IO_3)_2$ is calculated as:$Q_{sp} = [Cu^{2+}][IO_3^{-}]^2$$Q_{sp} = (0.001) 	imes (0.001)^2$$Q_{sp} = 10^{-3} 	imes 10^{-6} = 10^{-9}$Since $Q_{sp} (1 	imes 10^{-9}) Therefore,precipitation will not occur.

Equal volumes 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 $K_{sp} = 7.4 \times 10^{-8}$)

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Equal volumes 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 $K_{sp} = 7.4 \times 10^{-8}$)