Ncert Exercises & Solutions

7.69 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
Ksp = 7.4 × 10–8 ).

When equal volumes of sodium iodate and cupric chlorate solutions are mixed together, then the molar concentrations of both solutions are reduced to half i.e., 0.001 M. Then,

$Na 10_{3} \to {Na}^{+} + {10_{3}}^{-} 0.001 M 0.001 M Cu {({ClO}_{3})}_{2} \to {Cu}^{2 +} + 2 {ClO}_{3}^{-} 0.001 M 0.001 M$

Now the solubility equilibrium for copper iodate can be written as:

$Cu {(10_{3})}_{2} \to {Cu}^{2 -}_{(aq)} + {210_{3}}^{-}_{(aq)}$

Ionic product of copper iodate:

$= [{Cu}^{2 +}] {[{10_{3}}^{-}]}^{2} = (0.001) {(0.001)}^{2} = 1 \times 10^{- 9}$

Since the ionic product $(1 \times 10^{- 9})$ is less than $K_{s p} {(7.4 \times 10)}^{- 8}$ , precipitation will not occur.

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