Question

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 × $10^{-8}$.

Answer 1

${ K }_{ sp }\quad of\quad Cu(I{ O }_{ 3 }{ ) }_{ 2 }\quad =\quad 7.4\quad \times \quad 1{ 0 }^{ -7 }$

When equal volume of sodium iodate and copper chlorate are mixed the molar concentration of both the solutions would be reduced to half i.e., 0.001 M

$\begin{matrix} NaI{ O }_{ 3 }\\  0.001M \end{matrix}\quad \rightarrow \quad \begin{matrix} { Na }^{ + } \\ 0.001M \end{matrix}\quad +\quad I{ O }_{ 3 }^{ - }\begin{matrix} Cu(Cl{ O }_{ 3 }{ ) }_{ 2 } \\ 0.001M \end{matrix}\quad \rightarrow \quad \begin{matrix} C{ u }^{ 2+ }\\ 0.001M \end{matrix}\quad +\quad 2Cl{ O }_{ 3 }^{ - }$

$After\quad mixing\quad [I{ O }_{ 3 }^{ - }]\quad =\quad [NaI{ O }_{ 3 }]\quad =\quad 0.001\quad M$

$[C{ u }^{ 2+ }]\quad =\quad [Cu(I{ O }_{ 3 }{ ) }_{ 2 }]$

= 0.001M

$Cu(I{ O }_{ 3 }{ ) }_{ 2(s) }\quad \rightleftharpoons \quad C{ u }^{ 2+ }(aq)\quad +\quad 2I{ O }_{ 3(aq) }^{ - }$

$Ionic\quad product\quad =\quad [C{ u }^{ 2+ }][I{ O }_{ 3 }^{ - }{ ] }^{ 2 }$

$=\quad (0.001)\quad (0.001{ ) }^{ 2 }$

$Ionic\quad product\quad =\quad 1\quad \times \quad { 10 }^{ -9 }$