Chapter 15: Q53E (page 875)

Question: Perform the following calculations involving concentrations of iodate ions:
(a) The iodate ion concentration of a saturated solution of \(La{\left( {I{O_3}} \right)_3}\)was found to be\(3.1 \times 1{0^{ - 3}}mol/L\). Find the\({K_{sp}}\).
(b) Find the concentration of iodate ions in a saturated solution of \(Cu{\left( {I{O_3}} \right)_2}\left( {{K_{5p}} = 7.4 \times 1{0^{ - 8}}} \right)\).

Short Answer

Expert verified

(a).\({K_{sp}} = 3.0 \times {10^{ - 11}}\)

(b). The concentration of iodate ions is \(\begin{array}{l}\left( {C{u^{2 + }}} \right) = 2.64 \times {10^{ - 3}}{\rm{M}};\\\left( {IO_3^ - } \right) = 5.3 \times {10^{ - 3}}{\rm{M}}\end{array}\)

Step by step solution

Find \({K_{sp}}\).:

\(\begin{array}{l}{K_{sp}} = \left( {L{a^{3 + }}} \right){\left( {IO_3^ - } \right)^3}\\ = \left( {\frac{1}{3} \times 3.1 \times {{10}^{ - 3}}} \right){\left( {3.1 \times {{10}^{ - 3}}} \right)^3}\\ = 3.0 \times {10^{ - 11}}\end{array}\)

Find the concentration of iodate ions:

\(\begin{array}{l}{K_{sp}} = \left( {C{u^{2 + }}} \right){\left( {IO_3^ - } \right)^2}\\ = x{(2x)^2}\\ = 7.4 \times {10^{ - 8}}\end{array}\)

Where x,

\(\begin{array}{l}x = \sqrt(3){{\frac{{7.4 \times {{10}^{ - 8}}}}{4}}}\\ = 2.64 \times {10^{ - 3}}\end{array}\)

\(\left( {C{u^{2 + }}} \right) = 2.64 \times {10^{ - 3}}M\)

\(\begin{array}{l}\left( {IO_3^ - } \right) = 2 \times 2.64 \times {10^{ - 3}}\\ = 5.3 \times {10^{ - 3}}{\rm{M}}\end{array}\)

The solutions are \(\begin{array}{l}\left( {C{u^{2 + }}} \right) = 2.64 \times {10^{ - 3}}{\rm{M}};\\\left( {IO_3^ - } \right) = 5.3 \times {10^{ - 3}}{\rm{M}}\end{array}\)