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NCERT Class XI Chemistry Equilibrium Solutions

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Question : 69 of 73
Marks: +1, -0
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 KspK_{sp} = 7.4 × 10−810^{-8}).
Solution:  
2NaIO3+CuCrO42\mathrm{NaIO}_3 + \mathrm{CuCrO}_4 ⇌ Na2CrO4+Cu(IO3)2\mathrm{Na}_2\mathrm{CrO}_4 + \mathrm{Cu}(\mathrm{IO}_3)_2
After mixing, [NaIO3][\mathrm{NaIO}_3] = [IO3−][\mathrm{IO}_3^{-}] = 2×1032\frac{2 \times 10^3}{2} = 10−3 M10^{-3}\,\mathrm{M}
[CuCrO4][\mathrm{CuCrO}_4] = [Cu2+][\mathrm{Cu}^{2+}] = 2×103−2\frac{2 \times 10^{3-}}{2} = 10−3 M10^{-3}\,\mathrm{M}
Ionic product of Cu(IO3)2\mathrm{Cu}(\mathrm{IO}_3)_2 = [Cu2+][IO3−]2[\mathrm{Cu}^{2+}][\mathrm{IO}_3^{-}]^2 = (10−3)(10−3)2(10^{-3})(10^{-3})^2 = 10−910^{-9}
As ionic product is less than Ksp, no precipitation will occur.
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