NCERT Class XI Chemistry Equilibrium Solutions

Question : 69

Total: 73

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).

Solution:

2NaIO3+CuCrO4 ⇌ Na2CrO4+Cu(IO3)2
After mixing, [NaIO3] = [IO3–] =

2×103

= 10−3M
[CuCrO4] = [Cu2+] =

2×103−

= 10−3M
Ionic product of Cu(IO3)2 = [Cu2+][IO3–]2 = (10–3)(10–3)2 = 10–9
As ionic product is less than Ksp, no precipitation will occur.

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