Cupric chlorate and sodium iodate having equal volume are mixed together, then the molar concentration of cupric chlorate and sodium iodate will reduce to half. So, the molar concentration of cupric chlorate and sodium iodate in a mixture is 0.001M.
\( Na(IO_3)_2 → Na^{+} + IO_{3}^{-}\)
0.0001M 0.001M
Cu(ClO3)2→ Cu2+ + \( 2CIO_{3}^{-}\)
0.001M 0.001M
The Solubility for Cu(IO3)2 ⇒ Cu2+ (aq) + 2IO3– (aq)
Now, the ionic product of the copper iodate is:
=\( [Cu^{2+}] [IO_{3}^{-}]^{2}\)
\( = (0.001)(0.001)^2\)
\( = 1.0\times 10^{-9}\)
As the value of Ksp is more than Ionic product. Thus, precipitation will not occur.
Answered by Abhisek | 1 year agoThe concentration of sulphide ion in 0.1M HCl solution saturated with hydrogen sulphide is 1.0 × 10–19 M. If 10 mL of this is added to 5 mL of 0.04M solution of the following: FeSO4, MnCl2, ZnCl2 and CdCl2 . in which of these solutions precipitation will take place?
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