Question

in the laboratory you are given the task of separating ca²+ and cu²+ ions in aqueous solution.
for each reagent listed below indicate if it can be used to separate the ions. type \y\ for yes or
\ for no. if the reagent can be used to separate the ions, give the formula of the precipitate. if it cannot, type
o\

Explanation:

Step1: Analyze the reaction of \(Na_2CO_3\) with \(Ca^{2 + }\) and \(Cu^{2+}\)

When \(Na_2CO_3\) is added to a solution containing \(Ca^{2 + }\) and \(Cu^{2+}\), the following reactions occur:
\(Ca^{2+}+CO_3^{2 - }=CaCO_3\downarrow\) (white precipitate)
\(Cu^{2+}+CO_3^{2 - }=CuCO_3\downarrow\) (blue - green precipitate). Since both ions form precipitates, \(Na_2CO_3\) cannot be used to separate them.

Step2: Analyze the reaction of \(K_2SO_4\) with \(Ca^{2 + }\) and \(Cu^{2+}\)

When \(K_2SO_4\) is added:
\(Ca^{2+}+SO_4^{2 - }=CaSO_4\downarrow\) (slightly soluble, can be considered as a precipitate in this context)
\(Cu^{2+}\) does not react with \(SO_4^{2 - }\) under normal conditions. So \(K_2SO_4\) can be used to separate them. The formula of the precipitate is \(CaSO_4\).

Step3: Analyze the reaction of \(KOH\) with \(Ca^{2 + }\) and \(Cu^{2+}\)

When \(KOH\) is added:
\(Ca^{2+}+2OH^{-}=Ca(OH)_2\downarrow\) (slightly soluble, can be considered as a precipitate in this context)
\(Cu^{2+}+2OH^{-}=Cu(OH)_2\downarrow\) (blue precipitate). Since both ions form precipitates, \(KOH\) cannot be used to separate them.

Answer:

1. N
2. Y, \(CaSO_4\)
3. N