Asolution contains 0.021 m cl? and 0.017 m i? . a solution containing copper (i) ions is added to selectively precipitate one of the ions. at what concentration of copper (i) ion will a precipitate begin to form? what is the identity of the precipitate? ksp(cucl) = 1.0 × 10-6, ksp(cui) = 5.1 × 10-12.
a) 4 .8 × 10-5 m, cuclb) 3 .0 × 10-10 m, cuic) 3 .0 × 10-10 m, cucld) 4 .8 × 10-5 m, cuie) n o precipitate will form at any concentration of copper (i).

A precipitate will begin to form at [Cu+] = 3.0 *10^-10 M

The precipitate formed is CuI

Explanation:

Step 1: Data given

The solution contains 0.021 M Cl- and 0.017 M I-.

Ksp(CuCl) = 1.0 × 10-6

Ksp(CuI) = 5.1 × 10-12.

Step 2:  Calculate [Cu+]

Ksp(CuCl) = [Cu+] [Cl-]  

1.0 * 10^-6  = [Cu+] [Cl-]  

1.0 * 10^-6  = [Cu+] [0.021]  

[Cu+] = 1.0 * 10^-6 / 0.021

[Cu+] = 4.76 *10^-5 M

Ksp(CuI) = [Cu] [I]  

5.1 * 10^-12  = [Cu+] [I-]  

5.1 * 10^-12 =[Cu+] [0.017]  

[Cu+] = 5.1 * 10^-12 / 0.017

[Cu+] = 3.0 *10^-10 M

[Cu+]from CuI hast the lowest concentration

A precipitate will begin to form at [Cu+] = 3.0 *10^-10 M

The precipitate formed is CuI

answer: - 3.) the oxidation state of nitrogen in no changes from +2 to 0, and the oxidation state of carbon in co changes from +2 to +4 as the reaction proceeds.

solution: - as per the rules, oxidation number of oxygen in its compounds is -2. the sum of oxidation numbers of all the elements of a molecule or compound is zero.

also, the oxidation number in elemental form is zero.

in no, the oxidation number of o is -2 so the oxidation number of n is +2 so that the sum is zero.

similarly, the oxidation number of c in co is +2 and o is -2.

on product side the oxidation number of n is 0 as it is in the elemental form() .

oxidation number of c in carbon dioxide is +4 as there are two oxygen in it and we know that each oxygen is -2.

so, oxidation number of n is changing from +2 to 0 and the oxidation number of c is changing from +2 to +4.

since nitrogen is reduced and hence no is acting as an oxidizing agent. carbon is oxidized and so co is acting as a reducing agent.

so, the right choice is number 3.)the oxidation state of nitrogen in no changes from +2 to 0, and the oxidation state of carbon in co changes from +2 to +4 as the reaction proceeds.

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