Chemistry - A Molecular Science

Chapter 10 Solutions

combination of the cation in one compound with the anion in the other results in a compound that is insoluble. For example, consider mixing solutions of AgNO

and NaCl. 3

Both substances are strong electrolytes, so they are found as their separated ions in aqueous solution. Thus, we represent the tw

o solutions in different boxes as follows:

1

Ag

1+ + NO

1- + Na 3

1+ + Cl

1-^

We then designate the two new cation-anion co

mbinations that are possible in the mixed

solution by the lines labeled 1 and 2. The po

ssibilities for opposite charge interaction are

therefore

2

1) Ag

1+ + Cl

1-^

2) Na

1+ + NO

1-^3

Rule 3 states that AgCl is insoluble,

so it precipitates from solution, while NaNO

is 3

soluble (Rules 1 and 2) and does not precipitate. Example 10.9

Write the formula of the precipitate that forms when the following aqueous solutions are mixed or write none if no precipitate is expected.

a) Solutions of FeSO

and KOH 4

Express the solutions as the separated ions as shown in the margin and determine the new cation-anion combinations. (1) Fe

2+ + OH

1- and (2) K

1+ + SO

2- 4

. No precipitate results

from the K

1+ + SO

2- 4

combination (Rules 1 and 4), but neither Fe

2+ nor OH

1- are listed in

the solubility rules, so combination (1) produces an insoluble compound (Rule 5). The +2 charge on Fe

2+ requires 2OH

1-, so the formula of the precipitate is Fe(OH)

. 2

a) Fe

2+ + SO

2- + K 4

1+ + OH

1-

b) Pb

2+ + ClO

1- + Cu 4

2+ + SO

2-^4

c) K

1+ + PO

3- 4

+ Co

2+ + Cl

1-^

Example 10.9

1 2 1 2 1 2

b) Solutions of Pb(ClO

) 42

and CuSO

(^4)
As shown in the margin, the new cation-anion combinations that are possible upon mixing are (1) Pb
2+ + SO
2- and (2) Cu 4
2+ + ClO
1-. No precipitate results from Cu 4
2+ + ClO
1- 4
(Rule
2), but the combination of Pb
2+ + SO
2- 4
is insoluble (Rule 4). The formula of the precipitate
is PbSO

. 4

c) Solutions of K

PO 3

and CoCl 4

(^2)
Mixing introduces the following interactions: 1) K
1+ + Cl
1- and 2) Co
2+ + PO
3- 4
.
Combination (1) results in KCl, which is soluble (Rules 1 and 3). Neither Co
2+ nor PO
3- 4
is
listed in the solubility rules, so combination (2) produces a precipitate. The lowest common multiple of the +2 and -3 charges is 6, so three cations and two anions are required in the formula. The precipitate is cobalt(II) phosphate: Co
(PO 3
) 42
.
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