The anions to be tested in the lab are sulfate, carbonate, phosphate, chloride,

Question

1. The anions to be tested in the lab are sulfate, carbonate, phosphate, chloride, bromide and iodide. In Part A of the procedure, you will determine which of these ions form a precipitate with barium ion. Using the Ksp table on page A24 in your textbook, predict which of these anions should precipitate with barium ion. THEN calculate the minimum concentration of each of these anions which would be required for precipitation to occur if the [Ba2+] = 0.20 M
2. In Part C of this experiment, you will determine which of these ions form a precipitate with silver ion. Using the same anions listed in problem #1, determine which of the anions should precipitate with silver ion. THEN calculate the minimum concentration of each of these anions which would be required for precipitation to occur if the [Ag+] =0.40 M.

Explanation / Answer

part A

Solubility Rules

Whether or not a reaction will form a precipitate is dictated by the solubility rules. These rules provide us with the guidelines that tell which ions form solids and which remain in their ionic form in aqueous solution. The rules are to be followed from the top down, meaning that if something is insoluble (or soluble) due to rule 1, it has precedence over a higher-numbered rule, for example rule 4.

If the rules state that an ion is soluble, then it will remain in its aqueous ion form. If an ion is insoluble based on the solubility rules, then it will form a solid with an ion from the other reactant in an equation. If all the ions in a reaction are shown to be soluble, then no precipitation reaction occurs.

from rule 3 Bromides, chlorides, and iodides are soluble

chloride, bromide and iodide dont form a precipitate with barium

from rule 4 sulfates of barium are insoluble

so sulfate ions form a precipitate with barium ion

from rule 6 carbonates and phosphates are insoluble.

so carbonate, phosphate ions form a precipitate with barium ion

part B Ksp

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