If a solution containing 0.22 M Cl -, Br -, I -, and CrO42- is treated with Ag+, in what order will the anions precipitate?
To determine the order in which the anions will precipitate, you need to refer to the solubility rules. The solubility rules provide guidelines for predicting the solubility and precipitation of different ionic compounds.
1. Start by identifying the possible precipitates:
- AgCl: Silver chloride
- AgBr: Silver bromide
- AgI: Silver iodide
- Ag2CrO4: Silver chromate
2. According to the solubility rules:
- Chlorides (Cl-) are generally soluble, except when combined with silver, lead, or mercury(I) ions.
- Bromides (Br-) are generally soluble, except when combined with silver, lead, or mercury(I) ions.
- Iodides (I-) are generally soluble, except when combined with silver, lead, or mercury(I) ions.
- Chromates (CrO42-) are generally insoluble, except when combined with alkali metal cations or ammonium ion.
3. Based on the solubility rules, the precipitates will form in the following order:
- Chloride (Cl-) will precipitate first as AgCl because silver chloride is generally insoluble.
- Bromide (Br-) will precipitate second as AgBr because silver bromide is generally insoluble.
- Iodide (I-) will precipitate next as AgI because silver iodide is generally insoluble.
- Chromate (CrO42-) will not precipitate because silver chromate is generally insoluble, except when combined with alkali metal cations or ammonium ion.
Therefore, the order of precipitation is Cl-, Br-, and I-.