The table shows information about the results of chemical tests on solutions containing three different halide ions: chloride, bromide, and iodide.
| Halide ion in solution | Effect of adding silver nitrate solution (after acidifying with dilute nitric acid) | Effect of adding dilute ammonia solution |
|---|---|---|
| chloride, Cl−\text{Cl}^-Cl− | white precipitate forms | precipitate dissolves |
| bromide, Br−\text{Br}^-Br− | cream precipitate forms | precipitate remains |
| iodide, I−\text{I}^-I− | yellow precipitate forms | precipitate remains |
A student is provided with a sample of a potassium halide salt. The student dissolves some of the salt in water, acidifies with dilute nitric acid, and then adds silver nitrate solution. A precipitate forms. She then adds dilute ammonia solution and observes that the precipitate remains. She concludes that the sample contains iodide ions, I−\text{I}^-I−. Explain whether the student’s conclusion is valid.
Give a different chemical test to show that the salt contains iodide ions. Identify the test reagent(s) and the expected result.
A hydrated salt has the formula AB⋅xH2O\text{AB} \cdot x\text{H}_2\text{O}AB⋅xH2O where A\text{A}A is a positive metal ion and B\text{B}B is a negative ion. When the hydrated salt is heated, this reaction occurs:
AB⋅xH2O→AB+xH2O \text{AB} \cdot x\text{H}_2\text{O} \rightarrow \text{AB} + x\text{H}_2\text{O} AB⋅xH2O→AB+xH2OA scientist heats a sample of the hydrated salt until all the water has been lost. She records the mass of the salt before and after heating. The table shows her results.
| Mass of hydrated salt | Mass of salt after heating |
|---|---|
| 4.92 g | 2.40 g |
Describe how the scientist could make sure that all the water has been lost.
Use the scientist’s results to find the value of x x\,x in AB⋅xH2O\text{AB} \cdot x\text{H}_2\text{O}AB⋅xH2O. [Mr of AB=120,Mr of H2O=18][M_{\text{r}} \text{ of } \text{AB} = 120, \quad M_{\text{r}} \text{ of } \text{H}_2\text{O} = 18][Mr of AB=120,Mr of H2O=18]
Describe how the scientist could use a solution of the salt to find out if the negative ions are sulfate ions.
The test shows that the negative ions are sulfate ions (SO42−\text{SO}_4^{2-}SO42−). Calculate the relative atomic mass of metal A\text{A}A using the formula and MrM_{\text{r}}Mr value of the anhydrous salt, AB\text{AB}AB. [Ar of S=32,Ar of O=16][A_{\text{r}} \text{ of S} = 32, \quad A_{\text{r}} \text{ of O} = 16][Ar of S=32,Ar of O=16]
Identify metal A\text{A}A.
