The table shows information about the effect of adding sodium hydroxide solution and ammonia solution to solutions containing magnesium ions, aluminium ions or zinc ions.
| Ion in solution | Effect of adding sodium hydroxide solution until in excess | Effect of adding ammonia solution until in excess |
|---|---|---|
| magnesium, Mg2+\text{Mg}^{2+}Mg2+ | white precipitate forms; precipitate is insoluble in excess | white precipitate forms; precipitate is insoluble in excess |
| aluminium, Al3+\text{Al}^{3+}Al3+ | white precipitate forms; precipitate dissolves in excess | white precipitate forms; precipitate is insoluble in excess |
| zinc, Zn2+\text{Zn}^{2+}Zn2+ | white precipitate forms; precipitate dissolves in excess | white precipitate forms; precipitate dissolves in excess |
A student is provided with a sample of a white solid. The student dissolves some of the white solid in water and then adds sodium hydroxide solution until in excess. A white precipitate forms and then dissolves to leave a colourless solution. She concludes that the sample contains aluminium ions. Explain whether the student’s conclusion is valid.
Give a test to show that the white solid contains aluminium ions rather than zinc ions. Identify the test reagent and the expected result for aluminium ions.
A hydrated salt has the formula XY2⋅yH2O\text{XY}_2 \cdot y\text{H}_2\text{O}XY2⋅yH2O where X\text{X}X is a positive metal ion and Y\text{Y}Y is a negative ion. When the hydrated salt is heated, this reaction occurs:
XY2⋅yH2O→XY2+yH2O \text{XY}_2 \cdot y\text{H}_2\text{O} \rightarrow \text{XY}_2 + y\text{H}_2\text{O} XY2⋅yH2O→XY2+yH2OA chemist 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 |
|---|---|
| 14.60 g14.60\text{ g}14.60 g | 9.20 g9.20\text{ g}9.20 g |
Describe how the chemist could make sure that all the water has been lost.
Use the chemist's results to find the value of yyy in XY2⋅yH2O\text{XY}_2 \cdot y\text{H}_2\text{O}XY2⋅yH2O.
[Mr of XY2=184,Mr of H2O=18] [M_{\text{r}} \text{ of } \text{XY}_2 = 184, \quad M_{\text{r}} \text{ of } \text{H}_2\text{O} = 18] [Mr of XY2=184,Mr of H2O=18]Describe how the chemist could use a solution of the salt to find out if the negative ions are bromide ions.
The test shows that the negative ions are bromide ions. Calculate the relative atomic mass of metal X\text{X}X using the formula and MrM_{\text{r}}Mr value of the anhydrous salt, XY2\text{XY}_2XY2. Take the relative atomic mass of bromine (Br\text{Br}Br) to be 80.
Identify metal X\text{X}X.
