Tests for Anions
Tests for Anions
We often want to know which ions are present in solution. If we know which salts precipitate, we can derive tests to identify ions in solution. Given below are a few such tests.
Test for a chloride
Prepare a solution of the unknown salt using distilled water and add a small amount of silver nitrate solution. If a white precipitate forms, the salt is either a chloride or a carbonate.
\[\text{Cl}^{-}\text{(aq)} + \text{Ag}^{+}\text{(aq)} + \text{NO}_{3}^{-}\text{(aq)} \rightarrow \text{AgCl (s)} + \text{NO}_{3}^{-}\text{(aq)}\](AgCl is white precipitate)
\[\text{CO}_{3}^{2-}\text{(aq)} + 2\text{Ag}^{+}\text{(aq)} + 2\text{NO}_{3}^{-}\text{(aq)} \rightarrow \text{Ag}_{2}\text{CO}_{3}\text{(s)} + 2\text{NO}_{3}^{-}\text{(aq)}\](\(\text{Ag}_{2}\text{CO}_{3}\) is white precipitate)
The next step is to treat the precipitate with a small amount of concentrated nitric acid. If the precipitate remains unchanged, then the salt is a chloride. If carbon dioxide is formed and the precipitate disappears, the salt is a carbonate.
\[\text{AgCl (s)} + \text{HNO}_{3}\text{(l)} \rightarrow \text{ no reaction. precipitate is unchanged}\]\[\text{Ag}_{2}{CO}_{3}\text{ (s)} + 2\text{HNO}_{3}\text{(l)} \rightarrow 2\text{Ag}^{+}\text{(aq)} + 2\text{NO}_{3}^{-}\text{(aq)} + \text{H}_{2}\text{O (l)} + \text{CO}_{2}\text{ (g) precipitate disappears}\]Test for bromides and iodides
As was the case with the chlorides, the bromides and iodides also form precipitates when they are reacted with silver nitrate. Silver chloride is a white precipitate, but the silver bromide and silver iodide precipitates are both pale yellow. To determine whether the precipitate is a bromide or an iodide, we use chlorine water and carbon tetrachloride (\(\text{CCl}_{4}\)).
Chlorine water frees bromine gas from the bromide and colours the carbon tetrachloride a reddish brown.
\[2\text{Br}^{-}\text{(aq)} + \text{Cl}_{2}\text{(aq)} \rightarrow 2\text{Cl}^{-}\text{(aq)} + \text{Br}_{2}\text{(g)}\]Chlorine water frees iodine gas from an iodide and colours the carbon tetrachloride purple.
\[2\text{I}^{-}\text{(aq)} + \text{Cl}_{2}\text{(aq)} \rightarrow 2\text{Cl}^{-}\text{(aq)} + \text{I}_{2}\text{(g)}\]Test for a sulphate
Add a small amount of barium chloride solution to a solution of the test salt. If a white precipitate forms, the salt is either a sulfate or a carbonate.
\[\text{SO}_{4}^{2-}\text{(aq)} + \text{Ba}^{2+}\text{(aq)} + \text{Cl}^{-}\text{(aq)} \rightarrow \text{BaSO}_{4}\text{(s)} + \text{Cl}^{-}\text{(aq)}\](\(\text{BaSO}_{4}\) is a white precipitate)
\[\text{CO}_{3}^{2-}\text{(aq)} + \text{Ba}^{2+}\text{(aq)} + \text{Cl}^{-}\text{(aq)} \rightarrow \text{BaCO}_{3}\text{(s)} + \text{Cl}^{-}\text{(aq)}\](\(\text{BaCO}_{4}\) is a white precipitate)
If the precipitate is treated with nitric acid, it is possible to distinguish whether the salt is a sulphate or a carbonate (as in the test for a chloride).
\[\text{BaSO}_{4}\text{(s)} + \text{HNO}_{3}\text{(l)} \rightarrow \text{ no reaction. precipitate is unchanged}\]\[\text{BaCO}_{3}\text{(s)} + 2\text{HNO}_{3}\text{(l)} \rightarrow \text{Ba}^{2+}\text{(aq)} + 2\text{NO}_{3}^{-}\text{(aq)} + \text{H}_{2}\text{O (l)} + \text{CO}_{2}\text{(g) precipitate disappears}\]Test for a carbonate
If a sample of the dry salt is treated with a small amount of acid, the production of carbon dioxide is a positive test for a carbonate.
\[2\text{HCl} + \text{K}_{2}\text{CO}_{3}\text{(aq)} \rightarrow \text{CO}_{2}\text{(g)} + 2\text{KCl (aq)} + \text{H}_{2}\text{O (l)}\]If the gas is passed through limewater (an aqueous solution of calcium hydroxide) and the solution becomes milky, the gas is carbon dioxide.
\[\text{Ca}^{2+}\text{(aq)} + 2\text{OH}^{-}\text{(aq)} + \text{CO}_{2}\text{(g)} \rightarrow \text{CaCO}_{3}\text{(s)} + \text{H}_{2}\text{O (l)}\](It is the insoluble \(\text{CaCO}_{3}\) precipitate that makes the limewater go milky)
This lesson is part of:
Reactions in Aqueous Solution