<h2>Tests for Anions</h2>
<p>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.</p>
<section>
<h3 class="title">Test for a chloride</h3>
<p>Prepare a solution of the unknown <a data-lesson="acid-base-reactions-continued-2" data-subject="chemistry" data-tutorial="acid-base-redox-reactions" href="https://app.ulearngo.com/chemistry/acid-base-redox-reactions/acid-base-reactions-continued-2">salt</a> using distilled water and add a small amount of <strong>silver nitrate</strong> solution. If a white precipitate forms, the salt is either a chloride or a carbonate.</p>

\[\text{Cl}^{-}\text{(aq)} + \text{Ag}^{+}\text{(aq)} + \text{NO}_{3}^{-}\text{(aq)} \rightarrow \text{AgCl (s)} + \text{NO}_{3}^{-}\text{(aq)}\]

<p>(AgCl is white precipitate)</p>

\[\text{CO}_{3}^{2-}\text{(aq)} + 2\text{Ag}^{+}\text{(aq)} + 2\text{NO}_{3}^{-}\text{(aq)} \rightarrow \text{Ag}_{2}\text{CO}_{3}\text{(s)} + 2\text{NO}_{3}^{-}\text{(aq)}\]

<p>(\(\text{Ag}_{2}\text{CO}_{3}\) is white precipitate)</p>
<p>The next step is to treat the precipitate with a small amount of <strong>concentrated nitric acid</strong>. 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.</p>

\[\text{AgCl (s)} + \text{HNO}_{3}\text{(l)} \rightarrow \text{ no reaction. precipitate is unchanged}\]\[\text{Ag}_{2}{CO}_{3}\text{ (s)} + 2\text{HNO}_{3}\text{(l)} \rightarrow 2\text{Ag}^{+}\text{(aq)} + 2\text{NO}_{3}^{-}\text{(aq)} + \text{H}_{2}\text{O (l)} + \text{CO}_{2}\text{ (g) precipitate disappears}\]</section>
<section>
<h3 class="title">Test for bromides and iodides</h3>
<p>As was the case with the chlorides, the bromides and iodides also form precipitates when they are reacted with <a data-lesson="isolation-of-silver" data-subject="chemistry" data-tutorial="transition-metals" href="https://app.ulearngo.com/chemistry/transition-metals/isolation-of-silver">silver</a> 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 (\(\text{CCl}_{4}\)).</p>
<p>Chlorine water frees bromine <a data-lesson="non-ideal-gas-behavior" data-subject="chemistry" data-tutorial="gases" href="https://app.ulearngo.com/chemistry/gases/non-ideal-gas-behavior">gas</a> from the bromide and colours the carbon tetrachloride a reddish brown.</p>

\[2\text{Br}^{-}\text{(aq)} + \text{Cl}_{2}\text{(aq)} \rightarrow 2\text{Cl}^{-}\text{(aq)} + \text{Br}_{2}\text{(g)}\]

<p>Chlorine water frees iodine gas from an iodide and colours the carbon tetrachloride purple.</p>

\[2\text{I}^{-}\text{(aq)} + \text{Cl}_{2}\text{(aq)} \rightarrow 2\text{Cl}^{-}\text{(aq)} + \text{I}_{2}\text{(g)}\]</section>
<section>
<h3 class="title">Test for a sulphate</h3>
<p>Add a small amount of <a data-lesson="using-barium-sulfate-for-medical-imaging" data-subject="chemistry" data-tutorial="equilibria-other-reaction-classes" href="https://app.ulearngo.com/chemistry/equilibria-other-reaction-classes/using-barium-sulfate-for-medical-imaging">barium</a> chloride solution to a solution of the test salt. If a white precipitate forms, the salt is either a sulfate or a carbonate.</p>

\[\text{SO}_{4}^{2-}\text{(aq)} + \text{Ba}^{2+}\text{(aq)} + \text{Cl}^{-}\text{(aq)} \rightarrow \text{BaSO}_{4}\text{(s)} + \text{Cl}^{-}\text{(aq)}\]

<p>(\(\text{BaSO}_{4}\) is a white precipitate)</p>

\[\text{CO}_{3}^{2-}\text{(aq)} + \text{Ba}^{2+}\text{(aq)} + \text{Cl}^{-}\text{(aq)} \rightarrow \text{BaCO}_{3}\text{(s)} + \text{Cl}^{-}\text{(aq)}\]

<p>(\(\text{BaCO}_{4}\) is a white precipitate)</p>
<p>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).</p>

\[\text{BaSO}_{4}\text{(s)} + \text{HNO}_{3}\text{(l)} \rightarrow \text{ no reaction. precipitate is unchanged}\]\[\text{BaCO}_{3}\text{(s)} + 2\text{HNO}_{3}\text{(l)} \rightarrow \text{Ba}^{2+}\text{(aq)} + 2\text{NO}_{3}^{-}\text{(aq)} + \text{H}_{2}\text{O (l)} + \text{CO}_{2}\text{(g) precipitate disappears}\]</section>
<section>
<h3 class="title">Test for a carbonate</h3>
<p>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.</p>

\[2\text{HCl} + \text{K}_{2}\text{CO}_{3}\text{(aq)} \rightarrow \text{CO}_{2}\text{(g)} + 2\text{KCl (aq)} + \text{H}_{2}\text{O (l)}\]

<p>If the gas is passed through limewater (an aqueous solution of calcium hydroxide) and the solution becomes milky, the gas is carbon dioxide.</p>

\[\text{Ca}^{2+}\text{(aq)} + 2\text{OH}^{-}\text{(aq)} + \text{CO}_{2}\text{(g)} \rightarrow \text{CaCO}_{3}\text{(s)} + \text{H}_{2}\text{O (l)}\]

<p>(It is the insoluble \(\text{CaCO}_{3}\) precipitate that makes the limewater go milky)</p>
</section>

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Tests for Anions

Chemistry generally explores the substances of which matter is composed, the investigation of their properties and reactions, and the use of such reactions to form new substances. 

Chemistry

Explore how different substances dissolve and react with each other in water, including ionization and conductivity, precipitation reactions, tests for anions, and various types of reactions such as redox and ion exchange.


Tests for Anions

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