Tests for negative ions

\(\require{mhchem}\) The tests for negative ions we need to know about for GCSE are: halides (chloride \(\ce{Cl-}\), bromide \(\ce{Br-}\), iodide \(\ce{I-}\)), carbonate \(\ce{CO_3^2-}\), and sulfate \(\ce{SO_4^2-}\). The simplest test is for carbonate so let's start with that.

Carbonate ions

The basis for this test is that: $$\ce{acid + carbonate -> salt + water + carbon\ dioxide}$$ If we add acid to a substance and find that it produces \(\ce{CO2}\), we know we have a carbonate. We test for \(\ce{CO2}\) by bubbling the gas through our old friend limewater; if the limewater turns milky, the gas was \(\ce{CO2}\). So, by putting acid in our solution to be identified, and testing any gas produced with limewater, we can tell whether carbonate ions are present. (We can also use this reaction to test for an acid, by adding some carbonate and seeing if \(\ce{CO2}\) is produced. Examiners sometimes ask the question that way round.)

If we used hydrochloric acid and potassium carbonate, the full reaction would be: $$\begin{equation} \ce{2HCl(aq) + K2CO3(aq) -> 2KCl(aq) + H2O(l) + CO2(g)} \end{equation}$$ If we write out all the ions and remove the unchanged ones (the spectator ions), we are left with the ionic equation: $$\begin{equation} \ce{2H+(aq) + CO_3^{2-}(aq) -> H2O(l) + CO2(g)} \end{equation}$$

Sulfate ions

The test for sulfate ions is to acidify with hydrochloric acid and then add a solution of barium chloride. A dense white precipitate of barium sulfate forms if sulfate ions are present. The ionic equation for this reaction is simply: $$\begin{equation} \ce{SO_4^{2-}(aq) + Ba^{2+}(aq) -> BaSO4(s)} \end{equation}$$ Acid is used to get rid of any carbonate ions that would interfere with the test; hydrochloric acid is best because we already have chloride ions present from the barium chloride, so adding a few more doesn't cause problems.

Halide ions

To test for halide ions, we acidify with nitric acid then add silver nitrate solution. The product is a silver halide, and handily three of them are different colours. Chloride ions give a white precipitate, bromide give a cream precipitate, and iodide a yellow one. The colours get darker as we go down the group. Silver fluoride is soluble in water so this test can't be used for fluoride ions. An example equation using potassium chloride is: $$\begin{equation} \ce{KCl(aq) + AgNO3(aq) -> AgCl(s) + KNO3(aq)} \end{equation}$$ Removing the spectator ions to get the ionic equation gives: $$\begin{equation} \ce{Cl^{-}(aq) + Ag^+(aq) -> AgCl(s)} \end{equation}$$ Acid is used to get rid of any carbonate ions that would interfere with the test; here nitric acid is best because we already have nitrate ions present from the silver nitrate, so adding more doesn't cause any issues.

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