Question

What would happen when a solution of Potassium chromate is treated with an excess of dilute Nitric acid?

Answer 1

• When a solution of Potassium chromate (${\mathrm{K}}_2{\mathrm{CrO}}_4$) is treated with an excess of dilute Nitric acid (${\mathrm{HNO}}_3$), the following reaction will takes place to give ${\mathrm{Cr}}_2{{\mathrm{O}}_7}^{2-}$(dichromate ion) and ${\mathrm{H}}_2\mathrm{O}$(Water) as products.
• Reaction :$$\begin{gathered} 2{\mathrm{K}}_2{\mathrm{CrO}}_4\left(\mathrm{s}\right)+2{\mathrm{HNO}}_3\left(\mathrm{aq}\right)\to {\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7\left(\mathrm{s}\right)+2{\mathrm{KNO}}_3\left(\mathrm{s}\right)+{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right) \\ (\mathrm{Potassium}\mathrm{chromate})(\mathrm{Nitric}\mathrm{acid})(\mathrm{Potassium}\mathrm{dichromate})(\mathrm{Potassium}\mathrm{nitrate})\left(\mathrm{Water}\right) \end{gathered}$$.
• Hence, the products formed are ${\mathrm{Cr}}_2{{\mathrm{O}}_7}^{2-}$and ${\mathrm{H}}_2\mathrm{O}$.