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(B) The reaction between potassium chromate $(K_2CrO_4)$ and dilute nitric acid $(HNO_3)$ is an acid-base equilibrium reaction where the chromate ion

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$Cr_2O_7^{2-}$ and $H_2O$ are formed

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(B) The reaction between potassium chromate $(K_2CrO_4)$ and dilute nitric acid $(HNO_3)$ is an acid-base equilibrium reaction where the chromate ion $(CrO_4^{2-})$ is converted into the dichromate ion $(Cr_2O_7^{2-})$.The balanced chemical equation is:$2K_2CrO_4 + 2HNO_3 \rightarrow K_2Cr_2O_7 + 2KNO_3 + H_2O$In ionic form:$2CrO_4^{2-} + 2H^{+} \rightarrow Cr_2O_7^{2-} + H_2O$This is not a redox reaction; the oxidation state of chromium remains $+6$ in both $CrO_4^{2-}$ and $Cr_2O_7^{2-}$.

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

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