Question
What would happen when a solution of potassium chromate is treated with an excess of dilute nitric acid?
A.
$$C{r_2}O_7^{2 - }\,{\text{and}}\,{H_2}O\,{\text{are}}\,{\text{formed}}$$
B.
$$CrO_4^{2 - }\,{\text{is}}\,{\text{reduced}}\,{\text{to}}\, + 3\,{\text{state}}\,{\text{of}}\,Cr$$
C.
$$CrO_4^{2 - }\,{\text{is}}\,{\text{oxidized}}\,{\text{to}}\, + 7\,{\text{state}}\,{\text{of}}\,Cr$$
D.
$$C{r^{3 + }}\,{\text{and}}\,C{r_2}O_7^{2 - }\,{\text{are}}\,{\text{formed}}$$
Answer :
$$C{r_2}O_7^{2 - }\,{\text{and}}\,{H_2}O\,{\text{are}}\,{\text{formed}}$$
Solution :
When a solution of potassium chromate is treated with an excess of dilute nitric acid. Potassium dichromate and $${H_2}O$$ are formed.
$$\eqalign{
& 2{K_2}Cr{O_4} + 2HN{O_3} \to {K_2}C{r_2}{O_7} + 2KN{O_3} + {H_2}O \cr
& {\text{Hence}}\,C{r_2}O_7^ - \,{\text{and}}\,{H_2}O\,{\text{are}}\,{\text{formed}}{\text{.}} \cr} $$