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