In acidic medium, the rate of reaction between `[BrO_(3)^(-)]and Br^(-)` ions is given by the expression `-(d[BrO_(3)^(-)])/(dt)=k[BrO_(3)^(-)][Br^(-)][H^(+)]^(2)` It means

Options

A. Rate constant for the reaction depends upon the concentration of `H^(+)` ions
B. Rate of reaction is independent of the conc. Of acid added
C. The change in pH of the solution will affect the rate of reaction
D. doubling the conc. Of `H^(+)` ions will increase the reaction rate by 4 times

Answer

C, D

Explanation

As `[H^(+)]` comes in the rate law expression, a change in `[H^(+)]` i.e., a change in pH will change the rate of reaction. Here
rate `= k[BrO_(3)^(-)][Br^(-)][H^(+)]^(2)`
Therefore, doubling the `[H^(+)]` will increase the rate by `2^(2)` i.e., 4 times.