The `[CrO_(4)^(2-)]` ions are equilibrium for the reaction,
`Cr_(2)O_(7)^(2-)+H_(2)Oharr2CrO_(4)^(2-)+2H^(+), "at" pH=4` is:

Options

A. `10^(4)[Cr_(2)O_(7)^(2-).K_(c )]^(1//2)`
B. `10^(-8)[Cr_(2)O_(7)^(2-)].K_(c )`
C. `10^(-4)[Cr_(2)O_(7)^(2-)]^(1//2)`
D. `10^(-4)[Cr_(2)O_(7)^(2-)].K_(c _`

Answer

A

Explanation

`K_(c)=([Cr_(2)O_(7)^(2-)]xxK_(c ))/([H^(+)]^(2))=(K_(c )[Cr_(2)O_(7)^(2-)])/((10^(-4))^(2))`
`=10^(4)[K_(c )xxCr_(2)O_(7)^(2-)]^(1//2)`