Question

Consider the following reaction at 25 °C: CaCrO4s)-Ca2*la)+CrO,2(aq) Ko 7.1 104 What are the equilibrium concentrations of Ca2 and CrO42?

Answer 1

Answer: The equilibrium concentration of
`Ca^(2+)\text{ and }CrO_4^(2-)` are 0.0266 M.

Step-by-step explanation:

The chemical equation for the ionization of calcium chromate follows:

`CaCrO_4\rightleftharpoons Ca^(2+)+CrO_4^(2-)`

The expression for equilibrium constant is given as:

`K_c=([Ca^(2+)][CrO_4^(2-)])/([CaCrO_4])`

We are given:

`K_c=7.1* 10^(-4)`

The concentration of solid substances are taken to be 1. Thus, they do not appear in the equilibrium constant expression.

Let the equilibrium concentration for
`Ca^(2+)\text{ and }CrO_4^(2-)` be 'x'

Putting values in above equation, we get:

`7.1* 10^(-4)=x^2\\\\x=0.0266M`

Hence, the equilibrium concentration of
`Ca^(2+)\text{ and }CrO_4^(2-)` are 0.0266 M.