Respuesta :
Answer: The equilibrium concentration of [tex]Ca^{2+}\text{ and }CrO_4^{2-}[/tex] are 0.0266 M.
Explanation:
The chemical equation for the ionization of calcium chromate follows:
[tex]CaCrO_4\rightleftharpoons Ca^{2+}+CrO_4^{2-}[/tex]
The expression for equilibrium constant is given as:
[tex]K_c=\frac{[Ca^{2+}][CrO_4^{2-}]}{[CaCrO_4]}[/tex]
We are given:
[tex]K_c=7.1\times 10^{-4}[/tex]
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 [tex]Ca^{2+}\text{ and }CrO_4^{2-}[/tex] be 'x'
Putting values in above equation, we get:
[tex]7.1\times 10^{-4}=x^2\\\\x=0.0266M[/tex]
Hence, the equilibrium concentration of [tex]Ca^{2+}\text{ and }CrO_4^{2-}[/tex] are 0.0266 M.
