Question

In the system $Ca{F}_2\left(s\right)\rightleftharpoons C{a}^{2+}\left(aq\right)+2F^{-}\left(aq\right)$, if the concentration of $C{a}^{2+}$ is increased by four times, then the equilibrium concentration of $F^{-}$ will be changed to:

• A. one half of its initial value
• B. twice the initial value
• C. one fourth of its initial value
• D. thrice of its initial value

Answer 1

The correct option is A one half of its initial value

The equilibrium reaction is as follows:

$Ca{F}_2\left(s\right)\rightleftharpoons C{a}^{2+}\left(aq\right)+2F^{-}\left(aq\right)$.

The expression for the equilibrium constant is $K=\left[C{a}^{2+}\right][F^{-}{]}^2$.

Let $X$ and $Y$ be the initial concentrations of $\left[C{a}^{2+}\right]$ and $\left[F^{-}\right]$ ions.

The expression for the equilibrium constant will be $K=\left[C{a}^{2+}\right][F^{-}{]}^2=X{Y}^2$ ......$\left(1\right)$

When the concentration of the $C{a}^{2+}$ ions is increased four times, the expression for the new equilibrium constant will be $K=\left[C{a}^{2+}\right][F^{-}{]}^2=4X\times [F^{-}{]}^2$ ...... $\left(2\right)$

The magnitude of the equilibrium constant is not affected by the changes in concentrations of reactants and products.

Thus, from $\left(1\right)$ and $\left(2\right)$, we get $K=X{Y}^2=4X\times [F^{-}{]}^2$
$⟹\left[F^{-}\right]=\frac{Y}{2}$.

Thus, the equilibrium concentration of $F^{-}$ ions is reduced to one half its original value.

Hence, the correct option is $\text{A}$