Question

CdF2(s)⇄Cd2+(aq)+2F−(aq)A saturated aqueous solution of CdF2 is prepared. The equilibrium in the solution is represented above. In the solution, [Cd2+]eq=0.0585M and [F−]eq=0.117M. Some 0.90MNaF is added to the saturated solution. Which of the following identifies the molar solubility of CdF2 in pure water and explains the effect that the addition of NaF has on this solubility?a. The molar solubility of CdF2 in pure water is 0.0585M, and adding NaF decreases this solubility because the equilibrium shifts to favor the precipitation of some CdF2.b. The molar solubility of CdF2 in pure water is 0.0585M, and adding NaF has no effect on the solubility because only changes in temperature can increase or decrease the molar solubility of an ionic solid.c. The molar solubility of CdF2 in pure water is 0.117M, and adding NaF decreases this solubility because the equilibrium shifts to favor the precipitation of some CdF2.d. The molar solubility of CdF2 in pure water is 0.176M, and adding NaF increases this solubility because the Na+ ions displace the Cd2+ ions, causing the equilibrium to shift to favor the products.

Answer 1

Answer:

The correct answer is option a.

Explanation:

$CdF_2(s)\rightleftharpoons Cd^{2+}(aq)+2F^-(aq)$

Equilibrium concentration cadmium ions = $[Cd^{2+}]=0.0585 M$

Equilibrium concentration fluoride ions = $[F^{-}]=0.117 M$

Molar solubility is the maximum concentration of salt present in water in ionic form beyond that no more salt will exist in its ionic form and will settle down in bottom of the solution.

The molar solubility of the solid cadmium fluoride = 0.0585 M

$CdF_2(s)\rightleftharpoons Cd^{2+}(aq)+2F^-(aq)$..[1]

$NaF(s)\rightleftharpoons Na^{+}(aq)+F^-(aq)$

Due to addition of sodium fluoride will increase concentration of fluoride in the solution.And due to common ion effect the equilibrium will shift in backward direction in [1], that is precipitation of more cadmium fluoride.

Hence, decrease in solubility will be observed.