The solubility of CaF_2 in a solution of 0.1 | vedclass

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Question

(C) The dissolution of $CaF_2$ is represented by the equilibrium: $CaF_2(s) \rightleftharpoons Ca^{2+}(aq) + 2F^{\Theta}(aq)$.Let the solubility of $C

Vedclass · Accepted Answer

$\frac{[F^{\Theta}]}{2}$

Vedclass · Answer

(C) The dissolution of $CaF_2$ is represented by the equilibrium: $CaF_2(s) \rightleftharpoons Ca^{2+}(aq) + 2F^{\Theta}(aq)$.Let the solubility of $CaF_2$ be $s \ mol/L$.In the presence of $0.1 \ M \ Ca(NO_3)_2$,the concentration of $Ca^{2+}$ ions is $[Ca^{2+}] = 0.1 + s$.The concentration of $F^{\Theta}$ ions is $[F^{\Theta}] = 2s$.Since $s$ is very small compared to $0.1$,we can approximate $[Ca^{2+}] \approx 0.1 \ M$.From the stoichiometry of the reaction,for every $1 \ mol$ of $CaF_2$ that dissolves,$2 \ mol$ of $F^{\Theta}$ ions are produced.Therefore,the solubility $s = \frac{[F^{\Theta}]}{2}$.

The solubility of $CaF_2$ in a solution of $0.1 \ M \ Ca(NO_3)_2$ is.

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