Solid calcium fluoride is sparingly soluble in water according to the equation At ,

CHEMISTRY · Multi-part · CBSE · Chemistry · Equilibrium · Equilibrium

Problem

Solid calcium fluoride is sparingly soluble in water according to the equation

\mathrm{CaF_2(s) \rightleftharpoons Ca^{2+}(aq) + 2\,F^{-}(aq)}

25\,^{\circ}\!\text{C}

K_{\text{sp}} = 1.5 \times 10^{-10}

. A student mixes

100.\,\text{mL}

1.0 \times 10^{-3}\,\text{M}

\mathrm{Ca(NO_3)_2(aq)}

with

100.\,\text{mL}

2.0 \times 10^{-4}\,\text{M}

\mathrm{NaF(aq)}

. Assume volumes are additive and no precipitation has yet occurred at the instant of mixing. (1) Part 1: a. Determine the molar concentrations of

\mathrm{Ca^{2+}}

and

\mathrm{F^{-}}

immediately after mixing, before any solid forms. (2) Part 2: b. Compute the ion-product (the reaction quotient) for the dissolution process using the concentrations from part (a). (3) Part 3: c. Compare the ion-product with

K_{\text{sp}}

and predict whether a precipitate will form. Indicate the favored direction of the net reaction. (4) Part 4: d. State the sign of

\Delta G

for the dissolution (forward) direction at the instant of mixing and justify based on your comparison in part (c).

Related theory

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