A saturated solution of in pure water is compared with a second saturated solution…

CHEMISTRY · Single correct · AP · Chemistry · Equilibrium · Equilibrium

Problem

A saturated solution of

CaF_{2}

in pure water is compared with a second saturated solution that also contains 0.10 M NaF. Which statement best explains why the second solution contains a significantly lower concentration of dissolved

Ca^{2+}

ions? (a) The added

Na^{+}

ions precipitate as

NaF\text{(s)}

, thereby removing

F^{-}

and indirectly lowering the

Ca^{2+}

concentration. (b) The added

F^{-}

shifts the dissolution equilibrium toward solid

CaF_{2}

, decreasing the concentration of

Ca^{2+}

ions. (c) The heightened ionic strength created by

0.10\,\text{M NaF}

changes

K_{sp}

, making

CaF_{2}

less soluble and thus lowering

[Ca^{2+}]

. (d) Fluoride ions bind

Ca^{2+}

to form an insoluble complex, removing

Ca^{2+}

from solution and decreasing its concentration.

Related theory

Le Chatelier's Principle
Chemical Equilibrium
Equilibrium
Solubility product
Solubility Equilibria
Common-Ion Effect

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