At 25 °C the solubility–product constant for is

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

Problem

At 25 °C the solubility–product constant for CaF2\text{CaF}_2 is Ksp=1.5×10−10K_{sp}=1.5\times10^{-10}. Determine the molar solubility of CaF2\text{CaF}_2 in a 1.0×10−4 M1.0\times10^{-4}\,\text{M} NaF solution. (Do not assume that 2x2x is negligible relative to 1.0×10−4 M1.0\times10^{-4}\,\text{M}.) (a) 1.5×10−6 M1.5\times10^{-6}\,\text{M} (b) 3.0×10−4 M3.0\times10^{-4}\,\text{M} (c) 1.5×10−2 M1.5\times10^{-2}\,\text{M} (d) 8.7×10−6 M8.7\times10^{-6}\,\text{M}

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