The solubility product of CaF is and the acid-dissociation constant of HF is

CHEMISTRY · Single correct · AP · Chemistry · Acids and Bases · Acids and Bases

Problem

The solubility product of CaF2_2 is Ksp=3.9×10−11K_{sp}=3.9\times10^{-11} and the acid-dissociation constant of HF is Ka=6.6×10−4K_a=6.6\times10^{-4}. Calculate the molar solubility of CaF2_2 in a solution that is buffered at pH 2.00. (a) 2.1×10−4 M2.1\times10^{-4}\,\text{M} (b) 2.2×10−3 M2.2\times10^{-3}\,\text{M} (c) 1.3×10−3 M1.3\times10^{-3}\,\text{M} (d) 6.6×10−3 M6.6\times10^{-3}\,\text{M}

Related theory

Browse the problem library