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Chemistry
Chemistry
Problem
Calcium fluoride, CaF2(s), has
K
s
p
=
3.9
×
10
−
11
K_{sp}=3.9\times10^{-11}
K
s
p
=
3.9
×
1
0
−
11
at 25 °C. The acid dissociation constant for hydrofluoric acid is
K
a
=
6.6
×
10
−
4
K_a=6.6\times10^{-4}
K
a
=
6.6
×
1
0
−
4
. What pH must a solution have so that the molar solubility of CaF2 is exactly
1.0
×
10
−
3
M
1.0\times10^{-3}\,\text{M}
1.0
×
1
0
−
3
M
?
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