Answer:
6.4 × 10^-10 M
Explanation:
The molar solubility of the ions in a compound can be calculated from the Ksp (solubility constant).
CaF2 will dissociate as follows:
CaF2 ⇌Ca2+ + 2F-
1 mole of Calcium ion (x)
2 moles of fluorine ion (2x)
NaF will also dissociate as follows:
NaF ⇌ Na+ + F-
Where Na+ = 0.25M
F- = 0.25M
The total concentration of fluoride ion in the solution is (2x + 0.25M), however, due to common ion effect i.e. 2x<0.25, 2x can be neglected. This means that concentration of fluoride ion will be 0.25M
Ksp = {Ca2+}{F-}^2
Ksp = {x}{0.25}^2
4.0 × 10^-11 = 0.25^2 × x
4.0 × 10^-11 = 0.0625x
x = 4.0 × 10^-11 ÷ 6.25 × 10^-2
x = 4/6.25 × 10^ (-11+2)
x = 0.64 × 10^-9
x = 6.4 × 10^-10
Therefore, the molar solubility of CaF2 in NaF solution is 6.4 × 10^-10M
The molar solubility should be 6.4 × 10^-10 M
Here
CaF2 will dissociate should be
CaF2 ⇌Ca2+ + 2F-
where,
1 mole of Calcium ion (x)
2 moles of fluorine ion (2x)
Now
NaF will also dissociate like
NaF ⇌ Na+ + F-
Here
Na+ = 0.25M
F- = 0.25M
Now
Ksp = {Ca2+}{F-}^2
Ksp = {x}{0.25}^2
4.0 × 10^-11 = 0.25^2 × x
4.0 × 10^-11 = 0.0625x
x = 4.0 × 10^-11 ÷ 6.25 × 10^-2
x = 4/6.25 × 10^ (-11+2)
x = 0.64 × 10^-9
x = 6.4 × 10^-10
Therefore, the molar solubility of CaF2 in NaF solution is 6.4 × 10^-10M
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