Question

The solubility of SF2 in water is 1.2 = 19 g/L. Calculate the solubility product of the salt at room temperature ( molecular mass of SrF2 = 125.6)​

Answer 1

✪ Answer ✪

Let us consider the solubility be 's'.

SF$_2$ ⟶ S + 2F

s           s    2s

Then, solubility product K$_{sp}$ can be written as:

K$_{sp}$ = $\frac{(s)(2s)^{2} }{s}$

K$_{sp$ = $\frac{4s^{2} }{1}$

K$_{sp}$ = $4s^{2}$

Now, it is given that solubility of SF$_2$ is 19 g/L.

Substituting s = 19:

K$_{sp}$ =$4(19)^{2}$

=  1444 g/L

★  So, the solubility product will be 1444 g/L  ★