The K sp for barium fluoride, BaF2, is 2.45 × 10-5. What is the molar solubility of barium fluoride?

The molar solubility of barium fluoride is 0.0183 moles/liter.

Explanation:

The equation for the reaction will be as follows:

By Stoichiometry,

1 mole of gives 2 moles of and 1 mole of

Thus if solubility of is s moles/liter, solubility of  is s moles/liter and solubility of is 2s moles/liter

Therefore,  

Thus the molar solubility of barium fluoride is 0.0183 moles/liter.

when the volume of gas is   changed from   717.73 ml   to 852 ml   the temperature will   change   from 315   c to 452   c

    calculation

this is calculated using the charles law formula

that is,   v1/t1 = v2/t2

          v1=?

        t1 = 315 c   into   kelvin = 315 +273 = 588 k

        v2 = 852   ml

        t2 = 425c + 273 =698 k

make v1   the subject of the formula ( by   multiplying both side by t1)

  v1=( v2 xt1) / t2

v1= (852 ml   x 588 k)/ 698 k= 717.73 ml