Answer:
1. BaS Ksp = 0.169
2. CaSO₃ Ksp = 2.4 x 10⁻⁷
3. As₂O₃ Ksp = 6.7 x 10⁻⁴
3. Sr(IO₃)₂ Ksp = 1.7 x 10⁻⁷
Explanation:
The solubility constant product for a solid AB is given the expression for the equilibrium
AnBm (s) ⇄ A⁺(aq) + B⁻(aq) Ksp = [A⁺(aq)]^n x [B⁻ (aq)]^m
where the subscripts (s) and (aq) denote the solid and dissolved ions, [A⁺(aq)] and [B⁻ (aq)] are the concentrations of the dissolved ions in moles/Liter.
1. BaS
We are given the solubility in grams per 100 mL, since we need the solubility in moles per liter we will first convert the given solubility from g/ 100 mL to g/ L , and then divide this value by the molar mass to get molar solubility. From here it will be straight forward to calculate Ksp from the formula above
s = 6.97 g/ 100 mL = 6.97 g/0.1 L = 69.7 g/L
MW BaS = 169.39 g/mol
s BaS = (69.7 g/169.39 g/mol) / L = 0.41 mol/L = 0.41 M
Ksp = (0.41) x (0.41) = (0.41)² = 0.169
2. CaSO₃
s = 5.9 x 10⁻³ g / 100 mL = 5.9 x 10⁻³ g/ 0.1 L = 5.9 x 10⁻² g/L
MW CaSO₃ = 120.17 g/mol
s CaSO₃ = ( 5.9 x 10⁻² g / 5.9 x 10⁻² / 120.17 g/mol ) /L = 4.9 x 10⁻⁴ M
Ksp = (4.9 x 10⁻⁴) x ( 4.9 x 10⁻⁴) = (4.9 x 10⁻⁴)² = 2.4 x 10⁻⁷
3. As₂O₃
s = 1.80 g/ 100 mL = 1.80 g/ 0.100 L = 18.0 g /L
MW As₂O₃ = 197.84 g/mol
s = (18.0 g/ 197.84 g/mol) / L = 9.1 x 10⁻² mol/L = 9.1 x 10⁻² M
Ksp = (2 x 9.1 x 10⁻² )² x ( 3 x 9.1 x 10⁻² )³ = 6.7 x 10⁻⁴
Note we have to take into account that since the solubility of the compound As₂O₃ is 9.1 x 10⁻² mole/L we will have molar solubility concentrations of 2 x 9.1 x 10⁻² mol As³⁺ and 3 x 9.1 x 10⁻² . We did not need to do this in parts 1 and 2 since the ratio of ions were 1:1
4. Sr(IO₃)₂ 0.152 g/100 mL
s = 0.152 g / 100 mL = 0.152 g/0.1L = 1.52 g/L
MW Sr(IO₃)₂ = 437.43 g/mol
s = (1.52 g/437.43 g/mol)/L = 3.5 x 10⁻³ M
Ksp = (3.5 x 10⁻³ ) x ( 2 x 3.5 x 10⁻³ )² = 1.7 x 10⁻⁷
