Question

The Handbook of Chemistry and Physics gives solubilities of the following compounds in grams per 100 mL of water at 20°C. Because these compounds are only slightly to moderately soluble, assume that the volume does not change on dissolution and calculate the solubility product for each at this temperature. (Note that 100 mL in this context is a nominal value and not subject to measurement uncertainty.)
1. BaS 6.97 g/100 mL ___________

2. CaSO3 5.9 ✕ 10−3 g/100 mL ____________

3. As2O3 1.80 g/100 mL ______________

4. Sr(IO3)2 0.152 g/100 mL ____________

Answer 1

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