Question

1) Given the solubility in grams/100 mL of water determine the Ksp Values for a) BaSeO4, 0.0118 g/100 mL b) La2(MoO4)3. 0.00179 g/100 mL BaSeO4(s) Ba2 (aq) Se042-(aq) La2(MoO4)3(s)2La3+(aq) 3M0042(aq)

Answer 1

a)
Molar mass of BaSeO4,
MM = 1*MM(Ba) + 1*MM(Se) + 4*MM(O)
= 1*137.3 + 1*78.97 + 4*16.0
= 280.27 g/mol


Molar mass of BaSeO4= 280.27 g/mol
s = 0.0118 g/100 mL
=0.0118 g/0.1 L
= 0.118 g/L

To covert it to mol/L, divide it by molar mass
s = 0.118 g/L / 280.27 g/mol
s = 4.21*10^-4 mol/L

At equilibrium:
BaSeO4    <---->     Ba2+     +          SeO42-

                     s                  s       

Ksp = [Ba2+][SeO42-]
Ksp = (s)*(s)
Ksp = 1(s)^2
Ksp = 1(4.21*10^-4)^2
Ksp = 1.773*10^-7
Answer: 1.77*10^-7

b)
Molar mass of La2(MoO4)3,
MM = 2*MM(La) + 3*MM(Mo) + 12*MM(O)
= 2*138.9 + 3*95.95 + 12*16.0
= 757.65 g/mol


Molar mass of La2(MoO4)3= 757.65 g/mol
s = 0.00179 g/100 mL
= 0.00179 g / 0.100 L
= 1.79*10^-2 g/L
To covert it to mol/L, divide it by molar mass
s = 1.79*10^-2 g/L / 757.65 g/mol
s = 2.363*10^-5 mol/L

At equilibrium:
La2(MoO4)3<---->    2 La3+    +         3 MoO42-

                     2s                 3s      

Ksp = [La3+]^2[MoO42-]^3
Ksp = (2s)^2*(3s)^3
Ksp = 108(s)^5
Ksp = 108(2.363*10^-5)^5
Ksp = 7.95*10^-22
Answer: 7.95*10^-22