Question

Barium hydroxide, Ba(OH)_2 is reported to have Ksp of 1.5x10^-4 at 25 degrees Celsius.

a) What is the molar solubility of Ba(OH)_2

b) What is the solubility of Ba(OH)_2 in g/L

3. Barium hydroxide, Ba(OH), is reported to have a K, of 1.5 x 10 at 25 C. What is the molar solubility of Ba(OH), ? (Show your work.) a. b. What is the solubility of Ba(OH), in g/L? (Show your work.)

Answer 1

Pertinent equation for solubility of Ba(OH)₂ is given as,

Ba(OH)₂ ---------------> Ba²⁺ (aq.) + 2 OH^- (aq.).

Let solubility of Ba(OH)₂ be 'S' mole/L hence on ionization ionic concentrations are given as,

[Ba²⁺] = S mole/L

[OH^-] = 2S mole/L

Now solubility product Ksp has expresion as,

Ksp = [Ba²⁺][OH^-]².

Ksp = (S)(2S)².

Ksp = S x 4S².

Ksp = 4S³.

Ksp for Ba(OH)₂ = 1.5 x 10^-4.

  Ksp = 4S³ = 1.5 x 10^-4.

4S³ = 1.5 x 10^-4.

S³ = 1.5 x 10^-4 /4

S³ = 150 x 10^-6 / 4

S³ = 37.5 x 10^-6

S= 3.35 x 10^-2 mole/L

Molar Solubility of Ba(OH)₂ is 3.35 x 10^-2 mole/L

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b) Solubility in g/L :

Molar mass of Ba(OH)₂ = 171.34 g/mole

S = 3.35 x 10^-2 mole/L

On multiplying by molar mass to molar solubility we get S in g/L unit.

S = 3.35 x 10^-2 x Molar mass of Ba(OH)2 g/L

S = 3.35 x 10^-2 x 171.34 g/L

S = 5.74 g/L

Solubility of Ba(OH)₂ is 5.74 g/L.

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