Question

the solubility of SrSO4 in water is 0.107g in 1.0 L at 25 degrees celcius. what is the value for Ksp for SrSO4?

Answer 1

Ksp = 3.4 × 10⁻⁷

Step-by-step explanation:

First, we will express the solubility in molarity (molar mass = 183.68 g/mol).

`S=(0.107g)/((183.68g/mol).1.0L) =5.8 * 10^(-4) M`

Let's consider the solution of SrSO₄.

SrSO₄(s) ⇄ Sr⁺²(aq) + SO₄²⁻(aq)

To relate the solubility (S) to the solubility product (Ksp) we will use an ICE chart. We recognize 3 stages (Initial, Change, Equilibrium) and complete each row with the concentration or change in the concentration.

SrSO₄(s) ⇄ Sr⁺²(aq) + SO₄²⁻(aq)

I 0 0

C +S +S

E S S

Ksp = [Sr⁺²].[SO₄²⁻] = S² = (5. 8 × 10⁻⁴)² = 3.4 × 10⁻⁷

Answer 2

Ksp = 0.01145

Step-by-step explanation:

• SrSO4 ↔ Sr2+ + SO42-
• S S S...........................in the equilibrium

∴ S = 0.107 g/L = [ Sr2+ ] = [ SO42- ]

• Ksp = [ Sr2+ ] * [ SO42- ]

⇒ Ksp = S * S = S²

⇒ Ksp = ( 0.107 )²

⇒ Ksp = 0.01145