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3) One lab section accidentally used Ba(OH)2 instead of Ca(OH)2 when they performed this experiment. The TA told them that the Ksp of Ba(OH)2 is 2.55 x 10-4. What is the molar solubility of Ba(OH)2

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Answer:

The molar solubility of Ba(OH)2 is 0.03995 mol/L

Explanation:

Step 1: Data given

Ksp of Ba(OH)2 = 2.55 * 10^-4

Step 2: The balanced equation

Ba(OH)2(s)→ Ba^2+(aq) + 2OH-(aq)

Step 3: Define Ksp

Ksp = [Ba^2+] * [OH-]² = 2.55 *10^-4

[Ba^2+] = S

[OH-] = 2S

Ksp = S * (2S)²

Ksp = 4S³ = 2.55 * 10^-4

S³ = 0.00006375

S = 0.03995 mol/L

The molar solubility of Ba(OH)2 is 0.03995 mol/L

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