Barium sulfate (BaSO₄) is a slightly soluble salt, with K_sp = 1.1 * 10¯¹⁰. What mass of Ba²⁺ ions will be present in 1.0 L of a saturated solution of barium sulfate?

Question

Quizplus · Accepted Answer

To solve this problem, we need to use the equilibrium constant expression for the dissolution of BaSO4:

BaSO4(s) ⇌ Ba2+(aq) + SO42-(aq)

The equilibrium constant expression is written as:

Ksp = [Ba2+][SO42-]

We are given the value of Ksp as 1.1 x 10^-10, which means that the product of the concentrations of Ba2+ and SO42- ions in a saturated solution of BaSO4 is equal to this value.

Let x be the concentration of Ba2+ and SO42- ions in the saturated solution. Since we assume that BaSO4 is only slightly soluble, we can assume that x is very small compared to the initial concentration of BaSO4, which is usually given as the solubility of BaSO4. Therefore, we can also assume that the initial concentration of BaSO4 is equal to its solubility.

Let S be the solubility of BaSO4. We know that:

[Ba2+] = x
[SO42-] = x
[BaSO4] = S

Substituting these values into the equilibrium constant expression, we get:

Ksp = x2(S - x)

We are asked to find the mass of Ba2+ ions present in 1.0 L of a saturated solution of BaSO4. To do this, we need to convert the concentration of Ba2+ ions into mass using the molar mass of Ba2+.

The molar mass of Ba2+ is:

137.327 g/mol

The concentration of Ba2+ ions is equal to x. Substituting the expression for x from the equilibrium constant equation, we get:

Ksp = x2(S - x)
1.1 x 10^-10 = x2(S - x)

Assuming that x is much smaller than S, we can simplify the equation to:

1.1 x 10^-10 = x2S

Solving for x, we get:

x = sqrt(Ksp/S) = sqrt(1.1 x 10^-10/1) = 1.05 x 10^-5 M

The mass of Ba2+ ions present in 1.0 L of the saturated solution is:

mass = concentration x volume x molar mass
mass = 1.05 x 10^-5 mol/L x 1.0 L x 137.327 g/mol
mass = 0.0014 g

Therefore, the best choice is C.

Barium Sulfate (BaSO4) Is a Slightly Soluble Salt, with Ksp

Barium Sulfate (BaSO₄) Is a Slightly Soluble Salt, with K_sp