Question

Calculate the molar solubility of Barium carbonate (Ksp = 5.0 x 10⁻⁹) a) 2.2 x 10⁻³ M b) 2.5 x 10⁻³ M c) 2.8 x 10⁻³ M d) 3.1 x 10⁻³ M

Answer 1

To find the molar solubility of Barium carbonate, the square root of the solubility product constant (Ksp) is taken. For BaCO3 with a Ksp of 5.0 × 10−9, the molar solubility (s) is calculated to be 2.24 × 10−3M. Option a) is the closest to this value.

Step-by-step explanation:

The question asks to calculate the molar solubility of Barium carbonate (BaCO3). To find the molar solubility, we can set up an equilibrium expression based on the solubility product constant (Ksp). For BaCO3, the dissolution process in water is given by:

BaCO3(s) → Ba2+(aq) + CO32−(aq)

Let's denote the molar solubility of BaCO3 as 's'. At equilibrium, the concentration of Ba2+ ions and CO32− ions in the solution will both be 's' because the stoichiometry of the dissociation is 1:1. The Ksp expression can be written as:

Ksp = [Ba2+][CO32−] = (s)(s) = s2

Given that Ksp for BaCO3 is 5.0 × 10−9, we can solve for 's' by taking the square root of Ksp:

s = √(Ksp) = √(5.0 × 10−9)

After calculating, we find that s = 2.24 × 10−3M, which is the molar solubility of Barium carbonate in water at 25°C.

None of the given options a) 2.2 × 10−3M, b) 2.5 × 10−3M, c) 2.8 × 10−3M, d) 3.1 × 10−3M exactly match this result, but option a) is the closest to the correct value.