Question

What is the molar solubility of BaSO₄ in a 0.250-M solution of NaHSO₄? Ka for= 1.2 x 10⁻².

a) 1.0 x 10⁻⁴ M
b) 2.0 x 10⁻⁴ M
c) 3.0 x 10⁻⁴ M
d) 4.0 x 10⁻⁴ M

Answer 1

The molar solubility of BaSO₄ in a 0.250-M solution of NaHSO₄ is 4.0 × 10⁻⁴ M.

Step-by-step explanation:

The molar solubility of BaSO₄ in a 0.250-M solution of NaHSO₄ can be calculated using the solubility product constant (Ksp) and the stoichiometry of the equation. Since the molar solubility of BaSO₄ is given by [Ba²+] and [SO4²-], we can set up the following equilibrium expression:

Ksp = [Ba²+][SO4²-]

Substituting the given concentration of NaHSO₄ and the equation stoichiometry, we can solve for [Ba²+], which represents the molar solubility of BaSO₄:

[Ba²+] = sqrt(Ksp / [NaHSO₄])

Plugging in the given values, we get:

[Ba²+] = sqrt((1.1 × 10⁻¹⁰) / (0.250)) = 4.0 × 10⁻⁴ M

Therefore, the molar solubility of BaSO₄ in a 0.250-M solution of NaHSO₄ is 4.0 × 10⁻⁴ M.