Determine the molar solubility of BaF₂in pure water.K_sp for BaF₂ = 2.45 × 10^-5. A) 1.83 × 10^-2 M B) 1.23 × 10^-5 M C) 2.90 × 10^-2 M D) 4.95 × 10^-3 M E) 6.13 × 10^-6 M

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The molar solubility of BaF₂ in pure water can be calculated using the K_sp expression:K_sp = [Ba²⁺][F^-]²Assuming that the molar solubility of BaF₂ is x, then:[Ba²⁺] = x[F^-] = 2xSubstituting these values into the K_sp expression, we get:2.45 × 10^-5 = x(2x)²2.45 × 10^-5 = 4x³x³ = 6.13 × 10^-6x = 1.83 × 10^-2 MTherefore, the molar solubility of BaF₂ in pure water is 1.83 × 10^-2 M.

Determine the Molar Solubility of BaF2 in Pure Water

Determine the Molar Solubility of BaF₂in Pure Water