A sparingly soluble metal fluoride with formula MF₂, where m is an unknown metal, has a Ksp = 6.65 x 10⁻⁶. The concentration of F⁻ in solution 2.36 x 10⁻³M
Ksp is called the Molar solubility product and S is the Molar solubility of an ion in a solution.
According to given formula, the dissociation of metal fluoride MF₂ occurs as follows in aqueous solution:
MF₂ ------> M⁺² + 2F⁻
S 2S
Ksp = [S] [2S]²
Ksp = 4S³
Given, Ksp = 6.65 x 10⁻⁶
On substituting,
6.65 x 10⁻⁶ = 4S³
S³ = 1.66 x 10⁻⁶
S = 1.18 x 10⁻³
So, Solubility of F⁻ ion = 2S
Solubility of F⁻ ion = 2(1.18 x 10⁻³)
Solubility of F⁻ ion = 2.36 x 10⁻³
Since Molarity of the ions is equal to the solubility of the ion in aqueous solution.
Hence, the concentration of F⁻ ion is 2.36 x 10⁻³M.
Learn more about Molar solubility here, brainly.com/question/16243859
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