Question

A sparingly soluble metal fluoride with formula mf2, where m is an unknown metal, has a ksp = 6. 65 x 10-6. calculate the concentration of f- in solution

Answer 1

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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