Give the expression for the solubility product constant for baf2.

1 answer:

3 0

Solubility product constant (Ksp) is applied to the saturated ionic solutions which are in equilibrium with its solid form. The solid is partially dissociated into its ions.

For the BaF, the dissociation as follows;
BaF₂(s) ⇄ Ba²⁺(aq) + 2F⁻(aq)

Hence,
Ksp = [Ba²⁺(aq)] [F⁻(aq)]²

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