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Determine the molar solubility of agi in pure water. ksp (agi) = 8.51 × 10-17.

Ralph Lopez

in Chemistry

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Ashley Howard on February 18, 2019

When the salt of AgI dissolves in water dissociates in the following manner;AgI ---> Ag⁺ + I⁻solubility molar is the number of moles that are dissolved in 1 L of solution.If the solubility mole of AgI is x, then the solubility mole of Ag⁺ is x and I⁻ x.the formula of the constant of solubility product - ksp of AgI is as follows ksp = [Ag⁺][I⁻]ksp = (x)(x)ksp = 8.51 x 10⁻1⁷ therefore,x2 = 8.51 x 10⁻1⁷x = 9.22 x 10⁻⁹since solubility mole of AgI is x, then the solubility molar AgI is 9.22 x 10⁻⁹ M


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