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    Solubility at Equilibrium of AgOH.

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    10.0 mg of AgOH(s) is added to 350.0 mL of water with an initial pH of 11.00. The solubility equilibrium constant for the dissolution of AgOH (s) is Ksp = 1.5 x 10^-8. Determine the concentration of [Ag +] and the final pH at equilibrium.

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    AgOH(s) = Ag+ + OH- ----- (1)

    Let's assume that x mg of AgOH(s) was dissolved at equilibrium. According to the equation (1), 1 mol of AgOH(s) produces 1 mol of Ag+ and 1 mol of OH-. Therefore ...

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    This solution provides calculation and explanation in an attached Word document. Six steps of calculations and 113 words are used.