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A solution contains 1.36 times 10^{-2} M lead nitrate and 8.64 times 10^{-3} M silver acetate....

Question:

A solution contains {eq}\displaystyle \rm 1.36 \times 10^{-2} \ M {/eq} lead nitrate and {eq}\displaystyle \rm 8.64 \times 10^{-3} \ M {/eq} silver acetate. Solid ammonium chromate is added slowly to this mixture. What is the concentration of chromate ion when this precipitation first begins?

Solubility Product:

It can be evaluated from the product of the molar concentrations of the charged species present in the solution at an equilibrium state. It is expressed by an equilibrium constant for a substance dissolved in a liquid.

Answer and Explanation: 1

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Given data:

{eq}\left[ {{\rm{Pb}}{{\left( {{\rm{N}}{{\rm{O}}_{\rm{3}}}} \right)}_{\rm{2}}}} \right] = 1.36 \times {10^{ - 2}}\;{\rm{M}} {/eq}

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Solubility Equilibrium: Using a Solubility Constant (Ksp) in Calculations

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Chapter 11 / Lesson 5
17K

Learn about solubility product constant. Understand the definition of Ksp, the Ksp formula, how to calculate Ksp, and how to find molar solubility from Ksp.


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