How many moles of {eq}Ag_2CO_3 {/eq} ({eq}K_{sp} = 8.5 \times 10^{-12} {/eq}) will dissolve in {eq}500 \ mL {/eq} of {eq}0.010 \ M \ (NH_4)_2CO_3 {/eq} solution?

## Question:

How many moles of {eq}Ag_2CO_3 {/eq} ({eq}K_{sp} = 8.5 \times 10^{-12} {/eq}) will dissolve in {eq}500 \ mL {/eq} of {eq}0.010 \ M \ (NH_4)_2CO_3 {/eq} solution?

## Common Ion Effect:

A saturated aqueous solution of a salt compound will contain the compound's reversible dissociation reaction in a solubility equilibrium. The solubility product of the constituent ion molarity values will be controlled by the compound's {eq}K_{sp} {/eq} value, which only varies with temperature. If you increase one of these molarity values, then the dissociation reaction must shift left, even though the solubility product constant is unchanged. This type of reaction shift in a solubility equilibrium is termed the common ion effect.

## Answer and Explanation: 1

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View this answerSilver carbonate has the following solubility equilibrium for its saturated solution:

{eq}\rm Ag_2CO_3 (s) \leftrightharpoons 2Ag^+ (aq) +...

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Chapter 11 / Lesson 5Learn about solubility product constant. Understand the definition of Ksp, the Ksp formula, how to calculate Ksp, and how to find molar solubility from Ksp.