What is the silver ion concentration in a solution prepared by mixing 487 mL of 0.384 M silver...
Question:
What is the silver ion concentration in a solution prepared by mixing {eq}487\ mL {/eq} of {eq}0.384\ M {/eq} silver nitrate with {eq}379\ mL {/eq} of {eq}0.583\ M {/eq} sodium phosphate?
The {eq}K_{SP} {/eq} of silver phosphate is {eq}2.8 \times 10^{-18} {/eq}.
Solubility Product:
The solubility product {eq}K_{sp} {/eq} is a special equilibrium constant for partially soluble ionic solids. For a chemical reaction {eq}A_xB_{y(s)} \rightleftharpoons xA_{(aq)}^{y+}+ yB_{(aq)}^{x-} {/eq} the solubility product is equal to:
{eq}\displaystyle K_{sp}= \left[A^{y+} \right]^x \left[B^{x-}\right]^y \\ {/eq}
The presence of cation and anion must be great enough such that the reaction quotient is greater than the solubility product. That must be the condition for precipitation to occur.
Answer and Explanation: 1
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View this answerCalculate the concentration of silver and phosphate ions after mixing.
{eq}\displaystyle [Ag^+] = \frac{0.384 \ M \times 487 \ mL}{487 \ mL + 379 \...
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Solubility Equilibrium: Using a Solubility Constant (Ksp) in Calculations
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Chapter 11 / Lesson 5
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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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