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Chapter 14: Problem 38

The ionic substance EJ dissociates to form \(\mathrm{E}^{2+}\) and \(\mathrm{J}^{2-}\) ions. The solubility of EJ is \(8.45 \times 10^{-6} \mathrm{mol} / \mathrm{L}\) . What is the value of the solubility-product constant?

Short Answer

Expert verified
The solubility-product constant for the given ionic substance EJ is \(7.14 \times 10^{-11}\).

Step by step solution

01

Understand the Dissociation of EJ

First, the ionic compound EJ dissociates in water to form E^{2+} and J^{2-} ions. This can be represented as follows: \[EJ \rightarrow E^{2+} + J^{2-}\]
02

Define Solubility

The provided solubility tells us that \(8.45 \times 10^{-6}\) moles of EJ dissolve in a liter of solution. Thus, when EJ dissociates, it forms \(8.45 \times 10^{-6}\) moles of E^{2+} ions and \(8.45 \times 10^{-6}\) moles of J^{2-} ions.
03

Apply the Formula for Ksp

The solubility product constant (Ksp) is given by the product of the concentrations of the product ions, each raised to the power of its coefficient in the balanced chemical equation. In this case, the formula for Ksp would be as follows: \[ Ksp = [E^{2+}][J^{2-}]\]
04

Substitute and Calculate

Insert the values into the formula: \[ Ksp = (8.45 \times 10^{-6})^2 = 7.14 \times 10^{-11} \]

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