SOLUTION: When lithium fluoride is dissolved in water, the equation of the solubility constant (Ksp) is Ksp = [Li+] × [F-]. The value of Ksp for lithium fluoride is 0.00184 mol2/L2.

Question 1170219: When lithium fluoride is dissolved in water, the equation of the solubility constant (Ksp) is
Ksp = [Li+] × [F-].
The value of Ksp for lithium fluoride is 0.00184 mol2/L2.
If the concentration of fluorine, [F-], is 0.0060 mol/L more that the concentration of lithium, [Li+], what is the concentration of lithium, [Li+], in the solution?
Round your answer to 2 significant figures.

Answer by CPhill(1959) (Show Source): You can put this solution on YOUR website!
Let's break down this problem step by step.
**1. Set Up the Equations**
* We are given the solubility constant (Ksp) expression: Ksp = [Li+] × [F-]
* We are given the value of Ksp: Ksp = 0.00184 mol²/L²
* We are given the relationship between [F-] and [Li+]: [F-] = [Li+] + 0.0060 mol/L
**2. Substitute and Solve**
* Substitute the expression for [F-] into the Ksp equation:
0.00184 = [Li+] × ([Li+] + 0.0060)
* Expand the equation:
0.00184 = [Li+]² + 0.0060[Li+]
* Rearrange the equation into a quadratic equation:
[Li+]² + 0.0060[Li+] - 0.00184 = 0
**3. Use the Quadratic Formula**
* The quadratic formula is: x = (-b ± √(b² - 4ac)) / 2a
* In our equation, a = 1, b = 0.0060, and c = -0.00184
* Substitute these values into the quadratic formula:
[Li+] = (-0.0060 ± √((0.0060)² - 4 × 1 × (-0.00184))) / (2 × 1)
[Li+] = (-0.0060 ± √(0.000036 + 0.00736)) / 2
[Li+] = (-0.0060 ± √0.007396) / 2
[Li+] = (-0.0060 ± 0.0860) / 2
**4. Find the Two Possible Solutions**
* Solution 1: [Li+] = (-0.0060 + 0.0860) / 2 = 0.0800 / 2 = 0.0400 mol/L
* Solution 2: [Li+] = (-0.0060 - 0.0860) / 2 = -0.0920 / 2 = -0.0460 mol/L
**5. Choose the Valid Solution**
* Since concentration cannot be negative, we discard the second solution.
* Therefore, [Li+] = 0.0400 mol/L
**6. Round to 2 Significant Figures**
* 0.0400 rounded to 2 significant figures is 0.040 mol/L.
**Answer:**
The concentration of lithium, [Li+], in the solution is 0.040 mol/L.

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