Answer :
Answer:
6.1 × 10⁻⁸ M
Explanation:
Let's consider the solution of fluorapatite.
Ca₅(PO₄)₃F(s) ⇄ 5 Ca²⁺(aq) + 3 PO₄³⁻(aq) + F⁻(aq)
We can relate the solubility (S) with the solubility product constant (Ksp) using an ICE chart.
Ca₅(PO₄)₃F(s) ⇄ 5 Ca²⁺(aq) + 3 PO₄³⁻(aq) + F⁻(aq)
I 0 0 0
C +5 S +3 S +S
E 5 S 3 S S
The Ksp is:
Ksp = [Ca²⁺]⁵ × [PO₄³⁻]³ × [F⁻] = (5 S)⁵ × (3 S)³ × S = 84,375 S⁹
[tex]S = \sqrt[9]{\frac{Ksp}{84,375} } = \sqrt[9]{\frac{1.0 \times 10^{-60} }{84,375} } =6.1 \times 10^{-8} M[/tex]
The solubility of fluorapatite compound Ca₅(PO₄)₃F in water is 6.1 × 10⁻⁸M.
What is Ksp?
Ksp is the solubility product constant which tells about the relative solubility of the compound which is in equilibrium with their constitute ions.
Chemical reaction for the solubility of fluorapatite compound is:
Ca₅(PO₄)₃F(s) ⇄ 5Ca²⁺(aq) + 3PO₄³⁻(aq) + F⁻(aq)
Let at equilibrium formed concentration of Ca²⁺, PO₄³⁻ and F⁻ are 5x, 3x and x respectively.
Given value of Ksp for Ca₅(PO₄)₃F = 1.0 × 10⁻⁶⁰
Ksp equation for the given reaction will be represented as:
Ksp = [Ca²⁺]⁵[PO₄³⁻]³[F⁻]
On putting above values in the equation we get,
1.0 × 10⁻⁶⁰ = (5x)⁵. (3x)³. (x)
x = 6.1 × 10⁻⁸M
Hence, solubility of Ca₅(PO₄)₃F is 6.1 × 10⁻⁸M.
To know more about Ksp, visit the below link:
https://brainly.com/question/7185695