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Sagot :
Answer: [tex]K_{sp}=1.25\times 10^{-14}[/tex]
[tex]pK_{sp}=13.90[/tex]
Explanation:
Solubility product is defined as the equilibrium constant in which a solid ionic compound is dissolved to produce its ions in solution. It is represented as [tex]K_{sp}[/tex]
The equation for the ionization of magnesium phosphate is given as:
[tex]Mg_3(PO_4)_2\rightarrow 3Mg^{2+}+2PO_4^{3-}[/tex]
When the solubility of [tex]Mg_3(PO_4)_2[/tex] is S moles/liter, then the solubility of [tex]Mg^{2+}[/tex] will be 3S moles\liter and solubility of [tex]PO_4^{3-}[/tex] will be 2S moles/liter.
Thus S = 0.173 g/L or [tex]\frac{0.173g/L}{262.8g/mol}=0.00065mol/L[/tex]
[tex]K_{sp}=(3S)^3\times (2S)^2[/tex]
[tex]K_{sp}=108S^5[/tex]
[tex]K_{sp}=108\times (0.00065)^5=1.25\times 10^{-14}[/tex]
[tex]pK_{sp}=-log(K_{sp})=\log (1.25\times 10^{-14})=13.90[/tex]
The value of Ksp and pKsp for the given reaction is 1.25 × 10⁻¹⁴ and 13.90 respectively.
How we calculate the Ksp?
Solubility product constant (Ksp) is define as the product of the solubilities of the products, present in any chemical reaction.
Given chemical reaction will be represented as:
Mg₃(PO₄)₂ → 3Mg²⁺ + 2PO₄³⁻
Solubility of Mg₃(PO₄)₂ = 0.173 g/L or 0.173g/L / 262.8g/mole = 0.00065mol/L
Now for the given reaction let we consider that solubility of Mg₃(PO₄)₂ be x, solubility of Mg²⁺ be 3x and solubility of PO₄³⁻ will be 2x and equation for Ksp will be written as:
Ksp = [Mg²⁺]³[PO₄³⁻]²
Ksp = [3x]³[2x]²
Ksp = 108[x]⁵
Now we pot the solubility value of x as the solubility of Mg₃(PO₄)₂ & we get,
Ksp = 108(0.00065)⁵
Ksp = 1.25 × 10⁻¹⁴
Now we calculate the value of pKsp as:
pKsp = -log(Ksp)
pKsp = -log(1.25 × 10⁻¹⁴)
pKsp = -(-13.90) = 13.90
Hence value of pKsp is 13.90.
To know more about Ksp, visit he below link:
https://brainly.com/question/7185695
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