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The molar solubility of ba3(po4)2 is 8.89 x 10-9 m in pure water. calculate the ksp for ba3(po4)2. the molar solubility of ba3(po4)2 is 8.89 x 10-9 m in pure water. calculate the ksp for ba3(po4)2. 5.55 x 10-41 5.33 x 10-37 8.16 x 10-31 6.00 x 10-39 4.94 x 10-49

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Answer is: The molar solubility of ba3(po4)2 is 6.00 x 10-39.
Balanced chemical reaction: Ba₃(PO₄)₂(s) → 3Ba²⁺(aq) + 2PO₄³⁻(aq).
s(Ba₃(PO₄)₂) = 8.89·10⁻⁹ M.
[Ba²⁺] = 3s(Ba₃(PO₄)₂) = 3s.
[PO₄³⁻] = 2s.
Ksp = [Ba²⁺]³ · [PO₄³⁻]².
Ksp = (3s)³ · (2s)².
Ksp = 108s⁵.
Ksp = 108·(8.89·10⁻⁹ M)⁵.
Ksp = 108 · 5.55·10⁻⁴¹ = 6·10⁻³⁹.

Answer : The value of [tex]K_{sp}[/tex] is [tex]6.00\times 10^{-39}[/tex]

Explanation :

The solubility equilibrium reaction will be:

[tex]Ba_3(PO_4)_2\rightleftharpoons 3Ba^{2+}+2PO_4^{3-}[/tex]

Let the molar solubility be 's'.

The expression for solubility constant for this reaction will be,

[tex]K_{sp}=[Ba^{2+}]^3[PO_4^{3-}]^2[/tex]

[tex]K_{sp}=(3s)^3\times (2s)^2[/tex]

[tex]K_{sp}=108s^5[/tex]

Given:

Molar solubility of [tex]Ba_3(PO_4)_2[/tex] = s = [tex]8.89\times 10^{-9}M[/tex]

Now put all the given values in the above expression, we get:

[tex]K_{sp}=108\times (8.89\times 10^{-9})^5[/tex]

[tex]K_{sp}=6.00\times 10^{-39}[/tex]

Therefore, the value of [tex]K_{sp}[/tex] is [tex]6.00\times 10^{-39}[/tex]

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