Question

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

Answer 1

Answer : The value of $K_{sp}$ is $6.00\times 10^{-39}$

Explanation :

The solubility equilibrium reaction will be:

$Ba_3(PO_4)_2\rightleftharpoons 3Ba^{2+}+2PO_4^{3-}$

Let the molar solubility be 's'.

The expression for solubility constant for this reaction will be,

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

$K_{sp}=(3s)^3\times (2s)^2$

$K_{sp}=108s^5$

Given:

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

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

$K_{sp}=108\times (8.89\times 10^{-9})^5$

$K_{sp}=6.00\times 10^{-39}$

Therefore, the value of $K_{sp}$ is $6.00\times 10^{-39}$

Answer 2

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.