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* 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* (2s)^2`

`K_(sp)=108s^5`

Given:

Molar solubility of
`Ba_3(PO_4)_2` = s =
`8.89* 10^(-9)M`

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

`K_(sp)=108* (8.89* 10^(-9))^5`

`K_(sp)=6.00* 10^(-39)`

Therefore, the value of
`K_(sp)` is
`6.00* 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.