The solubility product for Mg3(PO4)2 is 6.3x10^-26. What is the solubility of Mg

Question

The solubility product for Mg3(PO4)2 is 6.3x10^-26. What is the solubility of Mg3(PO4)2 in pure water, in grams per liter?
A) 1.7x10^-23 g/L B) 3.4x10^-7 g/L C) 9.4x10^-4 g/L * D) 1.2x10^-3 g/L E) 2.4x10^-3 g/L The solubility product for Mg3(PO4)2 is 6.3x10^-26. What is the solubility of Mg3(PO4)2 in pure water, in grams per liter?
A) 1.7x10^-23 g/L B) 3.4x10^-7 g/L C) 9.4x10^-4 g/L * D) 1.2x10^-3 g/L E) 2.4x10^-3 g/L The solubility product for Mg3(PO4)2 is 6.3x10^-26. What is the solubility of Mg3(PO4)2 in pure water, in grams per liter?
A) 1.7x10^-23 g/L B) 3.4x10^-7 g/L C) 9.4x10^-4 g/L * D) 1.2x10^-3 g/L E) 2.4x10^-3 g/L

Explanation / Answer

Solution

Given data

Mg3(PO4)2 ksp =6.3E-26

Solubility of Mg3(PO4)2 = ?

Lets first write the dissociation equation for the Mg3(PO4)2

Mg3(PO4)2   ----------- > 3 Mg^2+   + 2PO4^2-

                                               3x                   2x

Ksp = [Mg^2+]3[PO4^3-]2

Lets put the values in the formula

6.3E-26 = [3x]3[2x]2

6.3E-26 = 27x3 * 4x2

6.3E-26 = 108x5

6.3E-26 / 108 = x5

5.833E-28=x5

(5.833E-28)^1/5 =x

3.57E-6 =x

Therefore the molar solubility of the Mg3(PO4)2 = 3.57E-6 M

Now lets calculate this molar solubility into g/L

Solubility of the Mg3(PO4)2 = (3.57E-6 mol per L )* (262.8577 g/ 1 mol ) = 9.4E-4 g/L

Therefore solubility of the Mg3(PO4)2 in pure water = 9.4E-4 g/L

Therefore option 'C is the correct answer.

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.