The solubility of limestone, CaCO3, at 25oC is 0.00067 g/100 mL. Write the chemi

Question

The solubility of limestone, CaCO3, at 25oC is 0.00067 g/100 mL. Write the chemical equation for the solubility equilibrium of this soluble salt in water. Then compute the molar solubility and the solubility product constant Ksp for CaCO3 at 25 degrees Celcius. The solubility of limestone, CaCO3, at 25oC is 0.00067 g/100 mL. Write the chemical equation for the solubility equilibrium of this soluble salt in water. Then compute the molar solubility and the solubility product constant Ksp for CaCO3 at 25 degrees Celcius.

Explanation / Answer

CaCO3 -------> Ca2+ + CO32-

Ksp = [Ca2+]*[CO32-]

Here [Ca2+] = [CO32-] = s

Molar Solubility is moles per 1000 ml

So

s = (0.00067*10/100) mol/L

s = 6.7*10^-5 mol/L

Therefore

Ksp = s^2

= (6.7*10^-5)^2

= 4.489*10^-9

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

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