Question

The Ksp of calcium carbonate in water at 25 °C is 2.25 x 10-8. CaCO3(s) <----> Ca2+ (aq) + CO3 2- (aq) What is favored at equilibrium?

Answer 1

Answer:

the solubility of CaCO3 is 0.015g/l 25 °C

is favored at equilibrium

Explanation:

The Ksp of calcium carbonate in water at 25 °C is 2.25 x 10-8. CaCO3(s) <----> Ca2+ (aq) + CO3 2- (aq) What is favored at equilibrium?

solubility is the property of a solute to dissolve in a solvent(liquid, gas ) to form a solution(soution can be saturated ,unsaturated, or supersaturated)

CaCO3(s) <----> Ca2+ (aq) + CO3 2- (aq)

in partial dissociation , we can say

2.25x 10^-8=$Ca^{2+} +CO^{2-}_{3} }$

let Ca^2+=CO3^-2=S

2.25x10^-8=S*S

S^2=2.25x10^-8

S=0.00015mol/L

Converting that to g/l

the relative molecular mass of CaCO3=100g/mol

0.00015*100g/mol

0.015g/l

the solubility of CaCO3 is 0.015g/l @room temperature

is favored   at equilibrium