If the solubility of CaCO_3 in water at 25°C is 5.84 ✕ 10^−3 g/L, calculate the solubility-product constant (Ksp) of CaCO_3, assuming complete dissociation of the CaCO_3 that has dissolved.
Convert 5.84E-3 g/L to mols/L.
mols/L = M = 5.84E-3/molar mass CaCO3 = ?
....................CaCO3 ==> Ca^2+ + [CO3]^2-
I....................solid..............0..............0
C...................solid..............x...............x
E....................solid..............x...............x
You know x = M from above. Plug that into the Ksp expression and solve for Ksp. Post your work if you get stuck.
To calculate the solubility-product constant (Ksp) of CaCO3, you need to know the molar mass of CaCO3 and the equation representing its dissociation in water.
The molar mass of CaCO3 can be calculated by adding the atomic masses of its constituent elements:
CaCO3 = 1 mole of Ca + 1 mole of C + 3 moles of O
= (40.08 g/mol) + (12.01 g/mol) + (3 * 16.00 g/mol)
= 100.09 g/mol
The equation representing the dissociation of CaCO3 in water is as follows:
CaCO3 (s) ⇌ Ca2+ (aq) + CO3^2- (aq)
Now we know the solubility of CaCO3 is given as 5.84 × 10^-3 g/L. To find the concentration in moles per liter (M), we need to convert grams to moles:
(5.84 × 10^-3 g/L) / (100.09 g/mol) = 5.837 × 10^-5 mol/L
Since Ca2+ and CO3^2- are present in a 1:1 ratio as per the balanced equation, their concentration is the same as the concentration of CaCO3. Thus, the concentration of Ca2+ and CO3^2- ions is also 5.837 × 10^-5 mol/L.
The solubility-product constant (Ksp) expression for CaCO3 is as follows:
Ksp = [Ca2+] [CO3^2-]
Since the concentration of Ca2+ and CO3^2- ions is the same, we can represent Ksp as:
Ksp = (5.837 × 10^-5 mol/L) × (5.837 × 10^-5 mol/L)
= 3.399 × 10^-9
Therefore, the solubility-product constant (Ksp) of CaCO3, assuming complete dissociation of the CaCO3 that has dissolved, is 3.399 × 10^-9.