Zinc carbonate dissolves in water to the extent of 1.12 x 10-4 g/L at 25 C. Calculate the solubility product Ksp for ZnCO3 at 25 C.

Ksp= 7.98 × 10^-13

Explanation:

According to the question, we are to calculate the solubility constant (Ksp) of Zinc carbonate (ZnCO3) in a dissolved solution.

The equilibrium of the reaction is:

ZnCO3 (aq) ⇌ Zn2+ (aq) + CO32- (aq)

According to this; 1 mole of Zinc carbonate (ZnCO3) dissolves to give 1 mole of Zinc ion (Zn2+) and 1 mole of carbonate ion (CO32-).

This illustrates that:

(Zn2+) = 1.12 x 10-4 g/L

(CO32-) = 1.12 x 10-4 g/L

However, 1.12 x 10-4 g/L is the solubility in mass concentration of ZnCO3, we need to convert it to molar concentration in mol/L by dividing by the relative molar mass of ZnCO3.

To calculate the molar mass of ZnCO3, we say:

Zn (65.4) + C (12) + 03 (16×3)

= 65.4+12+48

= 125.4g/mol.

Hence, molar concentration= 1.12 x 10-4 g/L / 125.4 g/mol

= 8.93 × 10^-7 mol/L.

Therefore;

Zn2+) = 8.93 x 10-7 mol/L

(CO32-) = 8.93 x 10-7 mol/L

Ksp = [Zn2+] [CO32-]

Ksp = (8.93 x 10-7) × (8.93 x 10-7)

Ksp = 7.98 × 10^-13

the answer is b. dilution

hey there!

mass = 6.75 g

volume =   9 cm³

therefore:

d = mass / volume

d   = 6.75 / 9

d = 0.7 g/cm³

answer b