Question

A river flows on a bedrock composed primarily of limestone (calcium carbonate – CaCO3), which partially dissolves in water into calcium (Ca2+) and carbonate (CO32-) ions. The equilibrium concentration of Ca2+ in the water is measured at 1.7 mg/L. The solubility product of calcium carbonate, Ks = 2.8 10-9.
Based on these data, calculate the equilibrium concentration of CO32- (in mg/L) in the water (write calculation details).

Answer 1

C = 1.7 m of Ca+2 g/L

Ks = 2.8*10^-9

calcualte CO3-2 n mg/L

First, the equilibrium expresion

CaCO3 ---> Ca+2 and CO3-2

Ksp = [Ca+2][CO3-2]

then , calculate Ca+2 ions in mol

[Ca+2] =1.7*10^-3 g / L

MW of Ca+2 = 40

mol = mass/MW = (1.7*10^-3)/(40 ) = 0.0000425 mol of Ca+2

then

[Ca+2] =1.7*10^-3 g / L = 0.0000425 mol/L = 0.0000425 M

Apply Ksp

Ksp = (0.0000425)[CO3-2]

2.8*10^-9= (0.0000425)[CO3-2]

[CO3-2] = (2.8*10^-9)/(0.0000425) = 0.00006588235 M

change to mg

MW CO3 = 60.01 g/mol

mass = mol*MW = 60.01*0.00006588235 = 0.00395359982 g

we need mg so

0.00395359982*1000 = 3.953 mg /L