Question

A 1.000 L river water sample contains [Ca2+] = 5.6 x 10-3 M, and no other hardness ions. (a) What is the hardness (in ppm CaCO3) of this sample? (b) Assuming the sample is saturated in CaCO3, what is [CO32-] in the sample? (c) Assuming the only other ion present is HCO3-, what is its concentration? (d) Assuming that all the Ca2+(aq) in this sample originally came from CaCO3 dissolution, what mass of CO2 dissolved in this sample, to give it this composition? (Hints: once you have answers to (b) and (c) above, this is simple stoichiometry. What reaction is involved?) Data: F. Mass of CaCO3 = 100.09 g/mol. K sp for CaCO3 = 4.5 x 10-9.

Answer 1

a) Concentration of Ca²⁺ ion = 5.6 x 10-3 M.

we know, 1 mol of Ca2+ ion is equivalent to 40.078 g. Thus amonut of Ca2+ ion in 1 litre river water is (40x5.6x10^-3) g or 0.224 g.

Now 40.078 gm water contains 100.09 g of CaCO3. Hence amonut of CaCO3 in the 1litre river water sample is 0.56 g or 560 mg.

Now hardness of water is expressed in terms of amount of CaCO3 present in a water sample. Thus, the hardness is 560 ppm.

b) Solubility product of CaCO3 is 4.5 x 10^-9. Let solubility of CaCO3 is S. The relation between solulibity and solubility product for sparing soluble salt CaCO3 is K_SP = S². Thus the solubility of CaCO3 is S = (K_SP)^1/2 and hence the solubility is found out to be 6.708 x 10^-5 M.

This shows that maximum 6.708 x 10^-5 M of CaCO3 get dissoved and concequently concentration of CO₃^-2 in the solution is 6.708 x 10^-5 M.

c) Given [Ca2+] = 5.6 x 10^-3 M. The concentration of HCO3- in the solution is 2 x 5.6 x 10^-3 M or 1.12 x 10^-3 M.

d) 40.078 gm Ca requires 44 gm of CO2. Hence to get 560 mg Ca2+ ion in solution, the amount of CO2 dissolved in this sample is 614.8 mg.

The reaction involved is CaO + CO2 = CaCO3.