Question

Please explain step by step how you get the answer!

(24) 2. Solve the following I. Calculate the Molarities, Normalities and mg/l as CaCO3 for the following (a) 200 mg/l HCI (b) 100 mg/1 Ca(HCO3)2 II. Calculate the mg/l and mg/l as CaCO, for the following (a)0.05 N H2CO3 (b) 0.001M CHCl3 III. Calculate the mg/l for the following (a) 100 mg/l as CaCO3 of HCO3(b) 100 mg/l as CaCO3 of 5042

Answer 1

The given concentrations have to expressed as CaCO₃ equivalent. This is done by mutiplying the mg/l of the substance by CaCO₃ equivalent factor. This factor is calculated as : (EW of CaCO₃ / EW of the substance)

Here, EW is the equivalent weight of the substance.

Now EW of CaCO₃ is = Molecular weight (MW)of CaCO₃ / 2        ... since it is a salt and total charge on each cation/anion =2

           = (100 g /2)

             = 50 g

(I) a) For 200 mg/l HCl,

     EW of HCl = (MW of HCl) / 1 = 36.5 g

The CaCO₃ equivalent factor = (50 g / 36.5 g) = 1.37

Therefore 200 mg/l HCl = 200 x 1.37 mg/l of CaCO₃

                                     = 274 mg /l of CaCO₃

For Molarity (M) of 200 mg/l HCl :

moles of HCl = 200 mg/ 36.5 g = 5.48 x 10 ^-3 mol

Therefore mole/litre (M) = 5.48 x 10 ^-3 M

And Normality = Molarity = 5.48 x 10 ^-3 N

Since, for HCl, each cation and anion has charge of 1 unit.

b) 100 mg/ l of Ca (HCO₃)₂

EW of Ca (HCO₃)₂ = (MW of Ca (HCO₃)₂ / 2 = 81 g

The CaCO₃ equivalent factor = (50 g / 81 g) = 0.62

Therefore 100 mg/l Ca (HCO₃)₂ = 100 x 0.62 mg/l of CaCO₃

                                     = 62 mg /l of CaCO₃

For Molarity (M) of 100 mg/l Ca (HCO₃)₂ :

moles of Ca (HCO₃)₂= 100 mg/ 162 g = 0.617 x 10 ^-3 mol

Therefore mole/litre (M) = 6.17 x 10 ^-4 M

And Normality = 2x Molarity = 2 x 0.617 x 10 ^-3 N = 1.23 x 10 ^-3 N

Since, for Ca (HCO₃)₂, total charge for each cation and anion = 2 units.

(II) a) 0.05 N H₂CO₃ = 0.05 x 2 M H₂CO₃ = 0.1 M H₂CO₃

Since total charge for each cation and anion = 2 units.

Now, 0.1 M = 0.1 mol / litre = 0.1 x 62 g /l = 6200 mg / l

And the    The CaCO₃ equivalent factor = (50 g / 31 g) = 1.61

Therefore 6200 mg / l  of H₂CO₃ = 6200 x 1.61 mg/l of CaCO₃

                                                            = 9982 mg /l of CaCO₃

b) 0.001 M CHCl₃ = 0.001 mol / litre = 0.1 x 119.4 g /l = 11940 mg / l

     And the    The CaCO₃ equivalent factor = (50 g / 119.4 g) = 0.42          .....since it is an acid with one acidic proton

Therefore 11940 mg / l of CHCl₃= 11940 x 0.42 mg/l of CaCO₃

                                                          = 5000 mg /l of CaCO₃

(III) Here 100 mg/l as CaCO₃ is given.

a) The conversion factor for HCO₃^- ion is = (50/ 61)           .....since it is an ion with charge =1

                                                              = 0.82

Hence concentration of HCO₃^- = (100 mg/ l) / (0.82) = 122 mg /l of HCO₃^-

b) The conversion factor for SO₄^2- ion is = {50/ (96/2)}          .....since it is an ion with charge = 2

                                                             = 1.04

Hence concentration of SO₄^2- = (100 mg/ l) / (1.04) = 96.15 mg /l of SO₄^2-