Question

A water sample was analyzed and was found to have the following constituents:

Ca+2, mg/L 123

HCO3 - , mg/L 82

Mg+2, mg/L 12

SO4 -2, mg/L 122

Na+ , mg/L 22.1

Cl- , mg/L 112

K+ , mg/L 3.8

CO3 -2, mg/L 6.0

Fe+2, mg/L 5.8

Mn+2, mg/L 1.5

Temperature 25°C

1. Calculate each of the concentrations as mg/L as CaCO3.

2. Using the equilibrium expression for carbonate and bicarbonate, calculate the hydrogen ion concentration:

a. as moles/L.

b. as mg/L.

c. as mg/L as CaCO3.

d. as pH.

3. Calculate the hydroxide ion concentration:

a. as moles/L.

b. as mg/L.

c. as mg/L as CaCO3.

d. as pOH.

4. Calculate the concentration of CO2 as mg/L as CaCO3 (use the equilibrium equations for the carbonate system and assume that H2CO3 concentration is equal to the CO2 concentration).

Answer 1

1. To calculate each of the concentrations as mg/L as CaCO₃, we have to convert all the species concentration in mmol/L divided by their molar masses and then multiplied by molar mass of CaCO₃.

Molar mass of CaCO₃ = 100 g/mol

ions	concentration (mg/L)	Molar mass of ion(g/mol)	concentration (mmol/L)	concentration as CaCO3 (mg/L)
Ca+2	123	40.078	3.069	306.9
HCO3-	82	61.017	1.344	134.4
Mg+2	12	24.305	0.494	49.4
SO4-2	122	96.06	1.270	127.0
Na+	22.1	23.00	0.961	96.1
Cl-	112	35.45	3.159	315.9
K+	3.8	39.10	0.097	9.7
CO3-2	6.0	60.00	0.100	10.0
Fe+2	5.8	55.845	0.104	10.4
Mn+2	1.5	54.938	0.027	2.7

2. The equation of equilibium of bicarbonate and carbonate

HCO₃^- = H⁺ +CO₃^2-

Keq = [H⁺][CO₃^2-]/[HCO₃^-] =  4.8 x 10^-11 [ at 25^o C]

Putting the concentration from the above table for each species. we get [H⁺]

[H⁺] = Keq [HCO₃^-] /[CO₃^2-] = (4.8 x 10^-11 x 1.344 mmol/L)/0.100 mmol/L

(a) [H⁺] = 64.512 x 10^-11 mol/L

(b) [H⁺]   = 64.512 x 10^-11 mol/L x 1 g/mol = 64.512 x 10^-11 g/L [H⁺ molar mass = 1 g/mol]

= 64.512 x 10^-8 mg/L [ 1 g = 1000 mg]

(c) [H⁺] = 64.512 x 10^-11 mol/L x 100 g/mol = 64.512 x 10^-9 g/L = 64.512 x 10^-6 mg/L (as CaCO₃)   

(d) pH = -log[H⁺]

pH = - log(64.512 x 10^-11)

= 9.19

pH = 9.19

3. We know Kw =[H⁺] [OH^-] = 10-¹⁴

(a) Using above relation we get [OH^-]

  [OH^-] = Kw/[H⁺] = 10^-14/ 64.512 x 10^-11 = 1.55 x 10^-5 mol/L

[OH^-] = 1.55 x 10^-5 mol/L

(b)   [OH^-] = 1.55 x 10^-5 mol/L x 17 g/mol = 26.4 x 10^-5 g/L = 26.4 x 10^-2 mg/L

(c)   [OH^-] = 1.55 x 10^-5 mol/L x 100 g/mol = 1.55 x 10^-3 g/L = 1.55 mg/L (as CaCO₃)

(d) pOH = 14 - pH = 14 - 9.19 = 4.81

pOH = 4.81

4. H₂CO₃ = HCO₃^- + H⁺

Keq = [HCO₃^-][H⁺]/[H₂CO₃] = 4.2 x 10^-7 [at 25^o C]

=>   [H₂CO₃] = [HCO₃^-][H⁺]/Keq = (1.344 x 10^-3 mol/L x 64.512 x 10^-11 mol/L)/4.2 x 10^-7

   [H₂CO₃]   = 20.644 x 10^-7 mol/L

Given, [H₂CO₃] = [CO₂] = 20.644 x 10^-7 mol/L

[CO₂] = 20.644 x 10^-7 mol/L x 100 g/mol = 2.064 x 10^-4 g/L (as CaCO₃)

[CO₂] = 2.064 x 10^-4 x 10³ mg/L = 2.064 x 10^-1 mg/L = 0.2064 mg/L (as CaCO₃)