A water contains 10.0 mg/L CO^2-_3 and 75.0 mg/L HCO^-_3 at a pH of 8. Note for

Question

A water contains 10.0 mg/L CO^2-_3 and 75.0 mg/L HCO^-_3 at a pH of 8. Note for carbonic acid, pK_a1 = 6.35; PK_a2 = 10.33. Write down all relevant reaction equations and expressions for equilibrium constants. Calculate the alkalinity exactly at 25 degree C. What is the approximate alkalinity by ignoring (OH) and (H^+)? If the water contains 80 mg/L, calcium, 40 mg/L of magnesium, 20 mg/L chloride. and 5 mg/L CO_2. what amount of lime and/or soda ash. in mg/L as CaCO_3, is required to soften the water to 80.0 mg/L hardness as CaCO_3?

Explanation / Answer

(a)

Reactions are:

When carbon dioxide enter in the water then following reaction occur:

CO2(g)+H2O = H2CO3

Then Carbonic acid will dissociate and form Bicarbonate.

H2CO3 =H++HCO3-

So equlibrium constant Ka1 = [H+][HCO3-]/[H2CO3 ]

Here equilibrium constant for formation of Carbonic acid is given:pKa1 =- 6.35

So for dissociation of carbonic acid is Ka1 = [H+][HCO3-]=10-6.35

Now bicarbonate dissociate and form carbonate:

HCO3-=H++CO32-

So Ka2 = [H+][CO32-]/[HCO3-]=10-10.33

(b)

Alkalinity = [HCO3-]+[CO32-]+[OH-]-[H-]

[HCO3-]=75 mg/L = 1.23*10-3 mol/L

(for conversion of mole divided to molecular weight.Molecular weight of HCO3=61 )

[CO32-]=10 mg /L = 0.167*10-3 mol/L

molecular weight of CO3 = 60

So from reaction 2 :

Ka2 = [H+][CO32-]/[HCO3-]=10-10.33

So solving these equation

[H+]=10-10.33*1.23*10-3/0.167*10-3=3.445*10-10 mol/L

We know that PH+POH=14

PH=-log[H+] =9.46

POH=-log[OH-]=14-9.46=4.537

So [OH-]=2.90*10-5 mol/L

Alkalinity = [HCO3-]+[CO32-]+[OH-]-[H-]

So alkalinity = 1.23*10-3+ 0.167*10-3+2.90*10-5 -3.445*10-10=1.425*10-3 mol/L

(c)

If we ignore [OH-] and [H+]

Alkalinity = [HCO3-]+[CO32-]

So alkalinity = 1.23*10-3+ 0.167*10-3=1.397*10-3 mol/L

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