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= fractional dissociation of the hydroxide ion from the base provided

Let us derive the formulas for calculating the quantities of sulfuric acid, hydrochloric acid, and nitric acid and the formulas for calculating the quantities of caustic soda and soda ash that may be needed to lower and to raise the pH, respectively. To find the equivalent masses of the acids, they must be reacted with the hydroxyl ion. Reaction with this ion is necessary, since total alkalinity may be represented by the hydroxyl ion. Remember that the acids must first consume all the existing alkalinity represented in the overall by the OH- before they can lower the pH. Thus, proceed as follows:

From these equations, the equivalent masses are sulfuric acid = H2SO4/2 = 49.05, hydrochloric acid = HCl/1 = 36.5, and nitric acid = HN03/1 = 63.01.

To find the equivalent masses of the caustic soda and soda ash, they must be reacted with an acid, which may be represented by the hydrogen ion. Again, remember that the bases specified previously must first consume all the existing acidity of the water represented in the overall by H+ before they can raise the pH. Thus,

From the equations, the equivalent masses are caustic soda = NaOH/1 = 40 and soda ash = Na2CO3/2 = 2 x 23 + 12 + 48/2 = 53.

Let MH SO pH, MHClpH, and MHNO;,pH be the kilograms of sulfuric acid, hydrochloric acid, or nitric acid used to lower the pH from the current pH to pH(0. Gleaning from Equation (13.54) and the respective equivalent masses of H2SO4, HCl, and HNO3 and the cubic meters, V, of water treated,

MH2SO4pH = 49.05] [Acadd]geq = [Acur]geq + 110--10-W (13.61)

MHClpH = 36.5i (Acadd]geq = (Acur]geq + 10--10-(13.62)

M HNO3pH = 63.01 i (Acadd]geq = (AcurLeq + --- K (13.63)

And, also, gleaning from Equation (13.55) and the respective equivalent masses of NaOH and Na2CO3 and the cubic meters of water treated, ¥,

T 10—pHcur - 10 pH'° 1 MNaOHpH = 40.0i (Accur]geq + 10-T-10- ¥ (13.64)

10-pHcur -10-pH"

where MNaOHpH and MNa CO pH are the kilograms of caustic soda or soda ash, respectively, used to raise the pH from the current pH to pHto.

They are strong acids, therefore, the Ta s for H2SO4, HCl, and HNO3 are unity. The hydrogen ion resulting from the second ionization of sulfuric is very small so it can be neglected. Also, because NaOH is a strong base, its T is equal to unity. The Tb for Na2CO3 is not as straightforward, and we need to calculate it. Sodium carbonate ionizes completely, as follows:

The carbonate ion then proceeds to react with water to produce the hydroxide ion, as follows (Holtzclaw and Robinson, 1988):

Now, proceed to calculate the fractional dissociation of the hydroxide ion from sodium carbonate. Begin by assuming a concentration for the carbonate of 0.1 M.

This will then produce, also, 0.1 M of the carbonate ion. By its subsequent reaction with water, however, its concentration at equilibrium will be smaller. Let x be the — — 2— concentrations of HCO3 and OH at equilibrium. Then the concentration of CO3 Healthy Chemistry For Optimal Health

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