Question

A treatment plant operator would like to estimate the maximum amount of coagulant, alum Al_2(SO_4)_3 14H20, that she can add to the plant's source water (initial pH = 7.5 and initial ALK = 40 mg/L as CaCO_3) without causing the pH to drop below the minimum desirable pH value of 6.5. Assume that all added alum precipitates as Al(OH)_3(s) Al_2(SO_4)_3 14H2O rightarrow 2Al(OH)_3(s) + 3SO4^2- + 6H^+ + 8H_2O You can also assume that gas exchange during the various treatment steps is slow enough to that the solution can be treated as a closed system.

Answer 1

the max pH drop = 7.5-6.5 = 1

so..

40 mg/L alkalinity of CaCO3

Calculate concentration of CaCO3:

mass to mol = mass/MW = (40*10^-3)/100 = 0.0004 mol of CaCO3

this is per liter so

[CaCO3] = 0.0004 M

1 mol of CaCO3 --> 1 mol fo CO3-2

1 mol of CO3-2 can neutralize up to 2 mol of H+

2H+ + CO3-2 --> H2CO3

Then...

the pH is decreasing by 1

so

[OH-] concentration is decreasing by 10^-1, or by 0.1

0.0004 M of CaCO3 --> 0.1*0.0004 = 0.00004 M of CaCO3 must be left...

so we need:

difference = (0.0004 - 0.00004 ) = 0.00036 M

then

since ratio is:

1 mol of CaCO3 --> neutralizes 2 mol of H+

1 mol of Al2(SO4)3 --> yield sup to 6 mol of H+

then

ratio is 1:3

0.00036*1/3 --> = 0.00012 M of Al2(SO4)3 may be added