[sophiefell]

Each week an industry produces 2.0 × 10^3 m3/d of a strong acid waste (pH = 1.5) during a 48 hour production cycle (total 4.0 × 10^3 m3/week). The plant engineer would like to discharge the waste into a river that passes near the plant. The river has a 7-day, 10-year flow of 25 × 10^3 m3/d, an alkalinity of 150 mg/L as CaCO3, and a pH of 7.7. If the temperature of the mixture is 25oC, will it be possible to discharge all the waste into the river within a 7-day period if the river pH must be maintained above 7.0?

I'm trying to solve as if its a titration/buffer problem with the concentration of the [acid] being 10^-1.5 M and the CaCO3 of the river being 1.5 M (alkalinity/molar mass of CaCO3 which is 100). Then using the 2 volumes as 2.0 x 10^6 L of acid added a day to 25 x 10^6 L of river water.

Not really getting and answer that makes much sense though. Would appreciate any clarity anyone could provide! 

[Borek]

the CaCO3 of the river being 1.5 M (alkalinity/molar mass of CaCO3 which is 100)

Check it out - what are units of dissolved mass of CaCO₃ per L?

[Enthalpy]

Wiki cites 13mg/L for the solubility of CaCO₃ in water. Is it a convention to describe the alkalinity as a CaCO₃ equivalent? I imagine it would be dissolved as Ca²⁺ and 2HCO₃^- rather.

[sophiefell]

Feeling more lost now, hoping someone can help to solve!

[Borek]

Feeling more lost now, hoping someone can help to solve!

You started with an error that I pointed out - have you corrected it? Your approach sounds more or less reasonable, you just started with a simple mistake.

[sophiefell]

so correcting for the mistake that was pointed out:

Units of dissolved CaCO3 is 25x10^6 L of river water x 150mg/L (CaCo3 Alkalinity) / 1000
= 3.75x10^6 g

3.75x10^6 g x (mol/100)
=3.75x10^4 M of CaCO3

From there I tried to solve as if a titration/buffer problem:

[Acid]= 10^-1.5 M x 2x10^6 L (volume of acid added per day)
         = 6.3x10^4 mol

[CaCO3]= 3.75x10^4 M x 25x10^6 L (volume of river flow per day)
            = 9.375x10^11 mol

       CaCO3              +              Acid ->                    HCO3- + Ca
       9.375x10^11                  6.3x10^4                       0
       -6.3x10^4                      -6.3x10^4                   +6.3x10^4

       ^(This amount is essentially negligible and unchanged)

total volume is 6.3x10^4 + 9.375x10^11 = 9.375x10^11

[CaCO3] = 1

[HCO3-} = 6.3x10^4 / 9.375x10^11
=6.72x10^-8

pH = log [A-]/[HA]
     = log [1]/[6.72x10^-8]
     = 7.17
     =7.17

not at all sure I've done this right. 7.17 seems like a possible number, seeing as we would expect the pH to go down after adding acid a strong acid to the river, but probably by not that much considering you are adding much less than the volume of the river.