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### Topic: Water chemistry question (adding a strong acid to a river)  (Read 1286 times)

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#### sophiefell

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• Mole Snacks: +0/-0 ##### Water chemistry question (adding a strong acid to a river)
« on: November 06, 2019, 04:45:09 PM »
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 ##### Re: Water chemistry question (adding a strong acid to a river)
« Reply #1 on: November 06, 2019, 06:43:36 PM »
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?
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info, pH-meter.info

#### Enthalpy ##### Re: Water chemistry question (adding a strong acid to a river)
« Reply #2 on: November 07, 2019, 10:04:36 AM »
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

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• Mole Snacks: +0/-0 ##### Re: Water chemistry question (adding a strong acid to a river)
« Reply #3 on: November 07, 2019, 12:14:45 PM »
Feeling more lost now, hoping someone can help to solve!

#### Borek ##### Re: Water chemistry question (adding a strong acid to a river)
« Reply #4 on: November 07, 2019, 03:32:16 PM »
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.
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info, pH-meter.info

#### sophiefell

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• Mole Snacks: +0/-0 ##### Re: Water chemistry question (adding a strong acid to a river)
« Reply #5 on: November 11, 2019, 04:36:36 PM »
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 /[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.