Calculate CO2 for Water Equilibrium at pH 7.2

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alex5d1
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Hi,
I must calculate how much CO2 has to be removed from the water (pHs=7.2) to obtain equilibrium

Given concentrations:
HCO3-: 223 mg/L
CO2: 60 mg/L
pH: unknown
SI: unknown

What is meant by equilibrium? (ph=7?) Please advise how to tackle this problem
 

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My bet is that they want you to assume concentration of CO2 is equivalent to the concentration of H2CO3. Then it is just a matter of treating the solution as a buffer, finding the required final concentration of carbonic acid, and comparing it with the initial (60 mg/L) CO2 concentration.
 
Sorry, I forgot to mention that this is a water treatment problem. Temperature is 12C.
What formula is applicable?
 
Acid dissociation constant definition, or Henderson-Hasselbalch equation (which is actually the same, just rearranged to a more convenient form).
 
Well, after some research I came up with this, what do you think?

pHs = 7.2 = equilibrium of CO2 (= H2CO3 ) and HCO3-

pHs = pK1 – log ([CO2] / [HCO3-])

pK1 = 6.45 @12C

log ([CO2] / [HCO3-]) = -0.750

log ([CO2] / [HCO3-]) = 0.178

[HCO3-] = 223/61 = 3.656 mol/L

[CO2] = 0.178 * 3.656 = 0.650 mol/L

CO2 = 0.650 * 44 = 28.6 mg/L

We need to remove 60 – 28.6 = 31.4 mg/L
 
Actually I did a mistake reading the original question, and it doesn't make sense to me now. Sorry about that. I missed the part about initial pH being 7.2. If the initial pH is 7.2 and you are asked to change the pH of the solution to 7.0, you have to ADD acid, not remove it from the solution.

However, if you use concentrations given to calculate pH (which I did) you will find that pH of the initial solution is around 6.8, not 7.2.

alex5d1 said:
log ([CO2] / [HCO3-]) = 0.178

Youdon't mean a log here, but OK.

[HCO3-] = 223/61 = 3.656 mol/L

223 mg, not g. Besides, don't ignore units, as you force others to guess what is 61.
 
Sorry for my typo's in the units. The question as I understand it is to obtain Carbonic acid equilibrium given the equilibrium pH = pHs = 7.2

Then:

pHs = 7.2 = equilibrium of CO2 (= H2CO3 ) and HCO3-

pHs = pK1 – log ([CO2] / [HCO3-])

pK1 = 6.45 @12C

log ([CO2] / [HCO3-]) = -0.750

[CO2] / [HCO3-] = 0.178

[HCO3-] = 223/61 = 3.656 mmol/L

[CO2] = 0.178 * 3.656 = 0.650 mmol/L

CO2 = 0.650 * 44 = 28.6 mg/L

We need to remove 60 – 28.6 = 31.4 mg/L. Do you think this is correct?
 
At least logic behind looks OK, I have not checked the exact value (but it is definitely in a correct ballpark).
 
Hi Borek,
That calculation was correct!
many thanks for your assistance.

One more question, how do I calculate this one?:
Lime (calcium hydroxide) is used in softening. How much lime (mmol/L) needs to be dosed in the softening step?