Problem #100 - ChE 6-Min Solutions

[rudy]

In Problem #100, in the ChE 6-Min Solutions, why is the alkalinity, contributed from CO3, multiplied by 2? Is it because of the 2 negative ions?

Also, there a similar problem in CERM 6th ed., page 35-2, Example 35.2. This is EXACTLY the same, except for the values of the given concentrations (i.e. values of HCO3 and CO3). Here, the alkalinity, contributed from CO3, is NOT multiplied by 2.

Which approach is the correct one (multiply by 2 or NOT multiply by 2)?

 

[Guest]

^^ Can you post the problem? I don't have ChE Six Min Solutions but I am quite familiar with carbonate buffering (e.g. CO3, HCO3, and H2CO3)

JR

 

[rudy]

Here it goes:

A wastewater contains 125 mg/L CO3 ^2- and 82 mg/L HCO3 ^1- at a pH of 9. At 25 C, the ion product of water is 10 ^ -14 mol^2/L^2. At 25 C, the alkalinity as calcium carbonate (CaCO3) is most nearly?

 

[rudy]

.

 

[Guest]

Rudy --

You are correct about the factor of 2 that is multiplied onto the CO3 term - it is due to balancing the chemical dissociation for the bicarbonate represented by HCO3- ---> H+ + CO32- (e.g. there are 2 charges to balance). Also note that calculations of the carbonate cycle are based on a diprotic acid (H2CO3), so you actually have two separate acid-base equilibria working. However, since pH = 9.0, we can safely assume we are working within the second ionization of the equilibrium (e.g. carbonate and bicarbonate species).

As far as calculation:

Total Alkalinity = [HCO3-] + [CO32-] + [OH-] - [H+]

We are given pH = 9.0 which means [H+] = 10-9 mol/L and since we are told Kw = 10-14 that means [OH-] = 10-4 mol/L

For the purposes of this problem, [H+] falls out because the concentration is negligible relative to the other constituents.

So, focusing on [OH-] mol/L = 1.7 mg/L. If you want to find the CaCO3 equivalent = 1.7 mg/L * 2.94 = 4.998 mg/L as CaCO3

The carbonate species are a little easier to resolve:

HCO3- = 82 mg/L * 0.82 = 67.24 mg/L as CaCO3

CO32- = 125 mg/L * 1.67 = 208.75 mg/L as CaCO3

Taking the summation: Total Alkalinity (as CaCO3) ~ 280 mg/L

Special note: I found my equivalency factors in my CERM 8th Edition, Appendix 22.C - I assurem ChERM has a similar appendix.

I had go back on this one - been awhile since I have done the carbonate balances for water/wastewater analysis. Please post if you have any follow-up questions.

JR

 

[rudy]

thank you jregieng. i got 280mg/L, the same way you did, based on the ChERM 6th ed. (yes, the ChERM has the equivalency factors in the appendix). However, the solution in the ChE 6-min Soln. says the answer is 484 mg/L:

67 + 2 (208) = 484 mg/L. At this point, I'm going to say the ChE 6-min soln is wrong and that you and the ChERM are right. thank you again.