June 15, 2019, 09:42:11 PM
Forum Rules: Read This Before Posting


Topic: Mass and Charge Balancing  (Read 555 times)

0 Members and 1 Guest are viewing this topic.

Offline ubcdumbass

  • Regular Member
  • ***
  • Posts: 13
  • Mole Snacks: +0/-0
Mass and Charge Balancing
« on: February 08, 2019, 11:27:10 PM »
My professor in class literally didn't know how to do these equations that we are expected to know, and then moved on. I'm stuck. Any tips on how to do mass and charge balancing would be great. Here's a question in example:

Q: Choose the correct and full set of mass balance equation(s) for a solution prepared by mixing equal volumes of 0.050 M sodium oxalate, Na2C2O4 (aq), and 0.010 M ammonium chloride, NH4Cl (aq).
 
        (a) 0.005 M = [NH4+] + [NH3]
   (b) 0.010 M = [NH4+] + [NH3]
   (c) 0.020 M = [NH4+] + [NH3]
   (d) 0.025 M = [HC2O4–] + [C2O42–]
   (e) 0.025 M = [H2C2O4] + [HC2O4–] + [C2O42–]
   (f) 0.050 M = [HC2O4–] + [C2O42–]
   (g) 0.050 M = [H2C2O4] + [HC2O4–] + [C2O42–]
   (h) 0.025 M = [Na+]
   (i) 0.050 M = [Na+]
   (j) 0.005 M = [Cl–]
   (k) 0.010 M = [Cl–]
I chose b, g and k because of course due to the molarity above, [Cl-] would be = 0.010M, so would [NH4+], and [C2O42-] would disassociate into [H2C2O4], [HC2O4-] and [C2O4 2-], and altogether the concentration would be 0.050M from the original question.
It's not in the question, but would [Na+] = 0.10M?

Charge Balancing:
Q:Choose the correct charge balance equation(s) for a solution prepared by mixing equal volumes of 0.050 M Na2C2O4 (aq), and 0.010 M NH4Cl (aq), without making any assumptions about pH.

        (a) [NH4+] + [NH3] = [H2C2O4] + [HC2O4–] + [C2O42–]
   (b) [NH4+] + [NH3] = [H2C2O4] + [HC2O4–] + 2[C2O42–]
   (c) [NH4+] + [H3O+] = [HC2O4–] + [C2O42–] + [OH–]
   (d) [NH4+] + [H3O+] = [HC2O4–] + 2[C2O42–] + [OH–]
   (e) [NH4+] + [Na+] + [H3O+] = [HC2O4–] + [C2O42–] + [Cl–] + [OH–]
   (f) [NH4+] + [Na+] + [H3O+] = [HC2O4–] + 2[C2O42–] + [Cl–] + [OH–]
I chose f as I believe that [C2O4 2-] disassociation would make 2[C2O4 2-] in the product and [HC2O4-] but not [H2C2O4] (that'd be cancelled out).
Does that make sense? How would one calculate the coefficients for OH- and H30+? I know in the multiple choices none of them have coefficients but in the future when I was doing the equilibriums there seemed to be more than one OH-.

Offline AWK

  • Retired Staff
  • Sr. Member
  • *
  • Posts: 6458
  • Mole Snacks: +465/-79
  • Gender: Male
Re: Mass and Charge Balancing
« Reply #1 on: February 09, 2019, 01:54:42 AM »
As you noticed there is no answer with [Na+]= 0.10 hence all your guesses are wrong. Though... you answer for this concentration consists of other answers.
How "mixing equal volumes" influences on concentrations?

« Last Edit: February 09, 2019, 02:49:08 AM by AWK »
AWK

Offline ubcdumbass

  • Regular Member
  • ***
  • Posts: 13
  • Mole Snacks: +0/-0
Re: Mass and Charge Balancing
« Reply #2 on: February 09, 2019, 02:50:50 AM »
I completely forgot to consider the total volume vs the volume added. I think I got it now.
Onto charge balancing:

NH4+ is an acid (weak) and therefore would give off an H3O+, is that would you're trying to say?
Same as the C2O4 2- + HC2O4- + H2C2O4.

Would the concentrations of the Nh4+ and c2o42- affect the coefficients OH- and H30+?

Offline AWK

  • Retired Staff
  • Sr. Member
  • *
  • Posts: 6458
  • Mole Snacks: +465/-79
  • Gender: Male
Re: Mass and Charge Balancing
« Reply #3 on: February 09, 2019, 03:32:10 AM »
All equilibria (including ionic product of water) that exist in solution influence on the final concentration of H3O+ and OH-.

But you should consider your problem "without making any assumptions about pH".

Your answer for charge balance, where "concentrations of charges" are expressed by the concentration of ions existed in solutions in the equilibrium. is correct.
AWK

Sponsored Links