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Homework Help: Solution of NaCl and CO2. Name species and find concentrations of species.

  1. Nov 15, 2011 #1
    So, I'm assuming no Sodium Bicarbonate or Carbonate is formed during this process.

    I found the concentrations of the species by basic ICE diagram.

    Mostly, I need to know how to approach this problem conceptually . I'm assuming the effect of NaCl on the CO2-->H2CO3-->HCO3- --> CO3-- is negligible, but I really don't know. This is for a geochem class, and I have very little chemistry experience, so any explanations or theory would help tremendously.


    1. The problem statement, all variables and given/known data

    Problemset 3.

    1. Consider and NaCl solution into which we introduce CO2 by bubbling gas. Recipe: 10-3M NaCl, 10-4M [CO2]T.
    (a) What species are present?

    (b) What reactions take place?

    (c) What is the concentration of the species

    (d) Now add 10-4.5M [NaOH]T .What is the alkalinity of the solution?

    H2CO3*=H++HCO3-; pK1=6.3 (I)
    HCO3-=H++CO32-; pK2=10.3 (II)


    2. Relevant equations



    3. The attempt at a solution


    (a) What species are present?

    Species present are Na+ , Cl-, H+, OH-, HCO3- , CO32- , H2CO3 and CO2 (aq), Na2CO3, NaHCO3


    (b) What reactions take place?

    (c) What is the concentration of the species


    [HCO3-]= 2.07E-5
    [H+] ~ 2.07E-5
    [CO32-]=5.62E-11

    [Na+] = .001 M

    [Cl-] = .001 M

    [HCO3-]= 2.07E-5

    Using this below.


    H2CO3 HCO3- + H+ pKa1 (25 °C) = 6.37
    HCO3-  CO32- + H+ pKa2 (25 °C) = 10.25


    H2CO3 HCO3- + H+ pKa1 (25 °C) = 6.37 Ka1=4.3E-7
    HCO3-  CO32- + H+ pKa2 (25 °C) = 10.25 Ka2 =5.62E-11


    H2CO3 HCO3- H+
    I 1E-3 0 0
    C -x +x +x
    E .001-x x x

    Ka1= [HCO3][H]/[H2CO3] = x2/ .001-x = 4.3E-7

    [H+]=[HCO3-]=2.07E-5


    The second dissociation contributes very little [H+] =5.62E-11


    [H+] ~ 2.07E-5

    [CO32-]=5.62E-11
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Nov 15, 2011 #2

    Borek

    User Avatar

    Staff: Mentor

    NaCl presence changes ionic strength of the solution, so it slightly changes equilibrium concentrations of all ions present in the solution. Not enough to make it a problem.

    .001 or .0001?
     
  4. Nov 15, 2011 #3
    .0001. Sorry.
     
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