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Solution of NaCl and CO2. Name species and find concentrations of species.

 
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Nov15-11, 02:58 PM   #1
 

Solution of NaCl and CO2. Name species and find concentrations of species.


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
 
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Nov15-11, 03:19 PM   #2
 
Admin
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.

Quote by geostudent View Post
x2/ .001-x = 4.3E-7
.001 or .0001?
 
Nov15-11, 03:42 PM   #3
 
.0001. Sorry.
 
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