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

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SUMMARY

The discussion focuses on the interaction of a 10-3M NaCl solution with CO2, leading to the formation of various species including Na+, Cl-, H+, OH-, HCO3-, CO32-, H2CO3, and CO2(aq). The concentrations of the species were calculated using an ICE diagram, resulting in [HCO3-] = 2.07E-5 M, [H+] ~ 2.07E-5 M, and [CO32-] = 5.62E-11 M. The presence of NaCl was determined to have a negligible effect on the CO2 equilibria, although it slightly alters the ionic strength of the solution.

PREREQUISITES
  • Understanding of acid-base equilibria, specifically the dissociation of carbonic acid (H2CO3)
  • Familiarity with ICE (Initial, Change, Equilibrium) diagrams for chemical reactions
  • Knowledge of pKa values and their application in calculating concentrations
  • Basic concepts of ionic strength and its effect on chemical equilibria
NEXT STEPS
  • Study the dissociation constants (Ka) for carbonic acid and bicarbonate to deepen understanding of acid-base equilibria
  • Learn about the effects of ionic strength on chemical equilibria in solutions
  • Explore the use of ICE diagrams in more complex chemical equilibria
  • Investigate the role of NaCl in geochemical processes and its impact on solubility
USEFUL FOR

This discussion is beneficial for students in geochemistry, particularly those studying acid-base reactions in aqueous solutions, as well as anyone interested in the effects of ionic compounds on chemical equilibria

geostudent
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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.


Homework Statement



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)


Homework Equations





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
 
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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.

geostudent said:
x2/ .001-x = 4.3E-7

.001 or .0001?
 
.0001. Sorry.
 

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