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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 |
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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 |
| Nov15-11, 03:19 PM | #2 |
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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.
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| Nov15-11, 03:42 PM | #3 |
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.0001. Sorry.
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