science( chem)

The total Carbon dioxide content (HCO3- + CO2) in a blood sample is determined by acidifiying the sample and measuring the volume of CO2 evolved with a van slyke manometric apparatuses. The total concent was determined to be 28.5 mmol/L. The blood pH at 37o was determined to be 7.48. What are the concentrations of HCO3- and CO2 in the blood? pKa= 6.10

I looked up the eqzn for CO2 in blood
and it was

CO2 + H2O=> H2CO3
H2CO3=> H+ + HCO3-

why is there this conversion factor online for the H-H eqzn for this rxn??
pH = 6.1 + log [HCO3-/(0.03)(PCO2)]

I don't think that I have to use this b/c It wasn't in the problem Q but
anyways I'll post what I did and say what I can't figure out..

and another thing HCO3- is a base??
while CO2 is a acid??

tot concent= (28.5mmol/L)/1000= 0.0285mol/L

pH = pKa+ log(cHCO3-/cCO2-)

but since I like the other eqzn better...
[H30+]= 10^-pH= 10^-7.48= 3.31e-8
Ka= 10^-pKa= 10^-6.10= 7.94e-7

[H30+]= Ka(cCO2/cHCO3-)

3.31e-8= 7.94e-7(cCO2/cHCO3-)

0.04168= (cCO2/cHCO3-)

After this I don't know what to do with the total concentration of 0.0285mol/L

I think that I get this now but just to make sure...

since the total concent is 28.5mmol/L

I plugged it into the eqzn for the H3O+

[H30+]= Ka(cCO2/(28.5mmol/L-[CO2]))

0.04168= (cCO2/(28.5mmol/L-[CO2]))
1.187= 1.0416cCO2
cCO2= 1.139

Plugged into the eqzn for the total

[HCO3-]= 28.5mmol- 1.139= 27.36mmol

Using that found concent of the HCO3- & CO2 I plugged it into the eqzn to check the answer.

[H30+]= Ka(cCO2/cHCO3-)
= (7.94e-7)(1.139/27.36)
= 3.30e-8
pH= -log(3.30e-8)= 7.48

which was the pH given..but I had read online that usually the concentration of CO2 is larger than HCO3- but that wasn't the case here. Unless that information is wrong...
and CO2 IS the acid right?
while the HCO3- is the base?
b/c if that wasn't the case then the ratio would be flipped around.

I would like if someone looked at my work and see if it's correct...
thanks =)

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  1. Thanks Christina,this is 2019 here and I'm amazed at your work cuz it is difficult for me now .my only problem is that I think you have reversed the concentrations for CO2 and HCO3- respectively.I don't know if you could still look into it

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