1
$\begingroup$

Carbon dioxide is transported through blood via 3 methods : 1. Dissolved in plasma 2. As bicarbonate ion 3.through RBCs. The carbondioxide when transported as bicarbonate ion i.e HCO3- and H+. What i dont understand is that dissolved carbon dioxide in plasma should also yield H2CO3, so is writing co2(aq) the same thing as writing H2CO3? Further shouldnt this carbonic acid also dissociate and form bicarbonate ion and proton? Doesn't this make both the processes equivalent? What i found out is that H2CO3 has a low ionisation constant of the order 10-4, is this tbe reason why they are separate events? Implying that carbonic anhydrase catalyses the formation of bicarbonate ion while some part of H2CO3 is not ionised and that is what refers to co2(aq)

$\endgroup$
  • $\begingroup$ Please put more effort into presenting your questions in a readable manner. You can do this by: 1. Using paragraphs to separate different ideas or different points. 2. By making sure that you capitalize the letters representing chemical elements (CO not co), 3. More difficult subscripting or superscripting chemical formulae. There are two ways of doing the latter. I use HTML markup e.g. <sub>numeral</sub> and <sup>+</sup>. This will work in questions and answers (not in comments) 4. Capitalize "I" in English. $\endgroup$ – David Feb 17 '18 at 10:43
  • 1
    $\begingroup$ Im sorry for being sloppy, i typed it on my phone and didn't pay much attention to details, what do you think about the answer? $\endgroup$ – Anamika Ghosh Feb 17 '18 at 11:45
  • $\begingroup$ Well if you can find a computer and take the trouble to clean the question up as I suggested — and if nobody has answered in the meantime — tomorrow I might look out the lecture notes I used to use on carbon dioxide transport and bicarbonate buffers and provide a comprehensive answer. In SE the idea is not to provide a limited answer for one person, but to provide answers to biological questions that will be of use to many. $\endgroup$ – David Feb 17 '18 at 17:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.