Question

The blood value of patient were pH=7.03 and [CO2]=1.1mM. What is the patient's blood[HCO3^-]
and how much of the normal [HCO3^-] has been used in buffering the acid causing the condition?

Answer 1

So you have this equilibrium:

\(CO_2+H_2O \rightleftharpoons H_2CO_3+H_2O \rightleftharpoons HCO_3^-+H_3O^+ \)

We can ignore the water because it's concentration is much higher than the rest of the substances present. So we have:

\(K^1_{eq}=\dfrac{[H_2CO_3]}{[CO_2]}\) and \(K^2_{eq}=\dfrac{[H_3O^+][HCO_3^-]}{[H_2CO_3]}\)

We want just 1 equation with what we are given, so, do some algebra.

rearrange the first: \([H_2CO_3]=K^1_{eq}*[CO_2]\)

sub into second: \(K^2_{eq}=\dfrac{[H_3O^+][HCO_3^-]}{K^1_{eq}*[CO_2]}\)

we have: \(K^1_{eq}*K^2_{eq}=\dfrac{[H_3O^+][HCO_3^-]}{[CO_2]}\)

You're given pH, so you know \([H_3O^+]\)...so it seems that you can simply say: \([H_3O^+]=[HCO_3^-]\)...But i'm not sure, did anything else happen? What acid are they talking about?