# How to determine pressure and concentration of carbon dioxide in blood plasma?

A man suffering from untreated diabetes mellitus is admitted to a hospital. Doctors fear that his blood $$\mathrm{pH}$$ may have dropped because of ketoacidosis. Analysis of his blood reveals that $$[\ce{HCO3-}] = \pu{16 mM}$$ and $$p_\ce{CO2} = 30.$$ If $$\mathrm{p}K_\mathrm{a}$$ of $$\ce{HCO3-}$$ is $$6.1,$$ determine whether the patient runs a risk of acidotic coma. (Note: in plasma under physiologic conditions, concentration of $$\ce{CO2}$$ and $$p_\ce{CO2}$$ are related by the solubility constant for $$\ce{CO2}$$ in plasma which is $$\pu{0.03 mM/mm Hg}.$$

We use the Henderson-Hasselbalch equation, then

$$\mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log[\ce{HCO3−}][\ce{H2CO3}]$$

Moreover, here $$\mathrm{p}K_\mathrm{a}$$ of $$\ce{HCO3-}$$ is given whereas the acid is $$\ce{H2CO3},$$ and also if we consider second dissociation that is $$\mathrm{p}K_\mathrm{a2}$$ then also it is wrong because $$\ce{HCO3-}$$ will dissociate into $$\ce{H+}$$ and $$\ce{CO3^2-}$$ so in this dissociation there is no relation of $$\ce{CO2}.$$ And also why solubility constant is given, how can we find out $$\ce{CO2}$$ from solubility constant?

Since $$p_\ce{CO2}$$ is $$\pu{30 mmHg}$$, and we are given the $$0.03$$ of $$\frac{\pu{mM}~\ce{CO2}}{\pu{mmHg}~\ce{CO2}}$$ ratio, we can calculate the concentration of the $$\ce{CO2}$$ in plasma.

$$\pu{30mmHg}~\ce{CO2}\times 0.03\frac{\pu{mM}~\ce{CO2}}{\pu{mmHg}~\ce{CO2}}= \pu{0.9 mM}~\ce{CO2}$$

$$\ce{CO2 + H2O \xrightarrow{\text{carbonic anhydrase}} H2CO3}$$

$$\text{Thus, } \pu{0.9 mM}~\ce{CO2}= \pu{0.9 mM}~\ce{H2CO3}$$

$$\ce{H2CO3 <=> H+ + HCO3-}$$

$$\frac{\ce{[H+][HCO3-]}}{\ce{[H2CO3]}} = K_{\mathrm{a}_\ce{H2CO3}}$$

$$\log{\frac{\ce{[H+][HCO3-]}}{\ce{[H2CO3]}}}=\log{K_{\mathrm{a}_\ce{H2CO3}}}$$

$$\log{\ce{[H+]}}+\log{\frac{\ce{[HCO3-]}}{\ce{[H2CO3]}}}=\log{K_{\mathrm{a}_\ce{H2CO3}}}$$

$$-\left(\log{\ce{[H+]}}+\log{\frac{\ce{[HCO3-]}}{\ce{[H2CO3]}}}\right)=-\log{K_{\mathrm{a}_\ce{H2CO3}}}$$

$$\mathrm{pH}-\log{\frac{\ce{[HCO3-]}}{\ce{[H2CO3]}}}=\mathrm{p}K_{\mathrm{a}_\ce{H2CO3}}$$

$$\ce{[HCO3-]}=\pu{16 mM}$$

$$\ce{[H2CO3]}=\pu{0.9 mM}$$

$$\mathrm{p}K_{\mathrm{a}_\ce{H2CO3}}=6.1$$

(PS: It is actually $$\mathrm{p}K_\mathrm{a}$$ of $$\ce{H2CO3}$$, not that of $$\ce{HCO3-}$$. Note that $$\mathrm{p}K_\mathrm{a}$$ of $$\ce{HCO3-}$$ is $$10.3$$.)

$$\mathrm{pH}-\log{\frac{\ce{[HCO3-]}}{\ce{[H2CO3]}}}=\mathrm{p}K_{\mathrm{a}_\ce{H2CO3}}$$ $$\mathrm{pH}-\log{\frac{16}{0.9}}=6.1$$

$$\mathrm{pH}-1.25=6.1$$

$$\mathrm{pH}=7.35$$

The $$\mathrm{pH}$$ of blood is maintained within extremely narrow limits of 7.36-7.4 for venous blood and 7.38-7.42 for arterial blood. (Source: Upadhyay - Biophysical Chemistry)

So, we can say the $$\mathrm{pH}$$ of the blood of the patient is slightly acidic than normal.