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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?

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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.

I hope I've answered your question.

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