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Buffer solutions are used by biological mammalian systems to maintain the $\mathrm{pH}$ of blood plasma within a narrow range. In these systems, the compound from which this solution is obtained is $\ce{CO2}$, produced in cell respiration, which is converted into $\ce{HCO3-}$ and $\ce{H2CO3}$ inside the red blood cells. Using the Henderson-Hasselbalch equation, answer: $\mathrm{p}K_\mathrm{a} = 6.35$. What is the ratio between $\ce{HCO3-}$ and $\ce{H2CO3}$ in a blood sample with $\mathrm{pH} = 7.4$?

The problem is I try to make the exercise in this way:

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

BUT in the solution they make $$\mathrm{pH} =\mathrm{p}K_\mathrm{a} - \log \dfrac{[\ce{HCO3-}]}{[\ce{H2CO3}]}$$

They make the inverse of $$\mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log \dfrac{[\ce{H2CO3}]}{[\ce{HCO3-}]}$$

And this don't make sense for me. I don't understand.

The solution given is in the white box on the left side. The white box is the solution who they give to me

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Recall the fundamental properties of the logarithm $$+\log \left(\frac{a}{b}\right)$$

is the same as

$$-\log \left(\frac{b}{a}\right)$$

so Henderson Hasselbach can have plus or minus signs. The correct equation is

$$\mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log \left(\frac{\text{conjugate base}}{\text{acid}}\right)$$

or

$$\mathrm{pH} = \mathrm{p}K_\mathrm{a} - \log \left(\frac{\text{acid}}{\text{conjugate base}}\right)$$

Now sort out whether the textbook solution is right or not.

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    $\begingroup$ I wish an image of the text would posted too in such cases. I'm guessing that by far the most probable explanation was that the OP just missed the flip in the log, but it is possible that the book was wrong. $\endgroup$ – MaxW May 10 at 21:43
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    $\begingroup$ Students can be further confused by the fact that $p=-log$. That is another headache. $\endgroup$ – M. Farooq May 10 at 21:58
  • $\begingroup$ Good night. The acid is H2CO3, right? And the conjugats base is HCO3-. In this reaction H2CO3 -> H+ + HCO3- $\endgroup$ – Liz May 10 at 22:12
  • $\begingroup$ I put the image now $\endgroup$ – Liz May 10 at 22:15
  • $\begingroup$ @Liz, The solution in the white box is correct. See the equations I wrote, with negative sign with log, acid will be in numerator. $\endgroup$ – M. Farooq May 10 at 22:30
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Important is the knowledge how $K_\mathrm{a}$ is defined:

$$K_\mathrm{a}=\frac{\ce{[H+][A-]}}{\ce{[HA]}}$$

If we perform logarithmization, we get:

$$\log{K_\mathrm{a}}=\log{\ce{[H+]}} + \log{\frac{\ce{[A-]}}{\ce{[HA]}}}$$

If we apply the operator $\mathrm{p}X = - \log{X}$:

$$\mathrm{p}K_\mathrm{a}=\mathrm{pH} -\log{\frac{\ce{[A-]}} {\ce{[HA]}}}$$

respectively

$$\mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log{\frac{\ce{[A-]}} {\ce{[HA]}}}$$

In our case, $\ce{HA = H2CO3}$ and $\ce{A- = HCO3-}$.

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