# Question a8f69

Dec 10, 2015

Here's what I got.

#### Explanation:

The idea here is that you need to use Henry's Law to determine the concentration of carbon dioxide, ${\text{CO}}_{2}$, in the blood if its partial pressure is equal to $\text{33 mmHg}$.

Once you know the concentration of carbon dioxide, you can use the equilibrium that exists between dissolved ${\text{CO}}_{2}$ and water on one side, and carbonic acid, ${\text{H"_2"CO}}_{3}$, on the other, to determine the concentration of carbonic acid in the blood.

Finally, you need to use the Henderson - Hasselbalch equation to find the concentration of the bicarbonate anion, ${\text{HCO}}_{3}^{-}$, the conjugate base of carbonic acid.

Now, Henry's Law tells you that gases dissolve in a liquid proportionally to the equilibrium that exists between the undissolved and the dissolved gas.

The proportionality constant for this equilibrium is called Henry's constant, ${k}_{H}$, and is defined as

color(blue)(k_H = P_"gas"/c_"gas")" " " "color(purple)((1))" ", where

${P}_{\text{gas}}$ - the partial pressure of the gas above the liquid
${c}_{\text{gas}}$ - the concentration of the gas dissolved in the liquid

I wasn't able to find the value of ${k}_{H}$ for carbon dioxide at the temperature of the human body, i.e. $\approx {36.8}^{\circ} \text{C}$, but I was able to find a value for ${k}_{H}$ at room temperature

In order to get the value of ${K}_{H}$ at ${36.8}^{\circ} \text{C}$, I used the equation

${k}_{H} = {k}_{H}^{\circ} \cdot \text{exp" [-2400 * (1/T - 1/T^@)]" }$, where

${k}_{H}^{\circ}$ - Henry's constant at $\text{298 K}$
$T$ - the equivalent in Kelvin for ${36.8}^{\circ} \text{C}$, i.e. $\text{309.95 K}$
${T}^{\circ}$ - the reference temperature, i.e. $\text{298 K}$

Plug in these values to get

${k}_{H} = {\text{40.12 atm L mol}}^{- 1}$

Now use equation $\textcolor{p u r p \le}{\left(1\right)}$ to find the concentration of carbon dioxide in the blood at this temperature - do not forget to convert the partial pressure from mmHg to atm

${k}_{H} = {P}_{C {O}_{2}} / {c}_{C {O}_{2}} \implies {c}_{C {O}_{2}} = {P}_{C {O}_{2}} / {k}_{H}$

c_(CO_2) = (33/760 color(red)(cancel(color(black)("atm"))))/(40.12color(red)(cancel(color(black)("atm")))" L mol"^(-1)) = "0.0010823 M"

Now you're ready to focus on the solution. You can think of carbonic acid as being in equilibrium with water and dissolved carbon dioxide

${\text{CO"_text(2(aq]) + "H"_2"O"_text((l]) rightleftharpoons "H"_2"CO}}_{\textrm{3 \left(a q\right]}}$

Notice that you have a $1 : 1$ mole ratio between carbon dioxide and carbonic acid. This tells you that the concentration of carbonic acid will be equal to that of dissolved ${\text{CO}}_{2}$.

$\left[{\text{H"_2"CO"_3] = ["CO}}_{2}\right]$

Next, look at the pH of the solution. Notice that its higher than the $p {K}_{a}$ of the acid. This should automatically tell you that you have more conjugate base than weak acid present in solution.

The equilibriu mreaction established in solution looks like this

overbrace("CO"_text(2(aq]) + "H"_2"O"_text((l]))^(color(red)("H"_2"CO"_3 - "acid")) + "H"_2"O"_text((l]) rightleftharpoons overbrace("HCO"_text(3(aq])^(-))^(color(green)("conjugate base")) + "H"_3"O"_text((aq])^(+)

The Henderson - Hasselbalch equation looks like this

color(blue)("pH" = pK_a + log ( (["conjugate base"])/(["weak acid"])))

In your case, the weak acid is carbonic acid and the conjugate base is the bicarbonate anion. Use this equation to find the concentration of the latter

"pH" = pK_a + log( (["HCO"_3^(-)])/(["H"_2"CO"_3]))

7.5 = 6.1 + log( (["HCO"_3^(-)])/"0.0010823 M")

This is equivalent to

10^1.4 = 10^log( (["HCO"_3^(-)])/"0.0010823 M")

10^1.4 = (["HCO"_3^(-)])/"0.0010823 M"

Therefore,

["HCO"_3^(-)] = 10^1.4 * "0.0010823 M" = "0.02719 M"#

Rounded to two sig figs, the answer will be

$\left[\text{HCO"_3^(-)] = color(green)("0.027 M}\right)$