Question

Question #a8f69

Answer 1

Here's what I got.

Explanation:

!! EXTREMELY LONG ANSWER !!

The idea here is that you need to use Henry's Law to determine the concentration of carbon dioxide, `"CO"_2`, in the blood if its partial pressure is equal to `"33 mmHg"`.

Once you know the concentration of carbon dioxide, you can use the equilibrium that exists between dissolved `"CO"_2` and water on one side, and carbonic acid, `"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, `"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_"gas"` - the partial pressure of the gas above the liquid
`c_"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. `~~36.8^@"C"`, but I was able to find a value for `k_H` at room temperature

![https://chemengineering.wikispaces.com/Henry's+Law]()

In order to get the value of `K_H` at `36.8^@"C"`, I used the equation

`k_H = k_H^@ * "exp" [-2400 * (1/T - 1/T^@)]" "`, where

`k_H^@` - Henry's constant at `"298 K"`
`T` - the equivalent in Kelvin for `36.8^@"C"`, i.e. `"309.95 K"`
`T^@` - the reference temperature, i.e. `"298 K"`

Plug in these values to get

`k_H = "40.12 atm L mol"^(-1)`

Now use equation `color(purple)((1))` 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_(CO_2)/c_(CO_2) implies c_(CO_2) = P_(CO_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

`"CO"_text(2(aq]) + "H"_2"O"_text((l]) rightleftharpoons "H"_2"CO"_text(3(aq])`

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 `"CO"_2`.

`["H"_2"CO"_3] = ["CO"_2]`

Next, look at the pH of the solution. Notice that its higher than the `pK_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

`["HCO"_3^(-)] = color(green)("0.027 M")`