Question #a8f69

1 Answer
Dec 10, 2015

Answer:

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")#