# Question #a8f69

##### 1 Answer

#### 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,

Once you know the concentration of carbon dioxide, you can use the equilibrium that exists between dissolved *carbonic acid*,

Finally, you need to use the **Henderson - Hasselbalch equation** to find the concentration of the *bicarbonate anion*,

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**,

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

I wasn't able to find the value of

In order to get the value of

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

Plug in these values to get

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

Now use equation **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 **equal** to that of dissolved

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

Next, look at the pH of the solution. Notice that its **higher** than the **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")#