Question #ca66b

1 Answer
Jan 29, 2015

The concentration of "HCO"_3^(-) is "0.151 mmol/L".

The first important aspect about the problem is the fact that you have another reaction that takes places prior to the one given

CO_(2(g)) rightleftharpoons CO_(2(aq))

Carbon dioxide gas is dissolved in water, which means that you can determine its concentration by using Henry's law

P_("gas") = k_H * C_("gas"), where

P_("gas") - the partial pressure of the gas;
k_H - Henry's constant - it has a specific value for every gas and it's temperature-dependent;
C_("gas") - the concentration of the dissolved gas.

Since your reaction takes place at "37"^@"C", you must calculate the value of k_H using a form of the van't Hoff equation (more here: https://chemengineering.wikispaces.com/Henry%27s+Law). I won't detail the calculations because I don't want the answer to become too long. The value you get for k_H at "37"^@"C" is "40.0 L" * "atm/mol".

This means that the concentration of the dissolved carbon dioxide will be (don't forget to convert mbar to atm)

[CO_2] = P_(CO_2)/k_H = "0.005181 atm"/("40.0 L" * "atm/mol") = "0.000130 mol/L"

Side note: at "25"^@"C", the value of k_H is "29.41 L atm/mol" for carbon dioxide; notice that an increase in temperature reduced the concentration of dissolved carbon dioxide - this is consistent with the idea that gas solubility drops with an increase in temperature.

You now know that you have

CO_(2(aq)) + H_2O_((l)) rightleftharpoons H_((aq))^(+) + HCO_(3(aq))^(-)

The expression for the equilibrium constant is

K = [[H^(+)] * [HCO_3^(-)])/([CO_2]) => [HCO_3^(-)] = (K * [CO_2])/([H^(+)])

[HCO_3^(-)] = (0.0000547 * 0.000130)/(0.0000471) = "0.000151 mol/L"

[HCO_3^(-)] = "0.151 mmol/L"

Notice that I've converted all the given concentrations into "mol/L" for the calculations.