# Question #ca66b

Jan 29, 2015

The concentration of ${\text{HCO}}_{3}^{-}$ is $\text{0.151 mmol/L}$.

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

$C {O}_{2 \left(g\right)} r i g h t \le f t h a r p \infty n s C {O}_{2 \left(a q\right)}$

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

${P}_{\text{gas") = k_H * C_("gas}}$, where

${P}_{\text{gas}}$ - the partial pressure of the gas;
${k}_{H}$ - Henry's constant - it has a specific value for every gas and it's temperature-dependent;
${C}_{\text{gas}}$ - the concentration of the dissolved gas.

Since your reaction takes place at $\text{37"^@"C}$, 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 $\text{37"^@"C}$ is $\text{40.0 L" * "atm/mol}$.

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

$\left[C {O}_{2}\right] = {P}_{C {O}_{2}} / {k}_{H} = \text{0.005181 atm"/("40.0 L" * "atm/mol") = "0.000130 mol/L}$

Side note: at $\text{25"^@"C}$, the value of ${k}_{H}$ is $\text{29.41 L atm/mol}$ 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

$C {O}_{2 \left(a q\right)} + {H}_{2} {O}_{\left(l\right)} r i g h t \le f t h a r p \infty n s {H}_{\left(a q\right)}^{+} + H C {O}_{3 \left(a q\right)}^{-}$

The expression for the equilibrium constant is

$K = \frac{\left[{H}^{+}\right] \cdot \left[H C {O}_{3}^{-}\right]}{\left[C {O}_{2}\right]} \implies \left[H C {O}_{3}^{-}\right] = \frac{K \cdot \left[C {O}_{2}\right]}{\left[{H}^{+}\right]}$

$\left[H C {O}_{3}^{-}\right] = \frac{0.0000547 \cdot 0.000130}{0.0000471} = \text{0.000151 mol/L}$

$\left[H C {O}_{3}^{-}\right] = \text{0.151 mmol/L}$

Notice that I've converted all the given concentrations into $\text{mol/L}$ for the calculations.