Question

The Henry’s law constant of CO$_2$ is $3.4 \times 10^{-2}$ M/atm at 25$^\circ$C.
Dissolved CO$_2$ undergoes:
CO$_2 \cdot$H$_2$O $\rightleftharpoons$ H$^+$ + HCO$_3^-$ ($K_1 = 4.3 \times 10^{-7}$ M)
HCO$_3^- \rightleftharpoons$ H$^+$ + CO$_3^{2-}$ ($K_2 = 4.7 \times 10^{-11}$ M)
If ambient CO$_2$ concentration is 300 ppm, the total dissolved CO$_2$ (in $\mu$M, rounded off to one decimal place) is __________.

Hint:

Total dissolved CO$_2$ = molecular CO$_2$ + bicarbonate + carbonate; bicarbonate dominates at neutral pH.

Answer 1

Correct Answer: 54

Ambient CO$_2$ partial pressure:
\[ P_{CO_2} = 300\ \text{ppm} = 300 \times 10^{-6}\ \text{atm} = 3 \times 10^{-4}\ \text{atm}. \]
Dissolved molecular CO$_2$:
\[ [\text{CO}_2(\text{aq})] = H\, P_{CO_2} = (3.4 \times 10^{-2})(3 \times 10^{-4}) = 1.02 \times 10^{-5}\ \text{M}. \]
At neutral pH, \([H^+] = 10^{-7}\) M.
First dissociation:
\[ \frac{[\text{HCO}_3^-]}{[\text{CO}_2]} = \frac{K_1}{[H^+]} = \frac{4.3\times 10^{-7}}{10^{-7}} = 4.3. \]
Second dissociation is negligible (very tiny):
\[ \frac{[\text{CO}_3^{2-}]}{[\text{HCO}_3^-]} = \frac{K_2}{[H^+]} = 4.7\times 10^{-4}. \]
Thus total dissolved inorganic carbon:
\[ [\text{Total}] = [\text{CO}_2] (1 + 4.3 + 4.3 \times 4.7\times 10^{-4}). \]
Compute:
\[ 1 + 4.3 + 0.002 = 5.302. \]
\[ [\text{Total}] = 1.02\times10^{-5} \times 5.302 = 5.41\times 10^{-5}\,\text{M}. \]
Convert to $\mu$M:
\[ 5.41\times10^{-5}\,\text{M} = 54.1\ \mu\text{M}. \]
Thus the answer is:
\[ \boxed{54.1\ \mu\text{M}} \quad (\text{acceptable range: } 54.0\text{–}54.2) \]