## Related questions with answers

The Henry's law constant for $CO_2$ in water at $25 ^ { \circ } \mathrm { C }$ is $3.1 \times 10 ^ { - 2 } M \mathrm { atm } ^ { - 1 }$ Assume that all of this $CO_2$ is in the form of $H_2CO_3$ produced by the reaction between $CO_2$ and $H_2O:$ $\mathrm { CO } _ { 2 } ( a q ) + \mathrm { H } _ { 2 } \mathrm { O } ( l ) \longrightarrow \mathrm { H } _ { 2 } \mathrm { CO } _ { 3 } ( a q )$ What is the pH of this solution?

Solution

VerifiedHere, we have the reaction:

$\mathrm{CO}_{2}(aq)+\mathrm{H}_{2} \mathrm{O}(l) \rightarrow \mathrm{H}_{2} \mathrm{CO}_{3}(aq)$

$\left[\mathrm{CO}_{2}\right]=\left[\mathrm{H}_{2} \mathrm{CO}_{3}\right]=1.16 \cdot 10^{-5} \mathrm{M}$ - these concentrations are the same because all of $\mathrm{CO}_{2}$ is in the form of $\mathrm{H}_{2} \mathrm{CO}_{3}$

The dissociation of $\mathrm{H}_{2} \mathrm{CO}_{3}$:

$\mathrm{H}_{2} \mathrm{CO}_{3}(aq) \rightleftharpoons \mathrm{H}^{+}(aq)+\mathrm{HCO}_{3}^{-}(aq)$

$\mathrm{K}_{a1}=4.3 \cdot 10^{-7}$

It is expressed as:

$\mathrm{K}_{a1}=\frac{\left[\mathrm{H}^{+}\right]\left[\mathrm{HCO}_{3}^{-}\right]}{\left[\mathrm{H}_{2} \mathrm{CO}_{3}\right]}$

Equilibrium concentrations:

$\mathrm{H}_{2} \mathrm{CO}_{3}=1.16 \cdot 10^{-5}-x$

$\mathrm{H}^{+}=x$

$\mathrm{HCO}_{3}^{-}=x$

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