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Question

The Henry's law constant for CO2CO_2 in water at 25C25 ^ { \circ } \mathrm { C } is 3.1×102Matm13.1 \times 10 ^ { - 2 } M \mathrm { atm } ^ { - 1 } Assume that all of this CO2CO_2 is in the form of H2CO3H_2CO_3 produced by the reaction between CO2CO_2 and H2O:H_2O: CO2(aq)+H2O(l)H2CO3(aq)\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?

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Here, we have the reaction:

CO2(aq)+H2O(l)H2CO3(aq)\mathrm{CO}_{2}(aq)+\mathrm{H}_{2} \mathrm{O}(l) \rightarrow \mathrm{H}_{2} \mathrm{CO}_{3}(aq)

[CO2]=[H2CO3]=1.16105M\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 CO2\mathrm{CO}_{2} is in the form of H2CO3\mathrm{H}_{2} \mathrm{CO}_{3}

The dissociation of H2CO3\mathrm{H}_{2} \mathrm{CO}_{3}:

H2CO3(aq)H+(aq)+HCO3(aq)\mathrm{H}_{2} \mathrm{CO}_{3}(aq) \rightleftharpoons \mathrm{H}^{+}(aq)+\mathrm{HCO}_{3}^{-}(aq)

Ka1=4.3107\mathrm{K}_{a1}=4.3 \cdot 10^{-7}

It is expressed as:

Ka1=[H+][HCO3][H2CO3]\mathrm{K}_{a1}=\frac{\left[\mathrm{H}^{+}\right]\left[\mathrm{HCO}_{3}^{-}\right]}{\left[\mathrm{H}_{2} \mathrm{CO}_{3}\right]}

Equilibrium concentrations:

H2CO3=1.16105x\mathrm{H}_{2} \mathrm{CO}_{3}=1.16 \cdot 10^{-5}-x

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

HCO3=x\mathrm{HCO}_{3}^{-}=x

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