Answers: 1
:![\displaystyle K = \frac{\left(a_{\mathrm{H_2CO_3\, (aq)}}\right)}{\left(a_{\mathrm{H^{+}}}\right)^2\, \left(a_{\mathrm{{CO_3}^{2-}\, (aq)}}\right)} \approx \frac{[\mathrm{H_2CO_3}]}{\left[\mathrm{H^{+}\, (aq)}\right]^{2} \, \left[\mathrm{CO_3}^{2-}\right]}](/tpl/images/0726/6258/84dbc.png)
.
, 
, and 
denote the activities of the three species, and ![[\mathrm{H_2CO_3}]](/tpl/images/0726/6258/0f634.png)
, ![\left[\mathrm{H^{+}}\right]](/tpl/images/0726/6258/81ca8.png)
, and ![\left[\mathrm{CO_3}^{2-}\right]](/tpl/images/0726/6258/1667f.png)
denote the concentrations of the three species.
is the only product of this reaction. Besides, its coefficient in the balanced reaction is one. Therefore, the numerator would simply be 
.
" on the product side of this reaction. 
is equivalent to 
. The species 
appeared twice among the reactants. Therefore, its activity should also appear twice in the denominator:
.![\displaystyle K = \frac{\left(a_{\mathrm{H_2CO_3\, (aq)}}\right)}{\left(a_{\mathrm{H^{+}}}\right)^2\, \left(a_{\mathrm{{CO_3}^{2-}\, (aq)}}\right)} \quad\begin{matrix}\leftarrow \text{from products} \\[0.5em] \leftarrow \text{from reactants}\end{matrix}](/tpl/images/0726/6258/9745b.png)
.
" species. Note that all the three species here are indeed aqueous. Hence, this equilibrium constant expression can be approximated as:![\displaystyle K = \frac{\left(a_{\mathrm{H_2CO_3\, (aq)}}\right)}{\left(a_{\mathrm{H^{+}}}\right)^2\, \left(a_{\mathrm{{CO_3}^{2-}\, (aq)}}\right)} \approx \frac{\left[\mathrm{H_2CO_3\, (aq)}\right]}{\left[\mathrm{H^{+}\, (aq)}\right]^2\cdot \left[\mathrm{{CO_3}^{2-}\, (aq)}\right]}](/tpl/images/0726/6258/9131e.png)
.
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Write the equilibrium constant expression for this reaction: 2H+(aq)+CO−23(aq) → H2CO3(aq)...
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