Heterogeneous equilibrium.${\mathbf{CO}}_{\mathbf{2}}$dissolves in water to give "carbonic acid" (which is mostly dissolved${\mathbf{CO}}_{\mathbf{2}}$, as described in Box 6-4).

${\mathbf{CO}}_{\mathbf{2}}\left(g\right)\mathbf{𝆏}\mathbf{}{\mathbf{CO}}_{\mathbf{2}}\left(\mathrm{aq}\right)\mathbf{}\mathbf{K}\mathbf{=}{\mathbf{10}}^{\mathbf{-}\mathbf{1}\mathbf{.}\mathbf{5}}$

(The equilibrium constant is called the Henry's law constant for carbon dioxide, because Henry's law states that the solubility of a gas in a liquid is proportional to the pressure of the gas.) The acid dissociation constants listed for "carbonic acid" in Appendix G apply to${\mathbf{CO}}_{\mathbf{2}}\left(\mathrm{aq}\right)$. Given that${\mathbf{P}}_{{\mathbf{CO}}_{\mathbf{2}}}$in the atmosphere is ${\mathbf{10}}^{\mathbf{-}\mathbf{3}\mathbf{.}\mathbf{4}}$atm, find the $\mathbf{pH}$of water in equilibrium with the atmosphere.

Heterogeneous equilibrium$C{O}_2$carbonic acid is produced when it dissolves in water :$$\begin{gathered} \left(\mathrm{K}={10}^{-1.5}\right) \\ {\mathrm{CO}}_2\left(\mathrm{g}\right)𝆏{\mathrm{CO}}_2\left(\mathrm{aq}\right) \end{gathered}$$

Because of this,$\mathrm{p}\left({\mathrm{CO}}_2\right)$there is in the atmosphererole="math" localid="1654939333790" ${10}^{-3.4}\mathrm{atm}$, the pH of water will be found to be in equilibrium with the atmosphere.

It'll start by looking for the:$\#c34632;>$Carbonic acid is formalised by the following formula:

$$\begin{gathered} \left[{\mathrm{H}}_2{\mathrm{CO}}_3\right]={\mathrm{CO}}_2\left(\mathrm{aq}\right)-\mathrm{Kp}\left({\mathrm{CO}}_2\right) \\ \left[{\mathrm{H}}_2{\mathrm{CO}}_3\right]={10}^{-15}\times {10}^{-3.4} \\ ={10}^{-4.9}\mathrm{M} \end{gathered}$$

Consider the reaction

${\mathrm{H}}_2{\mathrm{CO}}_3𝆏{\mathrm{H}}_2{\mathrm{CO}}_3^{-}+{\mathrm{H}}^{+}$

Then you should write something like this:

$$\begin{gathered} \left[{\mathrm{H}}_2{\mathrm{CO}}_3\right]={10}^{-4.9}-\mathrm{x} \\ \left[{\mathrm{H}}_2{\mathrm{CO}}_3\right]=\mathrm{x} \\ \left[{\mathrm{H}}^{+}\right]=\mathrm{x} \end{gathered}$$

It has a value of for carbonic acid.${\mathrm{K}}_{\mathrm{a}1}=4.40\times {10}^{-7}$

Calculate the value of x using the values from the previous step:

$$\begin{gathered} {\mathrm{K}}_{\mathrm{a}1}=\frac{{\mathrm{x}}^2}{{10}^{-4.9}-\mathrm{x}} \\ 4.46\times {10}^{-7}=\frac{{\mathrm{x}}^2}{{10}^{-4.9}-\mathrm{x}} \\ \mathrm{x}=2.16\times {10}^{-6}\mathrm{M} \end{gathered}$$

X is thought to have a value of $\mathrm{x}=\left[{\mathrm{H}}^{+}\right]$so that we can figure out what a solution's pH is:

$$\begin{gathered} \mathrm{pH}=-\log \left[{\mathrm{H}}^{+}\right] \\ \mathrm{pH}=-\log \left(2.16\times {10}^{-6}\right) \\ =5.67 \end{gathered}$$