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

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:

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

Consider the reaction

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

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.${\mathrm{K}}_{\mathrm{a}1}=4.40脳{10}^{-7}$

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

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

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:

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