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(a)

${\text{CO}}_{\text{2}}{\text{+H}}_{\text{2}}\text{O}鈬�{\text{H}}_{\text{2}}{\text{CO}}_{\text{3}}$

This reaction is an acid-base reaction as they form an adduct Lewis acid-base reaction form adducts, therefore, this is a Lewis acid-base reaction. The lone pair of electron of oxygen in${\text{H}}_2\text{O}$will be donated to the carbon of the${\text{CO}}_2$to form${\text{H}}_2{\text{CO}}_3.$ Therefore,${\text{H}}_2\text{O}$is a Lewis base and${\text{CO}}_2$is a Lewis acid.

${\text{H}}_2{\text{CO}}_3+{\text{H}}_2\text{O}鈬�{\text{HCO}}_3^{-}+{\text{H}}_3{\text{O}}^{+}$

This reaction is an acid-base reaction as they form${\text{H}}_3{\text{O}}^{+}.$The${\text{H}}_2{\text{CO}}_3$donates the${\text{H}}^{\text{+}}{\text{toH}}_{\text{2}}{\text{O to form HCO}}_{\text{3}}^{\text{-}}{\text{+H}}_{\text{2}}\text{O}$ . The proton transfer is governed by the Bronsted Lowry acid-base reaction.

Therefore, ${\text{H}}_2{\text{CO}}_3$is a Bronsted -Lowry acid and the ${\text{H}}_2\text{O}$is a Bronsted-Lowry base.

(b)

According to the Henry's Law,$[{\mathrm{CO}}_2={\mathrm{k}}_{{鈭�}_2}脳{\mathrm{P}}_{{\mathrm{CO}}_2}$

Solve for$\left[{\text{CO}}_2\right]$using the values given in the problem.

$$\begin{gathered} \left[{\mathrm{CO}}_2\right]={\mathrm{k}}_{{鈭�}_2}脳{\mathrm{P}}_{{鈭�}_2} \\ =0.033\frac{\mathrm{mol}}{\mathrm{L}脳\mathrm{atm}}脳3.2脳{10}^{-4}\mathrm{atm} \\ =1.056脳{10}^{-5}\frac{\mathrm{mol}}{\mathrm{L}} \end{gathered}$$

Write the overall reaction.

${\text{CO}}_{\text{2}}{\text{+H}}_{\text{2}}\text{O}鈬�{\text{H}}_{\text{2}}{\text{CO}}_{\text{3}}{\text{H}}_{\text{2}}{\text{CO}}_{\text{3}}{\text{+H}}_{\text{2}}\text{O}鈬�{\text{HCO}}_{\text{3}}^{\text{-}}{\text{+H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{CO}}_{\text{2}}{\text{+ 2H}}_{\text{2}}\text{O \&}鈬�{\text{HCO}}_{\text{3}}^{\text{-}}{\text{+H}}_{\text{3}}{\text{O}}^{\text{+}}$

Write the K expression for the overall reaction.

$\text{K}=\frac{\left[{\text{HCO}}_3^{-}\right]\left[{\text{H}}_3{\text{O}}^{+}\right]}{\left[{\text{CO}}_2\right]}$

Write the ICE table of the reaction so that we can solve for the$\left[{\text{H}}_3{\text{O}}^{+}\right]$.Therefore, the K is:

$$\begin{gathered} \mathrm{k}=\frac{\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]\left[{{\mathrm{HCO}}_3}^{-}\right]}{{\mathrm{COO}}_2} \\ =\frac{{\mathrm{x}}^2}{1.056脳{10}^{-5}-\mathrm{x}} \end{gathered}$$

We know that$\text{x}=\left[{\text{H}}_3{\text{O}}^{+}\right]=\left[{\text{HCO}}_3^{-}\right]$since the reaction produced one mole of each. Solve for.

_{$\mathrm{k}=\frac{{\mathrm{x}}^2}{1.056脳{10}^{-5}-\mathrm{x}}$}

$$\begin{gathered} (1.056脳{10}^{-5}-\mathrm{x})(4.5脳{10}^{-7})={\mathrm{x}}^2 \\ 4.75脳{10}^{-12}-4.5脳{10}^{-7}\mathrm{x}={\mathrm{x}}^2 \\ 2+4.5脳{10}^{-7}\mathrm{x}-4.75脳{10}^{-12}=0 \\ \mathrm{x}=1.97脳{10}^{-6} \end{gathered}$$

$$\begin{gathered} (1.056脳{10}^{-5}-x)(4.5脳{10}^{-7})=x^2 \\ 4.75脳{10}^{-12}-4.5脳{10}^{-7}x=x^2 \\ {x}^2+4.5脳{10}^{-7}x-4.75脳{10}^{-12}=0 \end{gathered}$$

Next, calculate the pH of the solution.

$$\begin{gathered} pH=-\log \left[H_3{O}^{+}\right] \\ =-\log (1.97脳{10}^{-8}) \\ =5.71 \end{gathered}$$

Hence, the pH of the non-polluted rain water in equilibrium with clean air is $pH=5.71$.

(c)

${\text{HCO}}_{\text{3}}^{\text{-}}$will further dissociate in water.

${\text{HCO}}_3^{-}+{\text{H}}_2\text{O}鈬�{\text{CO}}_3^{2-}+{\text{H}}_3{\text{O}}^{+}$

Write the${\text{K}}_{\text{a}}$expression for the overall reaction.

data-custom-editor="chemistry" ${\text{K}}_{\text{a}}\text{=}\frac{\left[{\text{CO}}_{\text{3}}^{\text{2 -}}\right]\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]}{\left[{\text{HCO}}_3\right]}$

Next, solve for data-custom-editor="chemistry" $\left[{\text{CO}}_{\text{3}}^{\text{2}}\right]$the Note that from the previous problem, we know that

data-custom-editor="chemistry" $$\begin{gathered} \left|{{\mathrm{HCO}}^{-}}_3\right|=\left|H_3{O}^{+}\right| \\ =x \\ =1.97脳{10}^{-6}M \end{gathered}$$

Therefore, the ${\text{K}}_{\text{a}}$is: C

data-custom-editor="chemistry" $$\begin{gathered} K_a=\frac{\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]\left[{{\mathrm{CO}}_3}^{2-}\right]}{\left[{{\mathrm{HCO}}_3}^{-}\right]} \\ =\frac{\mathrm{x}(1.97脳{10}^{-8}+\mathrm{x})}{1.97脳{10}^{-6}-\mathrm{x}} \end{gathered}$$

We know that_{data-custom-editor="chemistry" $\text{x}=\left[{\text{H}}_3{\text{O}}^{+}\right]=\left[{\text{CO}}_3^{2-}\right]$}since the reaction produced one mole of each. Since thedata-custom-editor="chemistry" ${\text{K}}_{\text{a}}$is very small, we can drop the added and subtracted x. Then solve for x.

$$\begin{gathered} K_a=\frac{x(1.97脳{10}^{-6})}{1.97脳{10}^{-6}}x \\ =\frac{(4.7脳{10}^{-11})(1.97脳{10}^{-6})}{1.97脳{10}^{-6}}x \\ =4.7脳{10}^{-11} \end{gathered}$$

Therefore, the data-custom-editor="chemistry" $\left[{\text{CO}}_{\text{3}}^{\text{2}}\right]$in rain water isdata-custom-editor="chemistry" $\left[{\text{CO}}_3^{2-}\right]=4.7脳{10}^{-11}\text{M.}$

(d)

Solve for the$\left[{\text{CO}}_2\right]$using the values given in the problem.

$$\begin{gathered} \left[{\mathrm{CO}}_2\right]=k_{c{o}_2}脳2P_{C{O}_2} \\ =0.33\frac{mol}{L脳atm}脳2脳3.2脳{10}^{-4}atm\left[C{O}_2\right] \\ =2.112脳{10}^{-5}\frac{mol}{L} \end{gathered}$$

$\left[{\mathrm{CO}}_2\right]={\mathrm{k}}_{{鈭�}_2}脳2{\mathrm{P}}_{{鈭�}_2}$

Write the overall reaction.

${\text{CO}}_{\text{2}}{\text{+H}}_{\text{2}}\text{O}鈬�{\text{H}}_{\text{2}}{\text{CO}}_{\text{3}}{\text{+H}}_{\text{2}}\text{O\&}鈬�{\text{HCO}}_{\text{3}}{\text{+H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{CO}}_{\text{2}}{\text{+ 2H}}_{\text{2}}\text{O \&}鈬�{\text{HCO}}_{\text{3}}{\text{+H}}_{\text{3}}{\text{O}}^{\text{+}}$

Write the K expression for the overall reaction.

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

Write the ICE table of the reaction so we can solve for $\left[{\text{H}}_3{\text{O}}^{+}\right]$

Therefore, the K is:

$$\begin{gathered} \mathrm{k}=\frac{\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]\left[{{\mathrm{HCO}}^{-}}_3\right]}{\left[{\mathrm{CO}}_2\right\}} \\ =\frac{{\mathrm{x}}^2}{2.112脳{10}^{-5}-\mathrm{x}} \end{gathered}$$

We know thatsince the reaction produced one mole of each. Solve foras shown below.

$$\begin{gathered} K=\frac{x^2}{2.112脳{10}^{-5}-x}(2.112脳{10}^{-5}-x)(4.5脳{10}^{-7}) \\ =x^2脳9.504脳{10}^{-7}x-9.504脳{10}^{-12} \\ =0 \\ x=2.87脳{10}^{-6} \end{gathered}$$

$x^2+4.5脳{10}^{-7}x-9.504脳{10}^{-12}=0$

Next, calculate the pH of the solution.

$$\begin{gathered} pH=-\log \left[H_3{O}^{+}\right] \\ =-\log (2.87脳{10}^{-6}) \\ =5.54 \end{gathered}$$

Hence, the pH of the rainwater when the partial pressure of in clean air doubles in the next few decades is pH= 5.54 .