Q18.149 CP - Chemistry: Molecular Nature Of Matter And Change

Question

When carbon dioxide dissolves in water, it undergoes a multistep equilibrium process, with $K_{overall} = 4.5\times {10}^{-7}$, which is simplified to the following:

$$\begin{gathered} {\text{CO}}_{\text{2}}{\text{(g) + H}}_{\text{2}}\text{O(l)}\rightleftharpoons {\text{H}}_{\text{2}}{\text{CO}}_{\text{3}}\text{(aq)} \\ {\text{H}}_{\text{2}}{\text{CO}}_{\text{3}}{\text{(aq) + H}}_{\text{2}}\text{O(l)}\rightleftharpoons {\text{HCO}}_{\text{3}}^{\text{ - }}{\text{(aq) + H}}_{\text{3}}{\text{O}}^{\text{ + }}\text{(aq)} \end{gathered}$$

(a) Classify each step as a Lewis or a Brønsted -Lowry reaction. 

(b) What is the pH of non-polluted rainwater in equilibrium with clean air (${\text{P}}_{{\text{CO}}_{\text{2}}}$in clean air $=3.2\times {10}^{-4}$atm; Henry’s law constant for  ${\text{CO}}_{\text{2}}$at 25C is $0.33$mol/L atm)?

(c) What is data-custom-editor="chemistry" $\left[{\text{CO}}_{\text{3}}^{\text{2}}\right]$ in rainwater ($K_0 of HC{O_3}^{-}=4.7\times {10}^{-11}$)? 

(d) If the partial pressure of  data-custom-editor="chemistry" ${\mathrm{CO}}_2$ in clean air doubles in the next few decades, what will the pH of rainwater become?

Step-by-Step Solution

Answer

(a) ${\text{H}}_2{\text{CO}}_3$ is the Bronsted-Lowry acid and  is the Bronsted-Lowry base.

(b) The pH of non polluted rain water in equilibrium with clean air is $\text{pH = } \text{5.71.}$

(c) The $\left[{\text{CO}}_{\text{3}}^{\text{2}}\right]$ in rain water is $\left[{\text{CO}}_3^{2-}\right]=4.7\times {10}^{-11}\text{M.}$

(d) The pH of rainwater if the partial pressure of ${\mathrm{CO}}_2$ in clean air doubles in the next few decades is $\text{pH = 5.54}$

Step 1: Classify each step as a Lewis or a Bronsted-Lowry reaction

(a)

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

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.

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

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

Step 2: Finding the pH of non polluted rainwater

(b)

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

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

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

Write the overall reaction.

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

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:

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

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 .

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

 

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

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

Next, calculate the pH of the solution.

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

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

Step 3: To find in rainwater

(c)

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

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

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

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

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

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 the data-custom-editor="chemistry" ${\text{K}}_{\text{a}}$ is very small, we can drop the added and subtracted x. Then solve for x.

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

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

x = H 3 O + = HCO 3 - Step 4: To find the pH of rainwater

(d)

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

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

$\left[{\mathrm{CO}}_2\right]={\mathrm{k}}_{{\infty }_2}\times 2{\mathrm{P}}_{{\infty }_2}$

Write the overall reaction.

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

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:

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

We know that  since the reaction produced one mole of each. Solve for  as shown below.

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

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

Next, calculate the pH of the solution.

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

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