Chapter 18, Problem 18.149P

(a)

Interpretation Introduction

Interpretation:

Carbon dioxide when dissolved in water undergoes multistep equilibrium process.  The reactions are,

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

Each step has to be classified as a Lewis or a Bronsted-Lowry reaction.

Concept Introduction:

Bronsted-Lowry reaction:

Any species that has the capability of accepting a proton, which requires a lone pair of electrons to bond to ${\text{H}}^{\text{+}}$ is said to be Bronsted-Lowry base.

Any species that has the capability of donating a proton (${\text{H}}^{\text{+}}$) is said to be Bronsted-Lowry acid.

According to Bronsted-Lowry theory, an acid-base reaction in which a proton is transferred from an acid to base is said to be Bronsted-Lowry reaction.

Example:

  ${\text{NH}}_{\text{3(g)}}{\text{+ HCl}}_{\text{(g)}}\to {\text{ NH}}_{\text{4}}{\text{Cl}}_{\text{(s)}}$

In the above reaction, $\text{HCl}$ acts as Bronsted-Lowry acid and donates proton which is accepted by ${\text{NH}}_{\text{3}}$ using its lone pair.  So, ${\text{NH}}_{\text{3}}$ is Bronsted-Lowry base.

Lewis reaction:

In an acid-base reaction, Lewis base donates electrons to the acid and Lewis acid accepts the electron pair to form a covalent bond between Lewis acid and Lewis base.

Example:

  $\text{A + B }\to \text{ A-B}$

In the above example, A is an acid and B is a base.  B donates electrons to A and forms an adduct.

Explanation of Solution

The given reactions are

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

Lewis reaction:

  ${\text{CO}}_{\text{2(g)}}{\text{ + H}}_{\text{2}}{\text{O}}_{\text{(l)}}\rightleftharpoons {\text{H}}_{\text{2}}{\text{CO}}_{\text{3(aq)}}$

In the formation of carbonic acid, ${\text{CO}}_{\text{2}}$ is the Lewis acid and ${\text{H}}_{\text{2}}\text{O}$ is Lewis base.  ${\text{H}}_{\text{2}}\text{O}$ donates electrons to ${\text{CO}}_{\text{2}}$ to form ${\text{H}}_{\text{2}}{\text{CO}}_{\text{3}}$.

Bronsted-Lowry reaction:

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

The above reaction is Bronsted-Lowry reaction and also Lewis reaction.

(b)

Interpretation Introduction

Interpretation:

The $\text{pH}$ value of non-polluted rainwater in equilibrium with clean air has to be calculated.  (${\text{P}}_{{\text{CO}}_{\text{2}}}$ in clean air = ${\text{4×10}}^{\text{-4}}\text{ atm}$; Henry’s law constant for ${\text{CO}}_{\text{2}}$ at ${\text{25}}^{\text{°}}\text{C}$ is $\text{0}\text{.033 mol/L}\text{.atm}$ and ${\text{K}}_{\text{overall}}\text{ = 4}{\text{.5×10}}^{\text{-7}}$).

Explanation of Solution

The reaction can be given as

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

The molarity of ${\text{CO}}_{\text{2}}$ can be calculated as

  $\begin{array}{l}{\text{Molarity of CO}}_{\text{2}}{\text{ = k}}_{\text{H}}{\text{P}}_{{\text{CO}}_{\text{2}}}\\ \\ {\text{Molarity of CO}}_{\text{2}}\text{ = (0}\text{.033 mol/L}{\text{.atm)(4×10}}^{\text{-4}}\text{atm)}\\ \\ \text{ = 1}{\text{.320×10}}^{\text{-5}}{\text{ M CO}}_{\text{2}}\end{array}$

Given: ${\text{K}}_{\text{overall}}\text{ = 4}{\text{.5×10}}^{\text{-7}}$,

  $\begin{array}{l}{\text{K}}_{\text{overall}}\text{ = }\frac{{\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{][HCO}}_{\text{3}}{}^{\text{-}}\text{]}}{{\text{[CO}}_{\text{2}}\text{]}}\\ \\ \text{4}{\text{.5×10}}^{\text{-7}}\text{ = }\frac{\text{[x][x]}}{\text{[1}{\text{.320×10}}^{\text{-5}}\text{-x]}}\end{array}$

Assume $\text{x}$ is small when compared to $\text{1}{\text{.320×10}}^{\text{-5}}$.

  $\begin{array}{l}\text{4}{\text{.5×10}}^{\text{-7}}\text{ = }\frac{\text{[x][x]}}{\text{[1}{\text{.320×10}}^{\text{-5}}\text{]}}\\ \\ \text{x = 2}{\text{.4372×10}}^{\text{-6}}\end{array}$

The value of $\text{x}$ calculated has to be compared to $\text{1}{\text{.320×10}}^{\text{-5}}$.

  $\frac{\text{2}{\text{.4372×10}}^{\text{-6}}}{\text{1}{\text{.320×10}}^{\text{-5}}}\text{ (100) = 18\% error}$

As the calculated error is more than $\text{5\%}$, the assumption of ‘$\text{x}$’ value smaller than the value of $\text{1}{\text{.320×10}}^{\text{-5}}$ is not acceptable.

Quadratic equation is necessary to use to calculate $\text{x}$.

  $\begin{array}{l}{\text{x}}^{\text{2}}\text{ = (4}{\text{.5×10}}^{\text{-7}}\text{)(1}{\text{.320×10}}^{\text{-5}}\text{-x)}\\ \\ {\text{x}}^{\text{2}}\text{ = 5}{\text{.940×10}}^{\text{-12}}\text{- 4}{\text{.5×10}}^{\text{-7}}\text{x}\\ \\ {\text{x}}^{\text{2}}\text{ + 4}{\text{.5×10}}^{\text{-7}}\text{x - 5}{\text{.940×10}}^{\text{-12}}\text{ = 0}\end{array}$

Consider $\text{a = 1, b = 4}{\text{.5×10}}^{\text{-7}}\text{, c = -5}{\text{.940×10}}^{\text{-12}}$.

  $\begin{array}{l}\text{x = }\frac{\text{-b ± }\sqrt{{\text{b}}^{\text{2}}\text{- 4ac}}}{\text{2a}}\\ \\ \text{x = }\frac{\text{-4}{\text{.5×10}}^{\text{-7}}\text{ ± }\sqrt{{\text{(4}{\text{.5×10}}^{\text{-7}}\text{)}}^{\text{2}}\text{- 4(1)(-5}{\text{.940×10}}^{\text{-12}}\text{)}}}{\text{2(1)}}\\ \\ \text{x = 2}{\text{.222575 × 10}}^{\text{-6}}{\text{ M H}}_{\text{3}}{\text{O}}^{\text{+}}\end{array}$

The $\text{pH}$ can be calculated as

  $\begin{array}{l}{\text{pH = -log[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}\\ \\ \text{pH = -log (2}{\text{.222575 × 10}}^{\text{-6}}\text{) }\\ \\ \text{pH = 5}\text{.6531}\end{array}$

The $\text{pH}$ of non-polluted rainwater in equilibrium with clean air is $\text{5}\text{.6}$.

(c)

Interpretation Introduction

Interpretation:

The ${\text{[CO}}_{\text{3}}^{\text{2-}}\text{]}$ in rain water has to be calculated.  (${\text{K}}_{\text{a}}{\text{ of HCO}}_{\text{3}}^{\text{-}}\text{ = 4}{\text{.7 × 10}}^{\text{-11}}$).

Explanation of Solution

The reaction can be given as

  ${\text{HCO}}_{\text{3}}^{\text{-}}{}_{\text{(aq)}}{\text{+ H}}_{\text{2}}{\text{O}}_{\text{(l)}}\rightleftarrows {\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\text{(aq)}}{\text{+ CO}}_{\text{3}}^{\text{2-}}{}_{\text{(aq)}}$

Given, ${\text{K}}_{\text{a}}{\text{ of HCO}}_{\text{3}}^{\text{-}}\text{ = 4}{\text{.7 × 10}}^{\text{-11}}$.

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

Using the $\text{x}$ value calculated in sub-part (b), $\text{x = 2}{\text{.222575 × 10}}^{\text{-6}}{\text{ M H}}_{\text{3}}{\text{O}}^{\text{+}}$,

  $\begin{array}{l}\text{4}{\text{.7 × 10}}^{\text{-11}}\text{ = }\frac{\text{[2}{\text{.222575 × 10}}^{\text{-6}}\text{+ x][x]}}{\text{[2}{\text{.222575 × 10}}^{\text{-6}}\text{- x]}}\\ \\ {\text{assume x is small compared to 2×10}}^{\text{-6}}\text{,}\\ \\ \text{4}{\text{.7 × 10}}^{\text{-11}}\text{ = }\frac{\text{[2}{\text{.222575 × 10}}^{\text{-6}}\text{][x]}}{\text{[2}{\text{.222575 × 10}}^{\text{-6}}\text{]}}\\ \\ {\text{[CO}}_{\text{3}}^{\text{2-}}\text{] = x = 4}{\text{.7×10}}^{\text{-11}}{\text{ = 5×10}}^{\text{-11}}{\text{ M CO}}_{\text{3}}^{\text{2-}}\\ \end{array}$

The value of $\text{x}$ is compared to ${\text{2×10}}^{\text{-6}}$.

  $\frac{\text{4}{\text{.7×10}}^{\text{-11}}}{\text{2}{\text{.222575×10}}^{\text{-6}}}\text{ (100) = 0}\text{.0021\% error}$

As the calculated error is less than $\text{5\%}$ ; the assumption made is valid.

The ${\text{[CO}}_{\text{3}}^{\text{2-}}\text{]}$ in rain water is calculated as ${\text{5×10}}^{\text{-11}}$.

(d)

Interpretation Introduction

Interpretation:

If the partial pressure of ${\text{CO}}_{\text{2}}$ in clean air doubles in the next few decades, the $\text{pH}$ of rainwater has to be calculated.

Explanation of Solution

Given, ${\text{P}}_{{\text{CO}}_{\text{2}}}$ in clean air = ${\text{4×10}}^{\text{-4}}\text{ atm}$; Henry’s law constant for ${\text{CO}}_{\text{2}}$ at ${\text{25}}^{\text{°}}\text{C}$ is $\text{0}\text{.033 mol/L}\text{.atm}$ and ${\text{K}}_{\text{overall}}\text{ = 4}{\text{.5×10}}^{\text{-7}}$.

The value of ${\text{P}}_{{\text{CO}}_{\text{2}}}$ is doubled.  The molarity of ${\text{CO}}_{\text{2}}$ can be calculated as

  $\begin{array}{l}{\text{Molarity of CO}}_{\text{2}}{\text{ = 2k}}_{\text{H}}{\text{P}}_{{\text{CO}}_{\text{2}}}\\ \\ {\text{Molarity of CO}}_{\text{2}}\text{ = 2(0}\text{.033 mol/L}{\text{.atm)(4×10}}^{\text{-4}}\text{atm)}\\ \\ \text{ = 2}{\text{.640 ×10}}^{\text{-5}}{\text{ M CO}}_{\text{2}}\end{array}$

Given: ${\text{K}}_{\text{overall}}\text{ = 4}{\text{.5×10}}^{\text{-7}}$,

  $\begin{array}{l}{\text{K}}_{\text{overall}}\text{ = }\frac{{\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{][HCO}}_{\text{3}}{}^{\text{-}}\text{]}}{{\text{[CO}}_{\text{2}}\text{]}}\\ \\ \text{4}{\text{.5×10}}^{\text{-7}}\text{ = }\frac{\text{[x][x]}}{\text{[2}{\text{.640×10}}^{\text{-5}}\text{-x]}}\end{array}$

Assume $\text{x}$ is small when compared to $\text{2}{\text{.640×10}}^{\text{-5}}$.

  $\begin{array}{l}\text{4}{\text{.5×10}}^{\text{-7}}\text{ = }\frac{\text{[x][x]}}{\text{[2}{\text{.640×10}}^{\text{-5}}\text{]}}\\ \\ \text{x = 3}{\text{.44674×10}}^{\text{-6}}\end{array}$

The value of $\text{x}$ calculated has to be compared to $\text{2}{\text{.640×10}}^{\text{-5}}$.

  $\frac{\text{3}{\text{.44674×10}}^{\text{-6}}}{\text{2}{\text{.640×10}}^{\text{-5}}}\text{ (100) = 13\% error}$

As the calculated error is more than $\text{5\%}$, the assumption ‘$\text{x}$’ smaller than the value of $\text{2}{\text{.640×10}}^{\text{-5}}$ is not acceptable.

Quadratic equation is necessary to use to calculate $\text{x}$.

  $\begin{array}{l}{\text{x}}^{\text{2}}\text{ = (4}{\text{.5×10}}^{\text{-7}}\text{)(2}{\text{.640×10}}^{\text{-5}}\text{-x)}\\ \\ {\text{x}}^{\text{2}}\text{ = 1}{\text{.1880×10}}^{\text{-11}}\text{- 4}{\text{.5×10}}^{\text{-7}}\text{x}\\ \\ {\text{x}}^{\text{2}}\text{ + 4}{\text{.5×10}}^{\text{-7}}\text{x - 1}{\text{.1880×10}}^{\text{-11}}\text{ = 0}\end{array}$

Consider: $\text{a = 1, b = 4}{\text{.5×10}}^{\text{-7}}\text{, c = -1}{\text{.1880×10}}^{\text{-11}}$.

  $\begin{array}{l}\text{x = }\frac{\text{-b ± }\sqrt{{\text{b}}^{\text{2}}\text{- 4ac}}}{\text{2a}}\\ \\ \text{x = }\frac{\text{-4}{\text{.5×10}}^{\text{-7}}\text{ ± }\sqrt{{\text{(4}{\text{.5×10}}^{\text{-7}}\text{)}}^{\text{2}}\text{- 4(1)(-1}{\text{.1880×10}}^{\text{-11}}\text{)}}}{\text{2(1)}}\\ \\ \text{x = 3}{\text{.229 × 10}}^{\text{-6}}{\text{ M H}}_{\text{3}}{\text{O}}^{\text{+}}\end{array}$

The $\text{pH}$ can be calculated as

  $\begin{array}{l}{\text{pH = -log[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}\\ \\ \text{pH = -log (3}{\text{.229 × 10}}^{\text{-6}}\text{) }\\ \\ \text{pH = 5}\text{.4909 = 5}\text{.5}\end{array}$

The value of $\text{pH}$ rainwater when partial pressure of ${\text{CO}}_{\text{2}}$ doubled is $\text{5}\text{.5}$.