Q19.148 CP - Chemistry: Molecular Nature Of Matter And Change

Question

Even before the industrial age, rainwater was slightly acidic due to dissolved ${\mathbf{\text{CO}}}_{\mathbf{2}}$. Use the following data to calculate $\mathbf{\text{pH}}$ of unpolluted rainwater at ${\mathbf{\text{25}}}^{\mathbf{\circ }}\mathbf{\text{C}}\mathbf{:}\mathbf{vol}\mathbf{\%}$in air of ${\mathbf{\text{CO}}}_{\mathbf{2}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{033}\mathbf{ }\mathbf{vol}\mathbf{\%}$; solubility of ${\mathbf{\text{CO}}}_{\mathbf{2}}$ in pure water at ${\mathbf{\text{25}}}^{\mathbf{\circ }}\mathbf{\text{C}}$and $\mathbf{1}\mathbf{atm}\mathbf{=}\mathbf{88}\mathbf{ }\mathbf{mL}\mathbf{ }{\mathbf{CO}}_{\mathbf{2}}\mathbf{/}\mathbf{100}\mathbf{ }\mathbf{mL}\mathbf{ }{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\mathbf{;}\mathbf{ }{\mathbf{K}}_{\mathbf{a}\mathbf{1}}$ of ${\mathbf{H}}_{\mathbf{2}}{\mathbf{CO}}_{\mathbf{3}}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{5}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{7}}$.

Step-by-Step Solution

Answer

The $\text{pH}$ of unpolluted rainwater ${25}^{\circ }C$at is $\mathrm{pH}=5.68$.

Step 1: Concept Introduction.

Solubility refers to the greatest amount of solute that can dissolve in a known quantity of solvent at a given temperature.

The phrase "solubility product" refers to salts that are only sparingly soluble. It is the maximal product of the molar concentration of the ions produced by dissociation of the molecule (raised to their proper powers).

Step 2: Dissociation of H 2 C O 3 .

The reactions involved in the acidification of rain water are as follows –

$$\begin{gathered} {\mathrm{CO}}_2\left(\mathrm{aq}\right)+{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)\rightleftharpoons {\mathrm{H}}_2{\mathrm{CO}}_3\left(\mathrm{aq}\right) \\ {\mathrm{H}}_2{\mathrm{CO}}_3\left(\mathrm{aq}\right)+{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)\rightleftharpoons {\mathrm{H}}_3{\mathrm{O}}^{+}\left(\mathrm{aq}\right)+{\mathrm{H}}_2{\mathrm{CO}}_3^{-}\left(\mathrm{aq}\right) \end{gathered}$$

The molar concentration of carbon dioxide depends on how much carbon dioxide is dissolved in pure water. At ${25}^{\circ }C$and $1 \mathrm{atm},88 \mathrm{mL}$ of $C{O}_2$can dissolve in $100 \text{mL}$of water.

The number of moles of ${\mathrm{CO}}_2$are calculated as follows –

$\begin{array}{c}\mathrm{n}=\frac{\mathrm{PV}}{\mathrm{RT}}\\ \mathrm{n}=\frac{1 \mathrm{atm}\times 88\times {10}^{-3} \mathrm{L}}{0.0821 \mathrm{L}·\mathrm{atm}/\mathrm{mol}·\mathrm{K}\times 298 \mathrm{K}}\times \frac{0.033\%}{100\%}\\ \mathrm{n}=1.18693\times {10}^{-6} \mathrm{mol}\end{array}$

The concentration of ${\mathrm{CO}}_2$in air –

$\begin{array}{c}\left[{\mathrm{CO}}_2\right]=\frac{1.18963\times {10}^{-6} \mathrm{mol}}{100\times {10}^{-3} \mathrm{L}}\\ =1.18963\times {10}^{-5} \mathrm{M}\end{array}$

The dissociation of ${\text{H}}_{2}C{O}_3$ is –

$\begin{array}{c}\left[{\mathrm{CO}}_2\right]=\left[{\mathrm{H}}_2{\mathrm{CO}}_3\right]\\ {\mathrm{K}}_{\mathrm{a}}=\frac{\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]\left[{\mathrm{HCO}}_3^{-}\right]}{\left[{\mathrm{H}}_2{\mathrm{CO}}_3\right]}\\ 4.5\times {10}^{-7}=\frac{\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]\left[{\mathrm{HCO}}_3^{-}\right]}{\left[{\mathrm{CO}}_2\right]}\end{array}$

Step 3: pH of unpolluted rain water.

The ICE table for the reaction can be drawn as –

	$\left[{\mathrm{CO}}_2\right]·\left[{\mathrm{H}}_2{\mathrm{CO}}_3\right]$	$\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]$	$\left[{\mathrm{HCO}}_3^{-}\right]$
Initial	$1.18693\times {10}^{-5}$	$0$	$0$
change	$\text{-x}$	$\text{+x}$	$\text{+x}$
Equilibrium	$1.18693\times {10}^{-5}-\mathrm{x}$	$\text{x}$	$\text{x}$

Solve for the value of $x$.

$\begin{array}{c}4.5\times {10}^{-7}=\frac{\left(\mathrm{x}\right)\left(\mathrm{x}\right)}{\left(1.18693\times {10}^{-5}\mathrm{m}-\mathrm{x}\right)}\\ =\frac{{\mathrm{x}}^2}{1.18693\times {10}^{-5}\mathrm{m}}\\ \mathrm{x}=2.0976596\times {10}^{-6}\mathrm{M}\end{array}$

Now, calculate $\text{pH}$the value –

$\begin{array}{c}\mathrm{pH}=-\log \left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]\\ \mathrm{pH}=-\log \left[{\mathrm{H}}^{+}\right]\\ \left[{\mathrm{H}}^{+}\right]=\mathrm{x}=2.0970596\times {10}^{-6}\mathrm{M}\\ \mathrm{pH}=-\log \left(2.0970596\times {10}^{-6}\mathrm{M}\right)\\ \mathrm{pH}=5.6783\\ \mathrm{pH}=5.68\end{array}$

 

Therefore, the value for $\mathrm{pH}$is obtained as $\mathrm{pH}=5.68$.