Q19.146 CP - Chemistry: Molecular Nature Of Matter And Change

Question

A lake that has a surface area of  $\mathbf{10}\mathbf{.}\mathbf{0}\mathbf{ }\mathbf{acres}\mathbf{(}\mathbf{1}\mathbf{ }\mathbf{acre}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{840}\mathbf{\times }{\mathbf{10}}^{\mathbf{3}}\mathbf{ }{\mathbf{yd}}^{\mathbf{2}}\mathbf{)}$ receives $\mathbf{1}\mathbf{.}\mathbf{00}\mathbf{ }\mathbf{in}\mathbf{.}$ of rain of $\mathbf{pH}\mathbf{ }\mathbf{4}\mathbf{.}\mathbf{20}$. (Assume that the acidity of the rain is due to a strong, monoprotic acid.) 

(a) How many moles of ${\mathbf{H}}_{\mathbf{3}}{\mathbf{O}}^{\mathbf{+}}$ are in the rain falling on the lake? 

(b) If the lake is unbuffered $\mathbf{(}\mathbf{pH}\mathbf{=}\mathbf{7}\mathbf{.}\mathbf{00}\mathbf{)}$and its average depth is $\mathbf{10}\mathbf{.}\mathbf{0}\mathbf{ }\mathbf{ft}$ before the rain, find the $\mathbf{\text{pH}}$ after the rain has been mixed with lake water. (Ignore runoff from the surrounding land.) 

(c) If the lake contains hydrogen carbonate ions ${\mathbf{(}\mathbf{HCO}}_{\mathbf{3}}^{\mathbf{-}}\mathbf{)}$, what mass of ${\mathbf{(}\mathbf{HCO}}_{\mathbf{3}}^{\mathbf{-}}\mathbf{)}$would neutralize the acid in the rain?

Step-by-Step Solution

Answer

(a) Number of moles of ${\mathrm{H}}_3{\mathrm{O}}^{+}$in the rain falling on the lake is $64.8565 \mathrm{mol}$.

(b) The after the rain has been mixed with lake water is $6.2828$.

(c) $3956{.2465 \mathrm{g} \mathrm{HCO}}_3^{-}$would neutralize the acid rain and maintain $\text{pH}$ a of $7$.

Step 1: Concept Introduction.

Any hydrogen-containing material capable of transferring a proton (hydrogen ion) to another chemical is classified as an acid. A base is a molecule or ion that can take a hydrogen ion from an acid and accept it.

Step 2: Moles of H 3 O + .

(a)

The volume of the rain must first be calculated from the provided data in order to compute the number of moles of ${\mathrm{H}}_3{\mathrm{O}}^{+}$.

The surface area of the lake is $10 \mathrm{acres}$.

Amount of rainwater is –

$\begin{array}{l}= (10 \mathrm{acres})\times \left(\frac{4.840\times {10}^3 {\mathrm{yd}}^2}{1 \mathrm{acre}}\right)\times {\left(\frac{36 \mathrm{in}}{1 \mathrm{yd}}\right)}^2\times (1 \mathrm{in})\times {\left(\frac{2.54 \mathrm{cm}}{1 \mathrm{in}}\right)}^3\times \left(\frac{1 \mathrm{mL}}{1 {\mathrm{cm}}^3}\right)\times \left(\frac{1 \mathrm{L}}{1000 \mathrm{mL}}\right)\\ = 1027901.531 \mathrm{L}\end{array}$

Identify the concentration of ${\mathrm{H}}_3{\mathrm{O}}^{+}$ from the $\text{pH}$.

$\begin{array}{c}\mathrm{pH}=-\log \left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]\\ \left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]={10}^{-\mathrm{pH}}\\ \left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]={10}^{-4.20}\\ \left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]=6.3096\times {10}^{-5} \mathrm{M}\end{array}$

Find the number of moles of ${\mathrm{H}}_3{\mathrm{O}}^{+}$.

$\begin{array}{l}\mathrm{Number} \mathrm{of} \mathrm{moles} {\mathrm{H}}_3{\mathrm{O}}^{+}= (1027901.531 \mathrm{L})\times \left(6.3096\times {10}^{-5} \mathrm{M}\right)\\ \mathrm{Number} \mathrm{of} \mathrm{moles} {\mathrm{H}}_3{\mathrm{O}}^{+}= 64.8565 \mathrm{mol}\end{array}$

 

Therefore, the value for moles is obtained as $64.8565 \mathrm{mol}$.

Step 3: Calculation for pH .

(b)

Determine the volume of the water in the lake in $\text{L}$.

$\begin{array}{l}= (10 \mathrm{acres})\times \left(\frac{4.840\times {10}^3 {\mathrm{yd}}^2}{1 \mathrm{acre}}\right)\times {\left(\frac{36 \mathrm{in}}{1 \mathrm{yd}}\right)}^2\times (10 \mathrm{ft})\times {\left(\frac{2.54 \mathrm{cm}}{1 \mathrm{in}}\right)}^3\times \left(\frac{1 \mathrm{mL}}{1 {\mathrm{cm}}^3}\right)\times \left(\frac{1 \mathrm{L}}{1000 \mathrm{mL}}\right)\\ = 123348183.8 \mathrm{L}\end{array}$

Calculate the total volume of the lake after it rains.

$\begin{array}{c}{\mathrm{V}}_{\mathrm{t}} = 123348183.8\mathrm{L}+1027901.531\mathrm{L}\\ {\mathrm{V}}_{\mathrm{t}} = 124376085.3\mathrm{L}\end{array}$

Identify the concentration of ${\mathrm{H}}_3{\mathrm{O}}^{+}$after it rains.

$\begin{array}{c}\left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right] = \frac{64.8565\mathrm{mol}}{124376085.3\mathrm{L}}\\ \left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right] = 5.2145\times {10}^{-7}\mathrm{M}\end{array}$

Find the $\text{pH}$ value.

$\begin{array}{c}\mathrm{pH} = -\log \left[{\mathrm{H}}_3{\mathrm{O}}^{+}\right]\\ \mathrm{pH} = -\log (5.2145\times {10}^{-7}\mathrm{M})\\ \mathrm{pH} = 6.2828\end{array}$

 

Therefore, the value for is obtained as $6.2828$.

 

Step 4: Mass of H 3 O + .

(c)

The reaction between the acid in rain and hydrogen carbonate is presented below.

${\mathrm{H}}_3{\mathrm{O}}^{+}{+\mathrm{HCO}}_3^{-}\to {\mathrm{H}}_2\mathrm{O}+{\mathrm{H}}_2{\mathrm{CO}}_3$

The mole ratio between the two starting material is $1:1$. As a result, the mass of ${\mathrm{HCO}}_3^{-}$ required is –

$\begin{array}{c}\mathrm{m} = \left(64.8565 \mathrm{mol} {\mathrm{H}}_3{\mathrm{O}}^{+}\right)\times \left(\frac{{1 \mathrm{mol} \mathrm{HCO}}_3^{-}}{1 \mathrm{mol} {\mathrm{H}}_3{\mathrm{O}}^{+}}\right)\times \left(\frac{{61 \mathrm{g} \mathrm{HCO}}_3^{-}}{{1 \mathrm{mol} \mathrm{HCO}}_3^{-}}\right)\\ \mathrm{m} = 3956{.2465 \mathrm{g} \mathrm{HCO}}_3^{-}\end{array}$

 

Therefore, the value of mass is obtained as $3956.2465 \mathrm{g}$.