Q18.175CP - Chemistry: Molecular Nature Of Matter And Change

Question

A site in Pennsylvania receives a total annual deposition of $\mathbf{\text{2.688}}$g/m2 of sulfate from fertilizer and acid rain. The ratio by mass of ammonium sulfate/ammonium bisulfate/sulfuric acid is.$\mathbf{\text{3.0/5.5/1.0}}$

 (a) How much acid, expressed as kg of sulfuric acid, is deposited over an area of ${\mathbf{\text{10.km}}}^{\mathbf{\text{2}}}$

(b) How many pounds of${\mathbf{\text{CaCO}}}_{\mathbf{\text{3}}}$ are needed to neutralize this acid?

 (c) If ${\mathbf{\text{10.km}}}^{\mathbf{\text{2}}}$is the area of an unpolluted lake$\mathbf{3}$ m deep and there is no loss of acid, what pH would be attained in the year? (Assume constant volume.) 

Step-by-Step Solution

Answer

a) The total acid deposited over an area of$10 {\text{km}}^2$ is.${\text{9460 kg\hspace{0.17em}\hspace{0.17em}H}}_{\text{2}}{\text{SO}}_{\text{4}}$

b) mass of ${\text{CaCO}}_{\text{3}}{\text{=\hspace{0.17em}2.12×10}}^{\text{4}}\text{\hspace{0.17em}lbs}$

c)$\text{pH =\hspace{0.17em}5.192}$

Step 1: To find how much acid

(a)

First, solve the mass of the annual sulfate deposition of different sulfate sources.

$\text{mass of }{\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}{\text{SO}}_{\text{4}}\text{ =\hspace{0.17em}}\frac{\text{3.0}}{\text{9.5}}\text{× }(2.688\text{\hspace{0.17em}g}/{\text{m}}^2){\text{=\hspace{0.17em}0.849 g/m}}^{\text{2}}$

$\begin{array}{l}{\text{mass of NH}}_{\text{4}}{\text{HSO}}_{\text{4}}\text{ =}\frac{\text{5.5}}{\text{9.5}}{\text{×2.688g/m}}^{\text{2}}{\text{\hspace{0.17em}=\hspace{0.17em}1.556 g/m}}^{\text{2}}\\ {\text{mass of H}}_{\text{2}}{\text{SO}}_{\text{4}}\text{ =}\frac{\text{1.0}}{\text{9.5}}{\text{×2.688g/m}}^{\text{2}}{\text{\hspace{0.17em}=\hspace{0.17em}\hspace{0.17em}0.283 g/m}}^{\text{2}}\end{array}$

${\left({\text{NH}}_4\right)}_2{\text{SO}}_4$is a weak acid so it will not significantly contribute to the acid in terms of sulfuric acid.

${\left({\text{NH}}_4\right)}_2{\text{SO}}_4$ produces${\text{HSO}}^{-}$ ions that will further dissociate to ${\text{SO}}_4^{2-}$

$\begin{array}{l}{\text{NH}}_{\text{4}}{\text{HSO}}_{\text{2}}\to {\text{NH}}_{\text{4}}^{\text{+}}{\text{+HSO}}_{\text{4}}^{\text{-}}\\ {\text{HSO}}_{\text{4}}^{\text{-}}{\text{+H}}_{\text{2}}\text{O}\rightleftharpoons {\text{SO}}_{\text{4}}^{\text{2-}}{\text{+H}}_{\text{3}}{\text{O}}^{\text{+}}\end{array}$

${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}{\text{\hspace{0.17em}will produce HSO}}^{\text{-}}{\text{ and H}}_{\text{3}}{\text{O}}^{\text{+}}$ions and will further dissociates to produce .${\text{SO}}^{2-}$

$\begin{array}{l}{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\to {\text{HSO}}^{\text{-}}{\text{+H}}_{\text{3}}{\text{O}}^{\text{+}}\\ {\text{HSO}}_{\text{4}}^{\text{-}}{\text{+H}}_{\text{2}}\text{O}\rightleftharpoons {\text{SO}}_{\text{4}}^{\text{2-}}{\text{+H}}_{\text{3}}{\text{O}}^{\text{+}}\end{array}$

As observed${\text{NH}}_{\text{4}}{\text{HSO}}_{\text{2}}$, only contributed half${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{ per mole of H}}_{\text{2}}{\text{SO}}_{\text{4}}$ (it did not produce any ${\text{H}}_3{\text{O}}^{+}$on its first dissociation).

Step 2: To solve for the total of acid as sulfuric acid

Now, solve for the total of acid as sulfuric acid.

Solve for the total sulfuric acid.

$\begin{array}{l}{\text{Total amount of acid = acid from NH}}_{\text{4}}{\text{SO}}_{\text{4}}{\text{+ acid from H}}_{\text{2}}{\text{SO}}_{\text{4}}\\ {\text{Total amount of acid\hspace{0.17em}=\hspace{0.17em}0.663 g/m}}^{\text{2}}{\text{+0.283 g/m}}^{\text{2}}\\ {\text{Total =\hspace{0.17em}0.946 g/m}}^{\text{2}}{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\end{array}$

Lastly, solve for the total acid deposited over an area of $10 {\text{km}}^2$

 in kilograms.

${\text{ Total\hspace{0.17em}amount of acid =\hspace{0.17em}10 km}}^{\text{2}}\frac{\text{0.946 g}}{{\text{ m}}^{\text{2}}}\frac{{\text{(1000 m)}}^{\text{2}}}{{\text{(1 km)}}^{\text{2}}}\text{×}\frac{\text{1 kg}}{\text{1000 g}}{\text{=\hspace{0.17em}\hspace{0.17em}9460 kg\hspace{0.17em}H}}_{\text{2}}{\text{SO}}_{\text{4}}$

Hence the total acid deposited over an area of  is .${\text{9460 kg\hspace{0.17em}H}}_{\text{2}}{\text{SO}}_{\text{4}}$

Step 3: To find the pounds

(b)

First, write the equation of the neutralization reaction between.

${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}{\text{ and CaCO}}_{\text{3}}$

${\text{CaCO}}_{\text{3}}{\text{+H}}_{\text{2}}{\text{SO}}_{\text{4}}\to {\text{CaSO}}_{\text{4}}{\text{+H}}_{\text{2}}{\text{CO}}_{\text{3}}$

${\text{H}}_{\text{2}}{\text{CO}}_{\text{3}}$will dissociate to water and carbon dioxide.

${\text{CaCO}}_{\text{3}}{\text{+H}}_{\text{2}}{\text{SO}}_{\text{4}}\to {\text{CaSO}}_{\text{4}}{\text{+H}}_{\text{2}}{\text{O+CO}}_{\text{2}}$

Now solve for the mass of ${\text{CaCO}}_{\text{3}}$needed to neutralize the acid.

$=21284.56606$

Hence mass of.${\text{CaCO}}_{\text{3}}{\text{\hspace{0.17em}=\hspace{0.17em}\hspace{0.17em}2.12×10}}^{\text{4}}\text{\hspace{0.17em}lbs}$

Step 4: To find pH would be attained in the year

(c)

Again, the formula of the dissociation of.${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$

$\begin{array}{l}{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\to {\text{HSO}}^{\text{-}}{\text{+H}}_{\text{3}}{\text{O}}^{\text{+}}\text{ H}\\ {\text{HSO}}_{\text{4}}^{\text{-}}{\text{+H}}_{\text{2}}\text{O}\rightleftharpoons {\text{SO}}_{\text{4}}^{\text{2-}}{\text{+H}}_{\text{3}}{\text{O}}^{\text{+}}\end{array}$

As observed, 2 moles of${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$ in total were produced from 1 mole${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}$. Solve for the concentration of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$in the lake.

First, solve for the number of moles.

${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{=\hspace{0.17em}9460 kg\hspace{0.17em}H}}_{\text{2}}{\text{SO}}_{\text{4}}\text{×}\frac{\text{1000 g}}{\text{1 kg}}\text{×}\frac{{\text{1 mol\hspace{0.17em}H}}_{\text{2}}{\text{SO}}_{\text{4}}}{{\text{98.1 g\hspace{0.17em}H}}_{\text{2}}{\text{SO}}_{\text{4}}}\text{×}\frac{{\text{2 mol\hspace{0.17em}H}}_{\text{3}}{\text{O}}^{\text{+}}}{{\text{1 mol\hspace{0.17em}HHO}}_{\text{4}}}{\text{moles\hspace{0.17em}of\hspace{0.17em}H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{=192864.4241moles\hspace{0.17em}H}}_{\text{3}}{\text{O}}^{\text{+}}$

Then solve for the volume of the lake.

${\text{V =\hspace{0.17em}lhw\hspace{0.17em}=\hspace{0.17em}Ahn=10 km}}^{\text{2}}\text{×}\frac{{\text{(1000 m)}}^{\text{2}}}{{\text{(1 km)}}^{\text{2}}}\text{3 m×}\frac{\text{1000 L}}{{\text{1 m}}^{\text{3}}}{\text{×V =3.0×10}}^{\text{10}}\text{ L}$

Now solve for $\left[{\text{H}}_3{\text{O}}^{+}\right]$

$\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\text{\hspace{0.17em}=\hspace{0.17em}}\frac{\text{mol}}{\text{L}}\text{\hspace{0.17em}=\hspace{0.17em}}\frac{{\text{192864.4241 molH}}_{\text{3}}{\text{O}}^{\text{+}}}{{\text{3.0×10}}^{\text{10}}\text{ L}}{\text{\hspace{0.17em}=\hspace{0.17em}6.4288}}^{\text{-6}}\text{\hspace{0.17em}M}$

Lastly, solve for the pH

$\begin{array}{l}\text{pH =\hspace{0.17em}-log}\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\\ \text{pH \hspace{0.17em}=\hspace{0.17em}-log}\left({\text{6.4288}}^{\text{-6}}\right)\\ \text{pH =\hspace{0.17em}5.192}\end{array}$

Hence$\text{pH =\hspace{0.17em}5.192}$