91影视

(a)

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

$\text{massof}{\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}{\text{SO}}_{\text{4}}\text{=鈥�}\frac{\text{3.0}}{\text{9.5}}\text{脳}(2.688\text{鈥塯}/{\text{m}}^2){\text{=鈥�0.849驳/尘}}^{\text{2}}$

$\begin{array}{l}{\text{massofNH}}_{\text{4}}{\text{HSO}}_{\text{4}}\text{=}\frac{\text{5.5}}{\text{9.5}}{\text{脳2.688驳/尘}}^{\text{2}}{\text{鈥�=鈥�1.556驳/尘}}^{\text{2}}\\ {\text{massofH}}_{\text{2}}{\text{SO}}_{\text{4}}\text{=}\frac{\text{1.0}}{\text{9.5}}{\text{脳2.688驳/尘}}^{\text{2}}{\text{鈥�=鈥夆赌�0.283驳/尘}}^{\text{2}}\end{array}$

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

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

${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}{\text{鈥墂颈濒濒辫谤辞诲耻肠别贬厂翱}}^{\text{-}}{\text{andH}}_{\text{3}}{\text{O}}^{\text{+}}$ions andwill further dissociates to produce.${\text{SO}}^{2鈭�}$

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

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

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

Solve for the total sulfuric acid.

$\begin{array}{l}{\text{Totalamountofacid=acidfromNH}}_{\text{4}}{\text{SO}}_{\text{4}}{\text{+acidfromH}}_{\text{2}}{\text{SO}}_{\text{4}}\\ {\text{罢辞迟补濒补尘辞耻苍迟辞蹿补肠颈诲鈥�=鈥�0.663驳/尘}}^{\text{2}}{\text{+0.283g/m}}^{\text{2}}\\ {\text{罢辞迟补濒=鈥�0.946驳/尘}}^{\text{2}}{\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}\end{array}$

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

in kilograms.

${\text{罢辞迟补濒鈥塧尘辞耻苍迟辞蹿补肠颈诲=鈥�10办尘}}^{\text{2}}\frac{\text{0.946g}}{{\text{m}}^{\text{2}}}\frac{{\text{(1000m)}}^{\text{2}}}{{\text{(1km)}}^{\text{2}}}\text{脳}\frac{\text{1kg}}{\text{1000g}}{\text{=鈥夆赌�9460办驳鈥塇}}_{\text{2}}{\text{SO}}_{\text{4}}$

Hence the total acid deposited over an area ofis.${\text{9460办驳鈥塇}}_{\text{2}}{\text{SO}}_{\text{4}}$

(b)

First, write the equation of the neutralization reaction between.

${\text{H}}_{\text{2}}{\text{SO}}_{\text{4}}{\text{andCaCO}}_{\text{3}}$

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

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

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

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

$=21284.56606$

Hence mass of.${\text{CaCO}}_{\text{3}}{\text{鈥�=鈥夆赌�2.12脳10}}^{\text{4}}\text{鈥塴产蝉}$

(c)

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

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

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

First, solve for the number of moles.

${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{\text{=鈥�9460办驳鈥塇}}_{\text{2}}{\text{SO}}_{\text{4}}\text{脳}\frac{\text{1000g}}{\text{1kg}}\text{脳}\frac{{\text{1尘辞濒鈥塇}}_{\text{2}}{\text{SO}}_{\text{4}}}{{\text{98.1驳鈥塇}}_{\text{2}}{\text{SO}}_{\text{4}}}\text{脳}\frac{{\text{2尘辞濒鈥塇}}_{\text{3}}{\text{O}}^{\text{+}}}{{\text{1尘辞濒鈥塇HO}}_{\text{4}}}{\text{尘辞濒别蝉鈥塷蹿鈥塇}}_{\text{3}}{\text{O}}^{\text{+}}{\text{=192864.4241尘辞濒别蝉鈥塇}}_{\text{3}}{\text{O}}^{\text{+}}$

Then solve for the volume of the lake.

${\text{痴=鈥塴丑飞鈥�=鈥碱丑苍=10办尘}}^{\text{2}}\text{脳}\frac{{\text{(1000m)}}^{\text{2}}}{{\text{(1km)}}^{\text{2}}}\text{3尘脳}\frac{\text{1000L}}{{\text{1m}}^{\text{3}}}{\text{脳痴=3.0脳10}}^{\text{10}}\text{L}$

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

$\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\text{鈥�=鈥�}\frac{\text{mol}}{\text{L}}\text{鈥�=鈥�}\frac{{\text{192864.4241molH}}_{\text{3}}{\text{O}}^{\text{+}}}{{\text{3.0脳10}}^{\text{10}}\text{L}}{\text{鈥�=鈥�6.4288}}^{\text{-6}}\text{鈥塎}$

Lastly, solve for the pH

$\begin{array}{l}\text{辫贬=鈥�-濒辞驳}\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]\\ \text{辫贬鈥�=鈥�-濒辞驳}\left({\text{6.4288}}^{\text{-6}}\right)\\ \text{辫贬=鈥�5.192}\end{array}$

Hence$\text{辫贬=鈥�5.192}$