Respuesta :
According to the given statement pH would be attained in the year = Â 5.192
What is pH scale?
A scale used to specify the acidity or basicity of an aqueous solution, originally denoting "potential of hydrogen." The pH of acidic solutions is lower than the pH of basic or alkaline solutions.
According to the given information:
The formula of the dissociation of H2SO4
          [tex]\begin{gathered}\mathrm{H}_2 \mathrm{SO}_4 \rightarrow \mathrm{HSO}^{-}+\mathrm{H}_3 \mathrm{O}^{+} \\\mathrm{HSO}_4^{-}+\mathrm{H}_2 \mathrm{O} \rightleftharpoons \mathrm{SO}_4^{2-}+\mathrm{H}_3 \mathrm{O}^{+}\end{gathered}\\[/tex]
As we can see ,
2 moles of [tex]\mathrm{H}_3 \mathrm{O}^{+}[/tex] in total were produced from 1 mole [tex]\mathrm{H}_2 \mathrm{SO}_4[/tex]
solve for the concentration of [tex]\mathrm{H}_3 \mathrm{O}^{+}[/tex] in the lake.
Now solve for the number of moles.
Mole of  [tex]\mathrm{H}_3 \mathrm{O}^{+}=9460 \mathrm{~kg} \mathrm{H} \mathrm{H}_2 \mathrm{SO}_4 \times \frac{1000 \mathrm{~g}}{1 \mathrm{~kg}} \times \frac{1 \mathrm{~mol} \mathrm{H}_2 \mathrm{SO}_4}{98.1 \mathrm{~g} \mathrm{H}_2 \mathrm{SO}_4} \times \frac{2 \mathrm{~mol} \mathrm{H}_3 \mathrm{O}^{+}}{1 \mathrm{~mol} \mathrm{H_{2 } \mathrm { SO } _ { 4 }}}[/tex]
Mole of [tex]\mathrm{H}_3 \mathrm{O}^{+}=192864.4241 \text { moles } \mathrm{H}_3 \mathrm{O}^{+}[/tex]
The solve for the volume of the lake:
V = lhw = Ah
  = 10 km² x ((1000 m)²/(1 km)²)x3mx(1000 L/1 m³)
V = 3.0 x 10^10
Now solving for [tex]\left[\mathrm{H}_3 \mathrm{O}^{+}\right][/tex]
[tex]\left[\mathrm{H}_3 \mathrm{O}^{+}\right]=\frac{\mathrm{mol}}{L}=\frac{192864.4241 \mathrm{~mol} \mathrm{H}_3 \mathrm{O}^{+}}{3.0 \times 10^{10} \mathrm{~L}}[/tex]
= Â 6.4288^-6 M
Solving for the pH
[tex]p H=-\log \left[H_3 O^{+}\right][/tex]
pH = -log(6.4288^-6)
pH = 5.192
pH would be attained in the year = 5.192
To know more about pH scale visit:
https://brainly.com/question/10825137
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