Respuesta :
Answer: The pH of the lake after accident is 5.44
Explanation:
To calculate the molarity of lake, we use the equation:
[tex]M_1V_1=M_2V_2[/tex]
where,
[tex]M_1\text{ and }V_1[/tex] are the molarity and volume of the water
[tex]M_2\text{ and }V_2[/tex] are the molarity and volume of hydrochloric acid
We are given:
Conversion factor used: [tex]1m^3=1000L[/tex]
[tex]M_1=?M\\V_1=(5\times 10^5m^3+150L)=[(5\times 10^8)+(150)]L\\M_2=12.0M\\V_2=150L[/tex]
Putting values in above equation, we get:
[tex]M_1\times(5\times 10^8+150)=12.0\times 150\\\\M_1=\frac{12.0\times 150}{(5\times 10^8+150)}=3.59\times 10^{-6}[/tex]
To calculate the pH of the solution, we use the equation:
[tex]pH=-\log[H^+][/tex]
We are given:
[tex][H^+]=3.59\times 10^{-6}[/tex]
Putting values in above equation, we get:
[tex]pH=-\log(3.59\times 10^{-6})\\\\pH=5.44[/tex]
Hence, the pH of the lake after accident is 5.44
Answer:
The pH of the lake following the accident is 5.43.
Explanation:
[tex]Molarity=\frac{Moles}{Voluem(L)}[/tex]
Concentration of HCL =12.0 M
Volume of HCL = 150 L
Moles of hydrogen ions= n
[tex]n=12 M\times 150 L=1800 mol[/tex]
The pH of the lake = 7
[tex]pH=-\log[H^][/tex]
[tex]7=-\log[H^+][/tex]
[tex][H^+]=10^{-7} M[/tex]
Concentration of hydrogen ions in lake = [tex]10^{-7} M[/tex]
Volume of water in lake = [tex]V_2=5\times 10^5 m^3 = 5\times 10^8 L[/tex]
Moles of hydrogen ions in lake = n'
[tex]n'=10^{-7} M\times 5\times 10^8 L=50 mol[/tex]
Concentration of hydrogen ions after release of HCL in lake: C
Total moles of HCl = n + n' = 1800 mol + 50 mol = 1850 mol
Volume of the lake = [tex]150 L + 5\times 10^8 L[/tex]
[tex]C=\frac{1850 mol}{150 L + 5\times 10^8 L}=3.6999\times 10^{-6} M[/tex]
The final pH of the lake:
[tex]pH=-\log[C][/tex]
=[tex]-\log[3.6999\times 10^{-6} M]=5.43[/tex]
The pH of the lake following the accident is 5.43.
