Question

Due to a highway accident, 150 L of concentrated hydrochloric acid (12.0 M) is released into a lake containing 5.0 ´ 105 m3 of water. If the pH of this lake was 7.0 prior to the accident, what is the pH of the lake following the accident?

Answer 1

The pH of the lake following the accident is 5.43.

Step-by-step explanation:

`Molarity=(Moles)/(Voluem(L))`

Concentration of HCL =12.0 M

Volume of HCL = 150 L

Moles of hydrogen ions= n

`n=12 M* 150 L=1800 mol`

The pH of the lake = 7

`pH=-\log[H^]`

`7=-\log[H^+]`

`[H^+]=10^(-7) M`

Concentration of hydrogen ions in lake =
`10^(-7) M`

Volume of water in lake =
`V_2=5* 10^5 m^3 = 5* 10^8 L`

Moles of hydrogen ions in lake = n'

`n'=10^(-7) M* 5* 10^8 L=50 mol`

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 =
`150 L + 5* 10^8 L`

`C=(1850 mol)/(150 L + 5* 10^8 L)=3.6999* 10^(-6) M`

The final pH of the lake:

`pH=-\log[C]`

=
`-\log[3.6999* 10^(-6) M]=5.43`

The pH of the lake following the accident is 5.43.

Answer 2

Answer: The pH of the lake after accident is 5.44

Step-by-step explanation:

To calculate the molarity of lake, we use the equation:

`M_1V_1=M_2V_2`

where,

`M_1\text{ and }V_1` are the molarity and volume of the water

`M_2\text{ and }V_2` are the molarity and volume of hydrochloric acid

We are given:

Conversion factor used:
`1m^3=1000L`

`M_1=?M\\V_1=(5* 10^5m^3+150L)=[(5* 10^8)+(150)]L\\M_2=12.0M\\V_2=150L`

Putting values in above equation, we get:

`M_1*(5* 10^8+150)=12.0* 150\\\\M_1=(12.0* 150)/((5* 10^8+150))=3.59* 10^(-6)`

To calculate the pH of the solution, we use the equation:

`pH=-\log[H^+]`

We are given:

`[H^+]=3.59* 10^(-6)`

Putting values in above equation, we get:

`pH=-\log(3.59* 10^(-6))\\\\pH=5.44`

Hence, the pH of the lake after accident is 5.44