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:

where,
are the molarity and volume of the water
are the molarity and volume of hydrochloric acid
We are given:
Conversion factor used:

![M_1=?M\\V_1=(5* 10^5m^3+150L)=[(5* 10^8)+(150)]L\\M_2=12.0M\\V_2=150L](https://img.qammunity.org/2020/formulas/chemistry/high-school/etudhcs6a751r70xiw9nbi5qfijjm5kol1.png)
Putting values in above equation, we get:

To calculate the pH of the solution, we use the equation:
![pH=-\log[H^+]](https://img.qammunity.org/2020/formulas/chemistry/middle-school/vz65x0ueuj8r8ibqa81zvsbzb2yaetlce4.png)
We are given:
![[H^+]=3.59* 10^(-6)](https://img.qammunity.org/2020/formulas/chemistry/high-school/2qwq5vf993ofpwe1sutux58a704qngpgrp.png)
Putting values in above equation, we get:

Hence, the pH of the lake after accident is 5.44