If I have storage tank containing 3.8 X 10E5 liters of water that is contaminated with NaOH and has a pH of 10.82. What volume of 0.1 M HCl is required to bring the pH= 7.0?

I should call your attention to the misuse of symbols. I assume you meant 3.8E5 or you could have written 3.89 x 10^5.

How many mols NaOH do you have.
pH = 10.82; therefore, pOH = 14.0-10.82 = 3.18 and (3.18 = log(OH^-); therefore,(OH^-) = 6.6E-4.
Then mL x M NaOH = mL x M HCl. Solve for mL HCl.

To solve this problem, we need to calculate the volume of 0.1 M HCl needed to neutralize the NaOH and bring the pH down to 7.0. Here's how you can do it:

Step 1: Calculate the number of moles of NaOH in the water:
We know the volume of the water is 3.8 x 10^5 liters. The concentration of NaOH can be determined from the pH value. Since the pH is 10.82, we know that the concentration of OH- ions is 10^-(14-10.82). So, the concentration of NaOH is that value multiplied by 2 since NaOH dissociates into one mole of Na+ and one mole of OH-.

Step 2: Calculate the number of moles of HCl needed:
To bring the pH down to 7.0, we need to add enough HCl to neutralize the NaOH in the water. Since NaOH and HCl react in a 1:1 ratio, we can calculate the number of moles of HCl needed using the number of moles of NaOH calculated in Step 1.

Step 3: Calculate the volume of HCl needed:
The volume of HCl needed can be calculated using the concentration of HCl (0.1 M) and the number of moles of HCl calculated in Step 2.

By following these steps, you can determine the volume of 0.1 M HCl required to bring the pH of the water to 7.0.