Question

You arrive at a crime scene and are told the body of the victim is in the 10 m deep pool. You walk over to what you are told is a 12 m in diameter, cylindrical pool. From the outside of the pool, you can see that the body looks badly burnt. Your partner says, “It looks like our victim had been burned alive and tried to put out the fire by jumping in the pool. The victim likely drowned to death.”
Something does not sit right with you though. If there was a fire, where did it start? There are no signs of combustion anywhere. You aren’t so sure and ask the crime scene investigator to run a sample of the pool water before letting anybody try to pull the body out.

The CSI comes back to you and tells you that normal pool water pH is roughly around 7.2, but the pool pH is actually highly basic at a level of 13 with a concentration of hydroxide ions at 1.0 x 10-1 mol/L. It becomes obvious to you that the body wasn’t burned before going in to the pool , but AFTER and there was no fire needed!

You order the body to be removed from the pool, but the CSI interjects, “It would be too dangerous with a pH that high. I suggest you get some vinegar from the store and pour it in to the pool before hand to drop the pH to 7. Draining the pool would take far too long and we need to examine the body as soon as possible.” She asks one of your constables to go to the store to purchase 5 – 4L jugs of vinegar (pH = 2) to pour in the pool, as she states it is enough to bring the pH to a safe level of 7.

It sounds like it could be enough vinegar based on your knowledge of acids and bases, but you want to double-check her estimate before sending your constable to the store. Verify whether or not she is correct using calculations. A diagram may help you with your calculations.

Hint 1: 1m3 = 1000L

Hint 2: Vcylinder = πr2h

Answer 1

Answer:

The vinegar is not enough to neutralize the pool.

Explanation:

The [OH⁻] in the pool is 1.0x10⁻¹mol / L. To know how many moles of OH⁻ are in the solution, you must calculate volume of the pool thus:

V(pool) = πr²h

Where r, radius is d/2 = 12m/2 = 6m  and h is deep of the pool = 10m

V(pool) = π(6m)²*10

V(pool) = 1131m³

As 1m³ = 1000L:

1131m³  × (1000L / 1m³) = 1131000L in the pool.

And moles of OH⁻ are:

1.0x10⁻¹mol / L ₓ 1131000L = 131100 moles of OH⁻ are in the pool

The neutralization of OH⁻ with H⁺ is:

OH⁻ + H⁺ → H₂O

That means to neutralize the pool you must add 131100 moles of H⁺.

The H⁺ concentration in a vinegar pH = 2 is:

pH = -log [H⁺]

2 = -log [H⁺]

1x10⁻²M = [H⁺]

4L are just 4x10⁻² moles of [H⁺]. As you need 131100 moles of H⁺:

The vinegar is not enough to neutralize the pool.