Question

Neutralizing an olympic size swimming pool is conceptually very similar to performaing a massive titration experiment. Suppose a 700 thousand gallon swimming pool has a pH 9.33 which is a bit too high for swimming. Calculate how many gallons of 10 M HCI (strong acid) it will take to neutralize the swimming pool to pH = 7.

Answer 1

6,97x10⁻³ gallons

Step-by-step explanation:

pH is defined as:

pH = -log [H⁺]

Thus, you need to have, in the end:

10⁻⁷ = [H⁺]

And you have, in the first:

`10^(-9,33)` = [H⁺]

The volume of swimming pool is:

700'000 galllons ×
`(3,78541 L)/(1 gallon)` = 2649787 L

Thus, the moles of H⁺ in the first and in the end are:

First:

`10^(-9,33)mol/L` × 2649787L = 1,24x10⁻³ moles

End:

`10^(-7)mol/L` × 2649787L = 0,265 moles

Thus, the moles of H⁺ you need to add are:

0,265 - 1,24x10⁻³ = 0,26376 moles

These moles comes from 10M HCl, thus, the volume in gallons you need to add are:

`0,26376moles*(1L)/(10moles)* (1gallon)/(3,78541L)` =

6,97x10⁻³ gallons

I hope it helps!