# Finding concentration and moles given final and initial pH

Questions:

How many moles of $$\ce{NaOH}$$ are needed to change $$500~\mathrm{L}$$ of solution with $$\mathrm{pH}\ 2$$ to $$\mathrm{pH}\ 11$$?

A swimming pool contains 2 million litres of water at $$\mathrm{pH}\ 7.80$$ and the $$\mathrm{pH}$$ needs to be changed to $$6.80$$. How many moles of $$\ce{HCl}$$ are needed?

Not quite sure how to tackle these types of questions.

Would I have to find the moles of hydroxide/hydrogen ions to neutralise the solution first (ie. moles of hydroxide ions needed to reach $$\mathrm{pH}\ 7$$) and then from there calculate the amount of hydrogen/hydroxide ions needed to get to final $$\mathrm{pH}$$?

Or do I just calculate the difference between the moles of ions in final $$\mathrm{pH}$$ with moles of ions in initial $$\mathrm{pH}$$?

I have tried both methods but I don't have the answer to the questions so I'm just trying to understand the concept.

The $[\ce{H+}]$ ion concentration changes from $\pu{10^-2~M}$ to $\pu{10^-11~M}$. Assuming water at $\pu{25^\circ C}$, this implies the change in $[\ce{OH-}]$ ion concentration from $\pu{10^-12~M}$ to $\pu{10^-3~M}$. Since the volume of the solution is fixed at $\pu{500L}$, the amount of $\ce{OH-}$ ions has changed from $500\times\pu{10^-12 moles}$ to $500\times\pu{10^-3 moles}$. This change is exactly the number of moles of $\ce{NaOH}$ that need to be added to this solution, since $\ce{NaOH}$ is a strong base and dissociates with $100\%$ yield.