This will produce water and a salt
Let's imagine we have two solutions: NaOH with a known concentration (let's say it is 1.0 M) and HCl of an unknown concentration.
Our goal is to determine what the concentration of the HCl solution is.
When the reaction is finished, the pH of the solution should be about 7, which is very close to phenolphthalein's working range.
So, if we add some of the phenolphthalein indicator to the HCl solution and add a bit of NaOH solution, drop by drop, we can see the endpoint or equivalence point of the reaction when the solution turns pink.
We are going to take advantage of the fact that we know that NaOH and HCl react in a one to one ratio.
NaOH + HCl --> H2O + NaCl
So, let's say that it took 50 mL of the 1.0 M NaOH to reach the equivalence point of the reaction when added to 100 mL of the HCl solution of unknown concentration. If we calculate the number of moles of NaOH that contains, we can figure out the number of moles of HCl that we had.
1.0 M is 1.0 mol/L.
We added 50 mL, which is 0.050 L.
So if we do (1.0 mol/L) * (0.050L), we determine that we added 0.050 mols of NaOH in our unknown sample.
Since NaOH and HCl react in a one to one ration, then we also had 0.050 mols of HCl in the original 100 mL (0.1 L) of the unknown HCl solution.
0.050 mols/0.1 L = 0.5 M HCl solution as the concentration of our unknown solution.
The video below shows the above calculation along with a verbal explanation of the math required.