Calculate the pH during the titration of 30.00 mL of 0.1000 M NaOH

Question

Calculate the pH during the titration of 30.00 mL of 0.1000 M NaOH(aq) with 0.1000 M HCl(aq) after 12 mL of the acid have been added

The answer is 12.63

Answer 1

So, a couple of things to note here. We're titrating a strong base with a strong acid. They have the same concentrations, so equivalence (equal numbers of moles of H+ and OH-) will happen after we've added the same volume of acid as the base we started with. We're not there, yet, so the pH is not 7, it's still very basic (from the strong base).

To find out how basic, we find out how much of the OH- remains (has not yet been neutralized). Again, since the concentrations are the same, if we know how many moles of H+ have been added (0.1000M=n/0.012L, so 0.0012mol), we can subtract that from how many moles of OH- we started with (0.1000M=n/0.0300L, or 0.00300mol). That means 0.0018mol of OH- remain.

Because we added the HCl to the original NaOH, we now need to find the new volume (0.0120L+0.030L) to get the new concentration of OH-. [OH-]=0.0018mol/0.0420L=0.043M.

Finally, we know that pOH=-log[OH-]=-log(0.043)=1.37, so pH = 14-pOH = 12.63.