If it takes 260 mL of 0.20 M NaOH to neutralize 74 mL of an HCl solution, what is the HCl concentration?

Answer 1

This is a great question! The key here is finding the number of moles of -OH and H+, because in the end they will match up, since it takes 1 -OH to neutralize 1 H+.

We can find the moles of OH- by first finding the moles of NaOH. 260 mL is the same as .26 L, so we multiply .26 L x .2 M NaOH. The L cancel out (since M is mol/L) and we're left with .052 moles of NaOH, and since there is one mole of OH- for every mole of NaOH, we have .052 moles of OH-.

We know that it takes one mole of OH- to neutralize 1 mol of H+, we know that there were .052 moles of H+ in the solution. We know that there are 74 mL of H+, or .074 L, so we divide .052 moles/.074 L and we get a concentration of .70 M for the HCl concentration!