HCl is a strong acid, so we can assume that all the HCl dissociates into its ions. We have 10mL of 1.0M HCl. Let's convert that to mol H+. I like to use dimensional analysis, where I make the units cancel out to what I want using conversion factors.
(1.0 mol HCl)/(1 L) * (1 mol H+)/(1 mol HCl) * (1L)/(1000mL) * 10mL = 0.01mol H+
Now let's see if we need to consider the mols of H+ that were already in the water. Pure water has a pH of 7, and pH is -log([H+]). So 7 = -log([H+]), and 10^-7 mol/L = [H+]. We have 1 L of water, so 10^-7 mol H+.
0.01 >> 10^-7, so we can just say total mol H+ ~= 0.01.
Now we find molarity, or moles per liter. We have 0.01 mol H+, and 1L + 10mL = 1.01L solution. So molarity = 0.01mol/1.01L = 0.0099M.
And finally, the pH equation. pH = -log([H+]) = -log(0.0099).
pH = 2.00
Check your intuition! We added an acid to water, so our solution should be acidic. Acidic solutions have a pH < 7. Is our answer less than 7? Yes! Looks good.
If you have access to a laboratory with strong acids and a pH meter, ask permission from the lab supervisor to set up a similar experiment or to show you a demonstration. It's fun to see this in real life.
