Unit 4, Lesson 22

Titration is the process of measuring the concentration of a strong acid in a water solution by using an indicator and adding a base to neutralize it. As stated before, neutralization reactions result in a salt and water being produced.  After titration, when a point of equality is reached, there are equal moles, equal volumes, and equal molarities of the acid and the base. Get this: if you really know what you're doing, you can completely neutralize an acid with a base (or vice versa) and get a solution safe enough to drink. Granted, no one would recommend you try that at home, or even in a chemistry classroom.

If the molarity of the acid or the base is known, the molarity of the other can be found. It's crucial to find out the concentration of H+ ions and know that that number is equal to the concentration of OH- ions. Those numbers can then be divided by the volume (in L) of the substance with the unknown molarity. (If that doesn't make much sense, see the problems below)

Problems:

3.) How many mL of 0.1 M NaOH would be needed to neutralize 2.0 L of 0.050 M HCl? 
First, know that 2.0 L = 2,000 mL.

0.050 M / 2,000 mL = 0.1 M / x mL
0.050 M( x mL) = 0.050x
2,000 mL( 0.1 M) = 200
200 / 0.050 = 4,000 mL

It would take 4,000 mL of NaOH to neutralize the HCl. 
Another way to look at this is to realize that the molarity of the HCl is 1/2 the molarity of the NaOH solution. This means you can convert the HCl volume into mL (2,000 mL) and multiply that by 2 to get 4,000 mL.

5.) A student mixes 100 mL of 0.20 M HCl with different volumes of 0.50 M NaOH. Are these final solutions acidic, basic, or neutral?
a.) 100 mL of 0.20 M HCl + 20 mL of 0.50 M NaOH Acidic (because there's more of the solution with the lower molarity and that overpowers the base)
b.) 100 mL of 0.20 M HCL + 40 mL of 0.50 M NaOH Neutral (there is enough base to neutralize the acid, despite their different molarities)
c.) 100 mL of 0.20 M HCl + 60 mL of 0.50 M NaOH Basic (if 40 is the neutral point, anything above that as far as the NaOH is concerned should be basic.)