Chemistry: Titrations


I found a bottle in the stockroom of my lab about a year ago. When I opened it, I knew from the burning smell that it was a bottle of nitric acid. However, I had absolutely no idea what the concentration of the acid was. Without knowing the concentration of the acid, it would be difficult to find a use for it.

Fortunately, there's a way to solve this problem. As I remembered from way back in Chemical Equations, acid-base reactions occur when an acid and base combine by the following general equation:

  • HA + BOH ⇔ BA + H2O
The Mole Says

In the lab, it's common to determine the point at which a solution has become neutralby using an indicator. When the indicator turns from the color for an acid to the colorfor a base, the titration has taken place. The use of indicators, however, does cause error because they don't change color at a pH of exactly 7.00. As a result, the point at which an indicator changes color in a titration is the "endpoint" (distinguishing it from the "equivalence point").

This sparked the following line of reasoning in my brain:

  • Every OH- I add to this solution will neutralize one of the H+ ions.
  • If I keep adding base to my nitric acid, it will eventually turn into neutral water with a pH of 7.00.
  • If I keep track of how much base I put into the acid, the quantity of acid when the pH is equal to 7.00 will be equal to the amount of base I've added.

When the solution is perfectly neutral (called the "equivalence point"), the number of moles of acid that I started with will be equal to the number of moles of base that I added to make them neutral. As a result, we get the very handy equation:

MaVa = MbVb

for neutralization reactions at a pH of exactly 7.00. Titration is the process in which neutralization reactions are used to determine the concentration of either an acidic or basic solution.

Here's how I did my experiment:

You've Got Problems

Problem 4: If it took 245 mL of 0.500 M HCl to neutralize 175 mL of a NaOH solution, what is the concentration of the NaOH solution?

I placed 175 mL of nitric acid into a beaker and added 1.00 M NaOH solution to it. I found that the solution was completely neutral after I had added 365 mL of sodium hydroxide to the acid.

Using the earlier equation, M1 = 175 mL, V1 is unknown, M2 = 1.00 M, V2 = 365 mL, we find that:

  • (175 mL)V1 = (1.00 M)(365 mL)
  • V1 = 2.09 M

The nitric acid had a molarity of 2.09 M. Why anybody would make a solution with this molarity, I have no idea, but that's what it was!

Excerpted from The Complete Idiot's Guide to Chemistry © 2003 by Ian Guch. All rights reserved including the right of reproduction in whole or in part in any form. Used by arrangement with Alpha Books, a member of Penguin Group (USA) Inc.

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