In my experience, a method of determining the concentration of an unknown substance by adding amounts of a standard solution, that is, one whose concentration is known. The only practical lab experience which I have had with titration is an acid-base titration. The idea is something like this: one has an acid of an unknown concentration and a base of a known concentration in two graduated burettes. One adds an amount of the acid and an amount of an indicator (the indicator should be provided by whoever set up the lab since only he or she will know what pH range is appropriate) to a solution. One then adds the base until whatever color change the indicator will undergo happens. One then attempts to return the solution to its original color. Once the solution can change color with one drop of either solution, the solution has reached the end point. The end point represents the point in the chemical reaction in which the concentrations of both substances are equally. This, obviously, does not mean that the volumes are the same, but they could be. By recording the volume of each substance added, one can then perform several stoichiometric calculations the find the concentration of the unknown solution. Please excuse me if this write up is inaccurate, I'm just a high school chemistry student.

Hey kids! Now you can perform a titration lab in your very own home!


The purpose of this lab is to determine the concentration of a reagent, using an indicator solution and a reagent whose concentration is known.



First, we must prepare an exact volume of our known solution, to which the unknown will be added. Fill your pipette (look under that node for proper pipetting procedure) and drain it completely into the Erlenmeyer flask. Add a few drops of your indicator solution—how many depends on how much solution you're titrating, but about 10 should do it—and you're ready to go.

Pour your unknown solution into the buret. You needn't fill the buret up to the top, as long as you record the reading, but if you fill it all the way (to 0, since the buret measures liquid dispensed) then the subtraction is easier. Look closely at the bottom of the buret (down where the stopcock is), and flick away any air bubbles that have formed. Clamp the buret over the flask and you're ready to begin.

Turn the stopcock so that the knob is parallel to the buret to begin dispensing solution, swirling the flask to ensure that the reagents are reacting. You will note that the solution in the flask will turn purple (if you're using phenolphthalein; I'm not sure what color you'll get with other indicators) briefly and then return to clear. This is because the phenolphthalein turns pale purple when present in a roughly neutral solution (pH=7 or thereabouts; any lower and the solution is clear, any higher and it turns red). But the purple rapidly vanishes as the flask is swirled, because the overall pH is still too low for the solution to change color.

As more and more titrant is added, the solution will remain pale pink for longer periods of time. Slow down the rate at which you add the solution—just a few drops at a time. When the solutions stays pink (the color you're looking for is somewhere between violet and plum) even after being swirled for a few seconds, you're done. Record the final volume in the buret.

Sample data

Note: these numbers have all been made up. You have been warned.

Known-concentration solution (1M HCl):

  • volume: 15 mL
  • pH: 0

Unknown-concentration solution (?M NaOH):

  • initial buret reading: 0.3 mL
  • final buret reading: 7.8 mL

Sample Calculations

total NaOH dispensed: 7.8-0.3 = 7.5 mL

reaction equation: NaOH(aq) + HCl(aq) --> NaCl(aq) + H2O(l)

Since final pH = 7, we see that

moles of HCl = moles of NaOH dispensed

moles of HCl = M· L = 1M· 0.015L = 0.015 mol HCl

0.015 mol NaOH = M· L = M· 0.0075L

M = 0.015 / 0.0075 = 2M NaOH

And, just for grins, let's find the pH (go there for a full explanation of these equations):

reaction for NaOH dissolving in water: NaOH(s) --> Na+(aq) + OH-(aq)

M(NaOH) = M(OH-) = 2

pOH = -log[OH-] = -log(2) = -0.3

pH = 14-pOH = 14.3

Source: my own lab work. This is only one procedure, of course; many variations are possible. If you've used a radically different procedure from mine (or if I'm flat wrong in my calculations, always a possibility), drop me a /msg.

Ti*tra"tion (?), n. Chem.



© Webster 1913.
