CK-12-Chemistry Intermediate

(Marvins-Underground-K-12) #1

21.4. Acid-Base Neutralization http://www.ck12.org


in an acidic or basic solution. A commonly used indicator for strong acid-strong base titrations is phenolphthalein.
Solutions in which a few drops of phenolphthalein have been added, turn from colorless to brilliant pink as the
solution turns from acidic to basic (Figure21.10). The steps in a titration reaction are outlined below.



  1. A measured volume of an acidic solution whose concentration is unknown is added to an Erlenmeyer flask.

  2. Several drops of an indicator are added to the acid and mixed by swirling the flask.

  3. A buret is filled with a basic solution of known molarity.

  4. The stopcock of the buret is opened and base is slowly added to the acid while the flask is constantly swirled
    to ensure mixing. The stopcock is closed at the exact point at which the indicator just changes color. If the
    color change does not remain after swirling, the neutralization reaction has not yet reached completion.


FIGURE 21.10


In the titration of an acid by a base, the
addition of the base causes the indicator
phenolphthalein to change from colorless
to pink.

Thestandard solutionis the solution in a titration whose concentration is known. In the titration described above,
the base solution is the standard solution. It is very important in a titration to add the solution from the buret slowly
so that the point at which the indicator changes color can be found accurately. Theend pointof a titration is the
point at which the indicator changes color. When phenolphthalein is the indicator, the end point will be signified by
a faint pink color.


Titration Calculations


At the equivalence point in a neutralization, the moles of acid are equal to the moles of base, assuming they react in
a 1:1 ratio according to the balanced neutralization equation.


moles acid = moles base

Recall that the molarity (M) of a solution is defined as the moles of the solute divided by the liters of solution (L).
The moles of solute are therefore equal to the molarity of a solution multiplied by the volume in liters.


moles solute = M×L
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