ABSE_Titration Lesson

Titration

Calculate the volume of a 0.200 M KOH solution that is needed to neutralize 25.00 mL of a 0.115 M HCI solution.
(Vacid) (Macid) (#H⁺) = (Vbase) (Mbase) (#OH-) (25.00 mL)(0.115 M)(1) = (Vbase) (0.200 M)(1) Vbase=14.4 mL
What volume of 0.075 M HCI is required to neutralize 100 ml of 0.01 M Ca(OH)2 solution?
(Vacid) (Macid) (#H⁺) = (Vbaso) (Mbase) (#OH-) (Vacid) (0.075 M) (1) (100.0 mL) (0.01 M)(2) Vacid = 26.7 mLTitration is a technique used often in acid/base reactions to determine a quantity, often the molarity, of an unknown solution. One solution is placed in the buret, the long tube pictured below. The solution in the buret is called the titrant. The titrant is delivered to the receiving flask by a valve at the bottom of the buret called the stopcock. The receiving flask contains the analyte and one or two drops of indicator which changes color when the appropriate amount of titrant is added. This color change can occur very quickly, so the titrant must be added very slowly. It is very important not to go past the color change. The titration should stop as soon as the color change lasts for 30 seconds, while the flask is swirled or stirred with a magnetic stirrer. The color change does not have to last past this time in order for the titration to be complete. When the color change occurs, the titration is said to have reached its endpoint. Ideally, this is the same as the equivalence point. The equivalence point is the point where the moles of acid and base are equal. This can also be referred to as the stoichiometric point. In order for the endpoint and equivalence point to be the same, the appropriate indicator must be chosen. We will discuss how to choose an indicator at the end of this lesson.

Strong Acid / Strong Base Titration

Titration Curve on a graphWhen a titration is performed, pH readings can be taken by a pH meter throughout the titration. A plot of pH versus volume of titrant added is called a titration curve. These curves give us lots of valuable information. Let's begin by looking at the titration of a strong acid with a strong base.

Consider what happens to the pH of 25 mL of 0.2 M HCl as it is titrated with 25 mL of 0.2 M NaOH. This result is shown in the titration curve shown to the right.

As you add base, [H+] decreases and pH increases. Near the equivalence point, the pH increases rapidly. For this titration, the equivalence point is at pH = 7. This is the characteristic endpoint of a titration of a strong monoprotic acid with a strong base. Links to an external site.

Titration Calculations - String Acid / Strong Base

If you did not have a titration curve for reference, you could calculate the pH at various points during the titration. Before the equivalence point, the pH is determined by the amount of H+ not neutralized. At the equivalence point, the moles of H+ equal the moles of OH-. After the equivalence point, the pH is determined by the amount of OH- in excess.

Since the method for determining pH depends upon the point in the titration relative to the equivalence point, we need to start by determining the equivalence point. This could be determined experimentally by actually performing the titration or it could be determined mathematically. A variation of the dilution equation we learned in module 3 can be used to determine the equivalence point for a strong/strong titration.   The dilution equation is: Vi x Mi = Vf x Mf. We can adapt this to acid/base neutralization (for strong acids and strong bases): Vacid x Macid = Vbase x Mbase. But, this equation does not take into consideration acids or bases that have more than one H+ or OH-. We can easily fix that by adding the number of H+ and OH­- to the equation.  

NEUTRALIZATION EQUATION
Vacid X Macid X #H+ = Vbase X Mbase X #OH-

Calculate the volume of a 0.200 M KOH solution that is needed to neutralize 25.00 mL of a 0.115 M HCI solution.
(Vacid) (Macid) (#H⁺) = (Vbase) (Mbase) (#OH-) (25.00 mL)(0.115 M)(1) = (Vbase) (0.200 M)(1) Vbase=14.4 mL
What volume of 0.075 M HCI is required to neutralize 100 ml of 0.01 M Ca(OH)2 solution?
(Vacid) (Macid) (#H⁺) = (Vbaso) (Mbase) (#OH-) (Vacid) (0.075 M) (1) (100.0 mL) (0.01 M)(2) Vacid = 26.7 mL

So, using this equation will allow you to determine the volume that represents the equivalence point for a strong/strong titration.

The following quantities of 0.100 M NaOH have been added to 50.0 mL of 0.100 M HCI. Calculate the pH of the resulting solutions.
Before titration begins:
This is asking us to calculate the pH of the acid before any base has been added, in other words, before the titration begins. To calculate pH of a strong acid, just use the pH equation.
pH = -log(0.100)
pH = 1
Now, we need to determine the equivalence point. Vacid X Mi++ X #H* = Voase X Mbase X #OH (50.0 mL)(0.100 M)(1) = (Vbase) (0.100 M)(1) Vbase = 50.0 mL
This is the equivalence point.
49.00 mL of 0.100 M NaOH added This is before equivalence point. This means that all the OH added will react and only the left over H will determine the pH.
IL
0.1 mol (0.1:
? mol H⁻= 50.00 mL(1L/1000mL)(0.1mol/1L)=5.00x10⁻³ mol present reacts with
? mol OH - =49.00 mL(1L/1000mL)(0.1mol/1L)=4.90x10⁻³ mol
H⁺ left = 5.00x10⁻³mol-4.90x10⁻³mol=1x1010⁻⁴mol H⁺
[H⁺] = 1 x 10⁻⁴/99 x 10³L 1 x 10⁻⁴ mol H⁺
pH = -log(1 x 10³)
pH = 3.00
(Note that this is just one way to solve this. You may have a different way that
you prefer.)
50.00 mL of 0.100 M NaOH added
This is at equivalence point.
pH = 7
(Note that this is only true for strong/strong titrations.)
50.10 mL of 0.100 M NaOH added This is after equivalence point.
All H+ reacts, OH- is in excess by 0.1 mL
? mol OH = 1L /1000 mL)(0.1 mol/1 L)
=1 x 10⁵ mol
[H] 1 x 10 mol 100.1 x 10⁵L 9.99 x 10'M
pOH = -log(9.99 x 10⁻⁵) 
pOH = 4.00, so pH = 10.00

Titration Calculations - Weak Acid / Strong Base

Calculating the pH of a weak acid or base at various points during a titration can be overwhelming. We have broken this down into the following to make it easier to understand. In addition, you can download a handout that goes along with the video:

As you can see, titration problems require a lot of practice. But, if you will first determine which point you are at in the titration (before, at, or after equivalence), you will know which type of problem you have and therefore how to approach it.

It is important not only to be able to do the math associated with titration, but also to understand it conceptually. 

Titration Calculations

It is interesting that although we describe the overall reaction in the titration as neutralization, at the equivalence point the solution is slightly basic, not neutral.

Another interesting and very useful concept is that at halfway to the equivalence point the concentration of weak acid is equal to the concentration of its conjugant base. Therefore the pH=pKa. This point is called the half-neutralization because half of the acid has been neutralized. You can use this fact to determine the half equivalence point, and therefore the equivalence point!

At Half the Equivalence Point
pH = pKa

Titration Calculations - Weak Base / Strong Acid

The titration of a weak base with a strong acid follows similar patterns as those we discussed above. 

Titration Curves

Click through the tutorial below on how to interpret titration curves.  

 

Test your understanding of the titration of a weak acid with a strong base by filling in this summary:

  1. The pH of a weak acid is initially higher than a strong acid of the same concentration This is because a weak acid ionizes less than a strong acid.
  2. The pH rises rapidly in the early part of a titration, but more slowly near the equivalence point because it becomes an effective buffer near the equivalency point.
  3. The equivalence point is not at pH=7 because the resulting salt solution contributes to the pH.

Watch this "Titration Roundup" video from Khan Academy for a review of interpreting titration curves.

Indicators

Ideally, in a titration, the endpoint (the point where the color changes) should equal the equivalence point (the point where moles of acid = moles of base).   In order for this to be true, we must choose an appropriate indicator. A useful equation for selecting an indicator is:    

pKin = pH (at the equivalence   point)

Most dyes that work as indicators are themselves weak acids. We represent this as HIn. The unionized HIn has one color, and its conjugate base, In-, has another color.  

HIn       H+   +   In-

LaTeX: k_{ln}=\frac{[H^+][]ln^-}{[Hln]}kln=[H+][]ln[Hln]

When in a very acidic solution, [H+] is high and equilibrium shifts left, the color therefore is that of HIn. If a solution becomes more basic (a decrease in H+) the equilibrium shifts right and the color is that of In-.

The color change occurs over a range of pH values. There is a large Δ pH from one drop of titrant added to another near the equivalence point, but the number of moles of titrant added does not dramatically change from one drop to the next. Because the pH change for strong/strong titrations is large near the equivalence point, the indicator need not change color at exactly pH = 7. Phenolphthalein is often the choice for strong/strong titrations. Its color change occurs during a pH change of 8.3 to 10, but just a slight excess (a few drops) of the titrant will cause this change.

Remember to work on the module practice problems as you complete each section of content.  

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