Chemistry 30

3.2 Acid-Base Titrations

Acid-base titrations are lab procedures used to determine the concentration of a solution. We will examine it's use in determining the concentration of acid and base solutions. Titrations are important analytical tools in chemistry.

During an acid-base titration, an acid with a known concentration (a standard solution) is slowly added to a base with an unknown concentration (or vice versa). A few drops of indicator solution are added to the base.

The indicator will signal, by colour change, when the base has been neutralized (when [H⁺] = [OH^-]).

At that point - called the equivalence point or end point - the titration is stopped. By knowing the volumes of acid and base used, and the concentration of the standard solution, calculations allow us to determine the concentration of the other solution.

It is important to accurately measure volumes when doing titrations. The instrument you would use is called a burette (or buret).

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Titration Procedure

Here's a brief summary of how you would carry out a titration in your chemistry lab. We'll use an example in which we want to determine the concentration of a base. Typically you will need two burettes - one for the acid solution and the other for the base.

See the "External Links" section for more information and simulations;
Google "titration" or "titration simulator" for more examples

1.	Rinse the burettes several times with the acid and base solutions, then fill them with the appropriate solution.
2.	Using the burette, accurately measure a volume of the base into an Erlenmeyer flask. How much you use is not important, as long as you know exactly how much is in the flask.
3.	Add a suitable indicator such as phenolphthalein to the flask. (see the note at the bottom of the page regarding indicator choice)
4.	The acid (with the known concentration - the titrant) is then added to the base. Release the acid from the burette relatively quickly at first, then much more slowly as the endpoint is neared.
	How do you know when you are reaching the endpoint? The indicator will begin to show a change in colour. Swirling the flask will cause the colour to disappear.
	ENDPOINT IS REACHED AS SOON AS THE COLOUR CHANGE IN PERMANENT. ONE DROP WILL DO IT - once the colour change has occurred, stop adding additional acid. Do NOT continue adding until you get a deep colour change - you just want to get a permanent colour change that does not disappear upon mixing.
	If a pH meter is used instead of an indicator, endpoint will be reached when there is a sudden change in pH.
5.	Record the volume of acid added to reach endpoint. Calculations will allow us to calculate the concentration of the base.
6.	The titration should be repeated several times and the results averaged. Often the first trial is done relatively quickly in order to get a general idea of the volume of titrant required. Subsequent trials are done more slowly.

 

Titration Calculations

To use titration information to calculate the concentration of the unknown solution, you must know the following information. Note the abbreviations that will be used in our calculations.

• The concentration of one of the solutions, the acid for example (M_A)
• The volume of acid used for the titration (V_A)
• The volume of base used for the titration (V_B)

What you will calculate:

• The concentration of the other solution, the base for example (M_B)

We will work through two examples. The first example will show the details of the theory behind the calculations. The second example will skip the theory and go more directly to the key formula you will use.

 

 

Example 1

During a titration 75.8 mL of a 0.100 standard solution of HCl is titrated to end point with 100.0 mL of a NaOH solution with an unknown concentration. What is the concentration of the NaOH solution?

Solution
	Begin with a balanced equation for the reaction: HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l) We see from the equation that the ratio between HCl and NaOH is 1:1.
	Moles of acid used during the titration can be calculated by multiplying concentration × volume: moles = concentration × volume moles Acid =   moles L × L       = MAVA Moles of base are calculated in the same way. moles Base =   moles L × L           = MBVB
	Since a 1:1 ratio exists between our acid and our base, equivalence will be reached when moles of acid = moles of base, or	MAVA = MBVB
	Clearly identify your variables and their values. Be careful - it is very common for students to mix up the numbers - acid numbers must stay with the acid, base values must match up with the base. Also, check the unit for volume - you may leave the volume in mL as long as you do so for both the acid and the base. MA = 0.100 M   MB = MB VA = 75.8 mL   VB = 100.0 mL Substitute in known values for the equation and solve for the unknown: MAVA = MBVB (0.100)(75.8) = MB(100.0) 7.58 100.0 = MB 0.0758 = MB Answer: The concentration of the NaOH solution is 0.0758 M or 7.58 ×10-2M

 

Example 2

Our second example will involve a reaction that does not involve a 1:1 ratio between the acid and the base.

A 20.0 mL solution of strontium hydroxide, Sr(OH)₂, is placed in a flask and a drop of indicator is added. The solution turns colour after 25.0 mL of a standard 0.0500M HCl solution is added. What was the original concentration of the Sr(OH)₂ solution?

Solution
	Write a balanced equation for the neutralization reaction: 2 HCl(aq) + Sr(OH)2 (aq) → SrCl2 (aq) + H2O(l) Notice that 2 moles of the acid HCl are required to neutralize 1 mole of the base Sr(OH)2.
	Define the variables to be used and the known values: MA = 0.050 M   MB = MB VA = 25.0 mL   VB = 20.0 mL   To balance this out we need to modify our formula: MAVA = 2 MBVB Notice that the "2" is on the base side of our formula even though it was in front of the acid side in the balanced equation - it "switches places". If you want to see a different approach to solving this question, click here. Substitute in known values for the equation and solve for the unknown: MAVA = 2MBVB (0.050)(25.0) = 2MB(20.0) 1.25 = 40.0MB 1.25 40.0 = MB 0.0312 = MB Answer: The concentration of the Sr(OH)2,solution is 3.12 ×10-2M.

This lesson concludes our unit on acids and bases. Work on the practice exercises (Set 3) before completing Assignment 3. Calculations are a critical objective of this unit, so be sure to understand them well.