The Burette's Measure: Mastering Titration Volume
The Anatomy of a Burette and How to Read It
A burette is a long, graduated glass tube with a precision tap, known as a stopcock, at its bottom end. It is designed to deliver variable, accurately measurable volumes of a liquid, called the titrant. The scale on a burette is unique. Unlike a graduated cylinder where you read from the bottom up, a burette's scale runs from top to bottom. This means the zero mark is at the top, and the numbers increase as you go down. This design allows you to directly read the volume of liquid that has been dispensed.
Reading a burette correctly is a skill that requires practice. The liquid inside forms a curved surface called a meniscus. To read the volume accurately, your eye must be level with the bottom of this meniscus. Parallax error, which occurs when the scale is viewed from an angle, is a common mistake. Always ensure the burette is vertical and read the volume at the meniscus's lowest point. The scale is typically marked in 0.1 mL increments, and you should estimate the reading to the nearest 0.01 mL.
Reading a Burette Correctly:
- Ensure the burette is perfectly vertical.
- Position your eye at the same level as the bottom of the meniscus.
- Record the volume where the bottom of the meniscus lines up with the scale markings.
- Estimate the final digit (e.g., if it is between 24.3 mL and 24.4 mL, you might record 24.35 mL).
Calculating the Titration Volume
The volume of titrant added during a titration is not a single reading but the difference between two readings: the initial and the final volume. The calculation is straightforward but must be done with care.
Titration Volume (V) = Final Burette Reading - Initial Burette Reading
For example, if you start with an initial reading of 0.45 mL and the titration ends at a final reading of 23.60 mL, the volume of titrant used is:
V = 23.60 mL - 0.45 mL = 23.15 mL
It is good practice to perform a titration at least two or three times (these are called trials) to ensure the results are consistent and reproducible. The volumes from these trials are then often averaged to get a more reliable value for the calculation.
| Trial | Initial Reading (mL) | Final Reading (mL) | Volume of Titrant Used (mL) |
|---|---|---|---|
| 1 | 0.45 | 23.60 | 23.15 |
| 2 | 0.50 | 23.70 | 23.20 |
| 3 | 0.40 | 23.55 | 23.15 |
| Average Volume | 23.17 mL | ||
From Volume to Concentration: The Stoichiometric Bridge
Once you have the average volume of titrant used, you can determine the concentration of the unknown solution. This is where chemistry and math meet. The core principle is that the reaction between the titrant and the analyte continues until the equivalence point1 is reached, the moment when the amount of titrant added is stoichiometrically equivalent to the amount of analyte in the flask.
Let's consider a classic example: titrating hydrochloric acid (HCl) of unknown concentration with a standard solution of sodium hydroxide (NaOH) of known concentration, say 0.100 M. The balanced chemical equation is:
$ HCl + NaOH -> NaCl + H_2O $
This equation shows a 1:1 mole ratio between HCl and NaOH. If the average volume of 0.100 M NaOH used to neutralize 25.00 mL of the HCl solution is 23.17 mL, we can calculate the concentration of HCl.
Step 1: Calculate moles of NaOH used.
Moles of NaOH = Concentration (M) × Volume (L) = 0.100 mol/L × 0.02317 L = 0.002317 mol
Step 2: Use the mole ratio to find moles of HCl.
From the equation, 1 mol NaOH reacts with 1 mol HCl.
So, Moles of HCl = Moles of NaOH = 0.002317 mol
Step 3: Calculate the concentration of HCl.
Concentration of HCl = Moles / Volume (L) = 0.002317 mol / 0.02500 L = 0.09268 M
Thus, the concentration of the hydrochloric acid solution is approximately 0.0927 M.
A Practical Example: Testing Vinegar's Acidity
Let's apply this to a real-world scenario. The concentration of acetic acid in vinegar can be found by titration. You titrate 10.00 mL of vinegar (the analyte) with a standardized 0.500 M NaOH solution (the titrant), using phenolphthalein as an indicator. The average titration volume from your trials is 16.40 mL of NaOH.
The balanced equation is: $ CH_3COOH + NaOH -> CH_3COONa + H_2O $
This is also a 1:1 mole ratio.
Step 1: Moles of NaOH = 0.500 mol/L × 0.01640 L = 0.00820 mol
Step 2: Moles of CH_3COOH = 0.00820 mol (from the 1:1 ratio)
Step 3: Concentration of CH_3COOH = 0.00820 mol / 0.01000 L = 0.820 M
This means the vinegar has an acetic acid concentration of 0.820 M. You could also express this as a mass/volume percentage, a common unit on food labels.
Common Mistakes and Important Questions
Q: What is the difference between the equivalence point and the endpoint?
The equivalence point is the theoretical point where the moles of titrant and analyte are stoichiometrically equal. The endpoint is the point we observe in the lab, usually signaled by a color change in an indicator. A good experiment uses an indicator whose endpoint is as close as possible to the true equivalence point. A small mismatch between the two is a potential source of error.
Q: Why is it important to record the initial burette reading, and not just assume it's zero?
It is a common mistake for students to fill the burette to the zero mark and assume the initial reading is 0.00 mL. However, the zero mark may not be perfectly accurate, and the meniscus might not be exactly at zero. To avoid a systematic error, you must always record the actual initial reading, whatever it is. The calculation (Final - Initial) automatically corrects for any offset, as long as you read the scale consistently.
Q: My titration volumes for different trials are not close. What could have gone wrong?
Inconsistent results between trials can stem from several sources: misreading the burette (parallax error), inaccurate detection of the endpoint (adding titrant too quickly near the end), an unclean burette or pipette that affects volumes, or a leaky stopcock. To improve consistency, perform the titration slowly, especially near the endpoint, swirl the flask continuously, and ensure all glassware is properly cleaned and rinsed.
Footnote
1 Equivalence Point: The point in a titration where the amount of titrant added is chemically equivalent to the amount of analyte present in the sample. It is the theoretical completion of the reaction.