Chapter 5 Acid-Base Volumetric Analysis ✓ Flashcards
(38 cards)
Define acid-base titration
It is an analytical procedure involving a reaction between an acid solution and a base solution
Define standard solution
a solution with a precisely known concentration
What is the purpose of a titration?
Purpose is to determine the concentration of the acid or base in the solution of unknown concentration
Define aliquot
The measured solution that comes out of the pipette
Define titre
the measured solution that is in the burette
Define equivalence point
It is the theoretical point in the titrations (when neither nor base remains)
What is used to to see the equivalence point of a titration?
An acid-base indicator
Define end point of a titration
The point where the indicator changes colour signalling the equivalence as occurred.
What two points are very close approximations in titration?
The end point and the equivalence point
Define primary standard
One of the accurately known concentration that can be used in a titration to find the concentration of other reagents.
How is a primary standard prepared?
It is prepared by carefully weighing the sample of the primary standard, dissolving it in distilled water and then making the solution up to a precise volume in a volumetric flask.
What features must a primary standard have?
It must:
-Be able to be obtained in a very pure form consistent with its chemical formula
-Be sufficiently stable so that on exposure to air, it doesn’t readily change its water content or react with other gases in the air
-Have a relatively high molar mass
These features allow the moles of a primary standard to be precisely calculated from its accurately measured mass and molar mass
Define secondary standard
A solution whose concentration is determined by titration against a primary standard
What problems come with finding the concentration of a secondary standard?
Finding the concentration of a secondary standard involves extra measurements and procedures which means its concentration will be subjected to greater uncertainty (error) than the concentration of a primary standard solution.
Describe the procedure of a titration
-In an acid-base titration a carefully measured volume, called an aliquot, usually 20mL of either the acid or BASE solution is measured using a pipette.
-This solution is transferred to a clean conical flask that has been pre-rinsed with distilled water. A few drops of a suitable pH indicator is also added to the conical flask
-The other solution is placed into a clean burette that has been pre-rinsed with some of the same solution with which it is to be filled.
-Reagents is then released in a controlled way from the burette into the aliquot present in the conical flask.
-An indicator colour changes signals the end point of the titration and no further reagent is added.
-The volume of solution added from the burette, known as the titre, can be found by subtracting the initial burette reading from the final burette reading.
Define meniscus
The curve that exists on the surface of a liquid when it is placed into a container.
20mL aliquots of HCl solution are titrated with 9.039 x 10-2 mol L-1 Na2CO3 (sodium carbonate). Several titrations were performed and the volumes of Na2CO3 solution used were 19.45mL, 18.90mL, 18.77mL and 18.85mL.
What is the concentration of the HCl?
Write equation
2HCl + Na2CO3 -> 2NaCl + H2O + CO2
Step 1: Find average titration volume
Note: When finding the average titration volume, discard any inconsistent data. Typically values should be within a range of 0.4mL.
(18.90 + 18.77 + 18.85)/3 = 18.84mL
Step 2: Find moles of Na2CO3
n(Na2CO3) = c x v -> (9.039 x 10-2) x (18.84 x 10-3)
Step 3: Find moles of HCl (by looking at the equation)
n(HCl) = 2 x n(NaCO3) = 2 x 1.703 x 10-3 = 3.406 x 10-3
Step 4: Find the concentration of HCl
c(HCl) = n/v =(3.406 x 10-3)/(20.00 x 10-3) = 1.703 x 10-1 mol L-1
What are the two types of errors that can occur in a titration?
Systematic error and random error
What can a systematic error cause in a titration and what is an example of a systematic error?
May causes results to be consistently lower or higher than the actual value.
An example of a systematic error is using the wrong indicator
How can random errors occur in a titration?
Random errors can be due to:
-Inconsistent rinsing techniques
-Careless attention to the colour change at the end point
-Or some other poor technique
Small random errors also occur due to limitations in the instruments used and the ability of the operator to use them correctly
How can you eliminate systematic errors and random errors?
Systematic errors: Using the correct technique
Random error: Can be reduced but never eliminated
What are some procedures that minimise error?
-Rinsing glassware: The final rinse for a pipette, burette or storage bottle is always with the the reagent they are fill in. For a volumetric flask and conical flask the final rinse is with distilled water
-Indicator: Use only a few drops and choose one with a distinct colour change close to the equivalence point pH.
-Reading scales: Avoid parallax error. (Parallax error is when the pointer of a device looks like it’s at a different reading when read to the side compared to when read face-on). Read all scale markings horizontal at eye level.
-Discharging a pipette: Allow suitable drainage time (20 seconds) and leave the small amount of liquid that remains in the pipette tip
-End point: When close to the equivalence point, use a wash bottle to rinse down the insides of the conical flask and the add reagent dropwise while swirling the flask contents until a suitable colour is achieved
What equation is used to find the dilution of a solution in a titration?
C1V1 = C2V2
Volume in litres
Concentration in mol L-1
A student analysed the ethanoic acid concentration of a commercial brand of vinegar. To do this a 25mL sample of the vinegar was diluted to 500mL in a volumetric flask. 20mL aliquots of the diluted vinegar were analysed by titration and found to have an ethanoic acid concentration of 0.0429 mol L-1. Determine the concentration of ethanoic acid in the original undiluted vinegar.
Equation: C1V1 = C2V2
C1 = ?
V1 = 25mL = 0.0025L
C2 = 0.0429 mol L-1
V2 = 500mL = 0.5L
C1 = (0.0429 x 0.5)/0.025
C1 = 0.858 mol L-1
