4.2 Acid-base Titrations

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AS - Level Chemistry (4 - Acids and Redox) Mind Map on 4.2 Acid-base Titrations, created by Bee Brittain on 11/04/2016.
Bee Brittain
Mind Map by Bee Brittain, updated more than 1 year ago
Bee Brittain
Created by Bee Brittain over 8 years ago
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Resource summary

4.2 Acid-base Titrations
  1. Titrations
    1. Technique used to accurately measure the volume of one solution that reacts exactly with another solution
      1. Titrations can be used for...
        1. ... finding the concentration of a solution
          1. ... Identification of unknown chemicals
            1. ... finding the purity of a substance
            2. Checking purity is an important aspect of quality control, especially for compounds manufactures for human used such as medicines
            3. Preparing a Standard Solution
              1. Solution of known concentration
                1. A volumetric flask is used to make up a standard solution very accurately
                  1. Measure volumes precisely
                  2. 1) The solid is first weighed accurately
                    1. 2) The solid is dissolved in a beaker using less distilled water than will be needed to fill the volumetric flask to the mark
                      1. 3) This solution is transferred to a volumetric flask. The last traces of the solution are rinsed into the flask with distilled water
                        1. 4) The flask is carefully filled to the graduation line by adding distilled water a drop at a time until the bottom on the meniscus lines up exactly with the mark. Care at this stage is essential - if too much water is added to solution will be too dilute and must be prepared again. You should view the graduation mark and meniscus at eye level for accuracy
                          1. 5) Finally, the volumetric flask is slowly inverted several times to mix the solution throughly. If this stage is omitted, titration results are unlikely to be consistent. You will be able to see the solution mixing when you invert the flask as the more dense origional solution moves through the solution
                          2. Acid-base Titrations
                            1. Solution of acid is titrated against a solution of a base using a pipette and a burette, which are typically manufactured to the tolerances below:
                              1. a 10cm3 pipette: -+ 0.04cm3
                                1. a 25cm3 pipette: -+ 0.06cm3
                                  1. a 50cm3 pipette: -+ 0.10cm3
                                    1. A burette reading is recorded to the nearest half division. Each reading is measured to the nearest -+ 0.05cm3 so the reading always has two decimal places, the last place being either 0 or 5
                                    2. 1) Add a measured volume of one solution to a conical flask using a pipette
                                      1. 2) Add the other solution to a burette and record the initial burette reading to the nearest 0.05cm3
                                        1. 3) Add a few drops of an indicator to the solution in the conical flask.
                                          1. 4) Add the solution in the burette into the solution in the conical flask a bit at a time, swirling the conical flask throughout to mix to the solutions. Eventually the indicator changes colour at the end point of the titration. The end point is used to indicate the volume of one solution that eactly reacts with the volume of the second solution
                                            1. 5) Record the final burette reading. The volume of solution added from the burette is called the titre, which is calculated by subtracting the initial reading from the burette reading
                                              1. 6) A quick, trial titration is carried out first to find the approximate titre.
                                                1. 7) The titration = repeated. Carries out until two accurate titres are concordant - agreeing within 0.10cm3.
                                                  1. The mean titre
                                                    1. Use only your closest accurate titres
                                                      1. By repeating titres until two agree within 0.10cm3, you can reject inaccurate titres
                                                        1. If you were to include all the titres in the mean, you have lost the accuracy of the titration technique
                                                    2. Titration Calculations
                                                      1. From the results of a titration you will know both the concentration and the reacting volume of one of the solution, and only one reacting volume of the other solution.
                                                        1. Step 1: Work out the amount, in mol, of the solute in the solution for which you know both the concentration and volume
                                                          1. Step 2: Use the equation to work out the amount, in mol, of the solute in the other solution
                                                            1. Step 3: Work out the unknown information about the solute in the other solution
                                                              1. WORKED EXAMPLE
                                                                1. Pipette: 25.00cm3 of 0.100moldm-3 KOH (aq)
                                                                  1. Mean titre from burette: 25.70cm3 of H2SO4 (aq)
                                                                    1. Unknown Information: The concentration of H2SO4 (aq)
                                                                      1. Step 1: From the titration results, calculate the amount of KOH
                                                                        1. n(KOH) = (c x V) / 1000 = (0.100 x 25) / 1000 = 0.00250mol
                                                                          1. Step 2: From the equation and step 1, determine the amount of H2SO4
                                                                            1. 2KOH (aq) + H2SO4 (aq) --> K2SO4 (aq) + 2H2O (l)
                                                                              1. 2 mol of KOH and 1 mole of H2SO4
                                                                                1. 2:1 Ratio
                                                                                  1. 0.00250mol of KOH so there must be 0.00125 mol of H2SO4
                                                                              2. Step 3: Work out the Unknown Information
                                                                                1. n = (c x V) / 1000 is rearranged to give c = 1000n / V
                                                                                  1. c(H2SO4) = (1000 x 0.00125) / 25.70 = 0.0486moldm-3
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