Acid-Base Titrations with Indicators (Edexcel International A Level Chemistry)

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Acid-Base Titrations with Indicators

  • Acid-base titrations are used to find the unknown concentrations of solutions of acids and bases
  • Acid-base indicators give information about the change in chemical environment
  • They change colour reversibly depending on the concentration of H+ ions in the solution
  • Indicators are weak acids and bases where the conjugate bases and acids have a different colour
  • Many acid-base indicators are derived from plants, such as litmus

Common Indicators Table

Common Indicators Table, downloadable IB Chemistry revision notes

  • A good indicator gives a very sharp colour change at the equivalence point
  • In titrations is it not always possible to use two colour indicators because of this limitation, so for example litmus cannot be used successfully in a titration
  • When phenolphthalein is used, it is usually better to have the base in the burette because it is easier to see the sudden and permanent appearance of a colour (pink in this case) than the change from a coloured solution to a colourless one

Volumes & concentrations of solutions

  • The concentration of a solution is the amount of solute dissolved in a solvent to make 1 dm3 of  solution
    • The solute is the substance that dissolves in a solvent to form a solution
    • The solvent is often water

Concentration (mol dm3) = fraction numerator number space of space moles space of space solute space left parenthesis mol right parenthesis over denominator volume space of space solution space left parenthesis dm cubed right parenthesis end fraction

  • A concentrated solution is a solution that has a high concentration of solute
  • A dilute solution is a solution with a low concentration of solute
  • When carrying out calculations involve concentrations in mol dm-3 the following points need to be considered:
    • Change mass in grams to moles
    • Change cm3 to dm

  • To calculate the mass of a substance present in solution of known concentration and volume:
    • Rearrange the concentration equation

number of moles (mol) = concentration (mol dm-3) x volume (dm3)

    • Multiply the moles of solute by its molar mass

mass of solute (g) = number of moles (mol) x molar mass (g mol-1)

Worked example

Neutralisation calculation

25.0 cm3 of 0.050 dm-3 sodium carbonate was completely neutralised by 20.00 cm3  of dilute hydrochloric acid. Calculate the concentration in mol dm-3 of the hydrochloric acid.  

Answer

Step 1: Write the balanced symbol equation

Na2CO3  +  2HCl  →  2NaCl  +  H2O  +  CO2

Step 2: Calculate the amount, in moles, of sodium carbonate reacted by rearranging the equation for amount of substance (mol) and dividing the volume by 1000 to convert cm3 to dm3

amount (Na2CO3) = 0.025 dm3 x 0.050 mol dm-3 = 0.00125 mol

Step 3: Calculate the moles of hydrochloric acid required using the reaction’s stoichiometry

1 mol of Na2CO3 reacts with 2 mol of HCl, so the molar ratio is 1 : 2

Therefore 0.00125 moles of Na2CO3 react with 0.00250 moles of HCl

Step 4: Calculate the concentration, in mol dm-3, of hydrochloric acid

Mole Calculations Worked example Neutralisation calculation equation 1

Mole Calculations Worked example Neutralisation calculation equation 2

concentration (HCl) (mol dm-3) = 0.125 mol dm-3

Worked example

Concentration in g dm-3

A student dissolved 10 g of sodium hydroxide, NaOH, in 2 dm3 of distilled water. Calculate the concentration of the solution.

Answer:

    • begin mathsize 16px style Concentration space equals space fraction numerator Mass space left parenthesis straight g right parenthesis over denominator Volume space left parenthesis dm cubed right parenthesis end fraction end style

    • begin mathsize 16px style Concentration space equals space 10 over 2 equals space 5 space straight g space dm to the power of negative 3 end exponent end style

Uncertainty - Calculations

Percentage Uncertainties

  • Percentage uncertainties are a way to compare the significance of an absolute uncertainty on a measurement
  • This is not to be confused with percentage error, which is a comparison of a result to a literature value
  • The formula for calculating percentage uncertainty is as follows:

percentage space uncertainty space open parentheses percent sign close parentheses equals fraction numerator total space uncertainty over denominator measured space value end fraction cross times 100

Adding or subtracting measurements

  • When you are adding or subtracting two measurements then you add together the absolute measurement uncertainties
  • For example,
    • Using a balance to measure the initial and final mass of a container
    • Using a thermometer for the measurement of the temperature at the start and the end
    • Using a burette to find the initial reading and final reading

  • In all of these examples, you have to read the instrument twice to obtain the quantity
  • If each time you read the instrument the measurement is ‘out’ by the stated uncertainty, then your final quantity is potentially ‘out’ by twice the uncertainty

Total experimental uncertainty

  • Some experiments use multiple pieces of equipment, which each have a measure of uncertainty in their use
  • For example, a titration against a standard solution will have the following uncertainties:
    • Using a balance to measure the mass of the solid used to make the standard solution
    • Using a volumetric flask to make the standard solution
    • Using a volumetric pipette to measure the sample that is being titrated against
    • Using a burette to find the initial reading and final reading that determine the titre volume

  • If each piece of equipment used in the experiment has its own associated uncertainty, you can determine the total experimental uncertainty by adding the total individual uncertainties

total space experimental space percentage space uncertainty space open parentheses percent sign close parentheses equals straight capital sigma open parentheses individual space uncertainties close parentheses

  • For example, a titration against a standard solution may have:
    • An uncertainty of ±0.5% for the use of the balance
    • An uncertainty of ±0.1% for the use of the volumetric flask
    • An uncertainty of ±0.2% for the use of the volumetric pipette
    • An uncertainty of ±0.4% for the use of the burette
    • Therefore, the total experimental uncertainty would be 0.5 + 0.1 + 0.2 + 0.4 = ±1.2%

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Stewart

Author: Stewart

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Stewart has been an enthusiastic GCSE, IGCSE, A Level and IB teacher for more than 30 years in the UK as well as overseas, and has also been an examiner for IB and A Level. As a long-standing Head of Science, Stewart brings a wealth of experience to creating Exam Questions and revision materials for Save My Exams. Stewart specialises in Chemistry, but has also taught Physics and Environmental Systems and Societies.