pH changes during acid-base titrations

• An acid of known concentration is added from a burette to a measured volume of base solution
• An indicator is used, upon changing colour it allows a visual representation that the mixture has been neutralised
• Similarly, a pH meter can be used to determine the solutions pH. This must be calibrated in a known pH solution; buffer solution.

Titration Curves • Each graph shows a curve which does not change in a linear fashion as the base is added
• Each graph has a horizontal section where a large volume of a base can be added without the pH changing much
• Each graph has a vertical section (known as the equivalence point) where a single drop of a base can change the pH several times (however not in the weak acid and weak base)
• The Equivalence point is the point at which sufficient base has been added to just neutralise the acid (or acid to neutralise the base)
• In all of the above graphs the equivalence point is found at the 25cm3 mark, however note that the pH is not neutral at 7
• If the acid is added to the base then the graph is flipped

Calculating concentrations of acid/base from the equivalence points

• To determine the concentration, the volume of the solutions involved must be known as well as the concentration of the opposite solution (i.e. if concentration for an acid is unknown, then the concentration for the base must be known)
• Monoprotic Acids

Number of moles (acid) = Concentration x volume (cm3)

1000

• Divide by 1000 to convert from cm3 to dm

e.g. If 25cm3 of 0.5 mol dm3 HCl was required to neutralise 35cm3 of NaOH, calculate the concentration of the NaOH

Number of moles (HCl) = (0.5 mol dm3)(25cm3) = 0.5 x 25 = 0.0125 moles

1000 1000

• Therefore for 0.125 moles of HCl is required to neutralise 0.125 moles of NaOH
• As the balanaced equation shows a 1:1 ratio

1HCl + 1NaOH à 1NaCl + 1H2O

Number of moles (NaOH) = 0.0125 x 1000 = 0.036 moles

35 cm3

• Diprotic Acids
• As Diprotic Acids release 2 protons when in a solution they have 2 equivalence points
• An example of this is ethanedioic acid which when it reacts with a base, e.g. Sodium Hydroxide, it is neutralised. However, this process occurs over two stages as the two protons are removed from the acid separately

1st stage) Equivalence point = pH 2.7

HOOC-COOH + OH à HOOC-COO- + H2O

The first proton is lost to the base

2nd stage) Equivalence point = pH 8.4

HOOC-COO + OH à OOC-COO + H2O

The second proton is lost to the base

• To calculate the concentration of a diprotic acid the moles of the base must be half of the moles of the acid

e.g. 25cm3 of ethanedioic acid is neutralised by 20cm3 of 0.1 mol dm-3 NaOH solution. Calculate the concentration of the ethanedioic acid solution

Number of moles (NaOH) = 0.1 mol dm3 x 20 cm3 = 0.002 moles

1000

• As the solution is diprotic, the number of moles is divide by 2, therefore there is (0.002 / 2 = 0.001) 0.001 moles of ethanedioic acid#

Number of moles (C2H2O4) = 0.001 moles x 1000 = 0.04 moles dm-3

25 cm3