Titration calculations

Titration calculations

• Work out the moles of acid and base at the start
• Work out the excess moles of acid or base (the rest will be neutralised)
• Work out the new [H⁺] or new [OH^-] and then the pH

Calculate the pH of the solution formed when 20 cm³ of 0.10 mol dm^-3 HCl is added to 30 cm^-3 of 0.04 mol dm^-3 NaOH

Moles HCl                = 0.10 x 20/1000

                              = 0.0020 mol

Moles NaOH            = 0.04 x 30/1000

                              = 0.0012 mol

Excess HCl               = 0.0020 – 0.0012

                              = 0.0008 mol

Total volume           = 50 cm³

New [H⁺]                = moles/volume

                              = 0.0008/(50/1000)

                              = 0.016 mol dm^-3

pH                          = -log(0.016)

                              = 1.80

4)      Buffer solutions

• A buffer solution minimises pH changes on addition of an acid or base
• Buffer solutions are important for controlling pH in blood (so that enzymes are not denatured) and shampoos (so that eyes do not sting and skin is not damaged)

a)   Acidic buffers

• An acidic buffer consists of a weak acid and the salt of a weak acid (e.g. ethanoic acid & sodium ethanoate)
• For ethanoic acid/sodium ethanoate, the following equilibrium exists:

CH₃COOH(aq) ¾ CH₃COO^-(aq) + H⁺(aq)

• If a small amount of acid is added, equilibrium will shift to the left to remove the added H⁺. The following reaction occurs: CH₃COO^- + H⁺ ® CH₃COOH
• If a small amount of base is added, the OH^- will react with H⁺ to form water. The equilibrium will shift to the right to replace the H⁺ that has been removed. The following reaction occurs: CH₃COOH ® CH₃COO^- + H⁺

b)   Calculating the pH of a buffer solution

• The pH of an acidic buffer can be calculated using the K_a expression for the weak acid e.g. for ethanoic acid/sodium ethanoate:

K_a = [CH₃COO^-][H⁺]

                                                      [CH₃COOH]

      Rearranging gives:

                                          [H⁺] = K_a x      [CH₃COOH]

                                                                  [CH₃COO^-]

Calculate the pH of a buffer solution containing equal volumes of 2.5 mol dm^-3 HCOONa and 1.0 mol dm^-3 HCOOH (K_a = 1.6 x 10^-4 mol dm^-3)

      [H⁺]      = K_a x [HCOOH]/[HCOONa]

                  = 1.6 x 10^-4 x 1.0/2.5

                  = 6.4 x 10^-5 mol dm^-3

      pH        = 4.19

Source:           OCR June 2003 paper

• The pH of a particular buffer depends on the value of K_a and the ratio of [CH₃COOH] to [CH₃COO^-]

d)   pH of Blood

• The pH of blood is kept in the range 7.35-7.45
• This is achieved by dissolved carbon dioxide acting as a buffer solution
• The relevant equations are:

CO₂ + H₂O ¾ H₂CO₃

H₂CO₃ ¾ HCO₃^- + H⁺

• H₂CO₃ is carbonic acid. HCO₃^- is the hydrogencarbonate ion
• If the blood pH is less than 7.35 (too acidic), more CO₂ is breathed out. Both equilibria shift to the left to produce more CO₂ and [H⁺] is reduced
• If the blood pH is more than 7.45 (too alkaline), less CO₂ is breathed out. Both equilibria shift to the right to remove CO₂ and [H⁺] is increased

5)      Enthalpy of neutralisation

• Enthalpy of neutralisation is defined as the change in enthalpy that occurs when an acid and base undergo a neutralisation reaction to form one mole of water i.e.

            H⁺(aq) + OH^-(aq) ® H₂O(l)

• This is a calorimetry calculation e.g.

Example

50 cm³ of 1.0 mol dm^-3 hydrochloric acid was added to 50 cm³ of 1.0 mol dm^-3 sodium hydroxide solution.   The temperature rose by 6.8°C.  Calculate the enthalpy of neutralisation for this reaction.  Assume that the density of the solution is 1.00 g cm^-3, the specific heat capacity of the solution is 4.18 J g^-1 K^-1.

                  HCl + NaOH ® NaCl + H₂O

m   = mass of solution = total of acid and alkali = 100g

q    = -mcDT/1000

      = -100 x 4.18 x 6.8/1000

      = -2.8424 kJ

n    = moles HCl = moles NaOH = 1.0 x 50/1000 = 0.05 mol

DH  = q/n

      = -2.8424/0.05

      = -56.8 kJ mol^-1