Question

Calculate the theoretical pH after 9.0 mL of the standard NaOH have been added to the acetic acid solution. Show all calculations and explain / justify any assumptions.

concentration of NaOH = 0.1067 M

volume of acetic acid = 20 mL

volume of NaOH to reach equivalent point =19.15 mL

Answer 1

ANSWER: pH = 4.7

( In case of weak acid strong base titration , before equivalence point system contain weak acid and its salt with strong base which a Buffer system and pH of solution is Calculated by using Henderson's equation)

First calculate Molarity of Acetic acid solution

Consider a titration CH₃COOH vs NaOH

CH₃COOH + NaOH ---> CH₃COONa + H2O

M _acid x V _acid = M _base x V _base

M _acid = M _base x V _base / V _acid

= 0.1067 M x 19.15 ml / 20 ml

= 0.1022 M

Calculate mmol of acid and base and salt

mmol of Acetic acid = concentration x volume = 0.1022 M x 20 ml = 2.044 mmol

mmol of sodium Hydroxide = 0.1067 M x 9.0 ml = 0.9603 mmol

mmol of salt produced = mmol of sodium hydroxide = 0.9603 mmol

mmol of acetic acid unreacted = mmol of acetic acid - mmol of NaOH added

= 2.044 - 0.9603

= 1.084 mmol

At this stage of titration, total volume of solution = 20 ml acetic acid + 9.0 ml sodium hydroxide

= 29.0 ml

[CH₃COOH ] = no of moles / volume of solution in L

= [1.084 / 1000 ] x [ 1000 / 29 ]

= 0.037 M

[CH₃COONa ] = no of moles / volume of solution in L

   = [0.9603  / 1000 ] x [ 1000 / 29 ]

= 0.033 M

Solution is a mixture of weak acid ( acetic acid ) and its salt with strong base.

This is a Buffer system.

We know that, pH of a Buffer system is calculated by using Henderson's equation

pH = pka + log [CH₃COONa ] / [CH₃COOH ]

= 4.75 + log 0.033 / 0.037

= 4.75 + log 0.8919

= 4.75 -0.04967

pH = 4.7