Calculating Kc for an equilibrium reaction

 

A)    Difficult bit – moles of CH3COOH at equilibrium: 

  • Your mixture contains 2 acids:

HCl  and CH3COOH         

  • Both of these will react with NaOH:

  • 1cm3 of your equilibrium mixture was removed. 

  • That is 1/100th of your original amount.

HCl  +  NaOH  à  NaCl  +  H2O

 

CH3COOH  +  NaOH  à  CH3COONa  +  H2O

  • Both react as 1:1 ratio. 

  • Overall NaOH : Acid is 1:1      Animation

Equilibrium mixture (100cm3)

Titrated mixture – 1cm3 of equilibrium mixture made up to 25 cm3 with distilled water then titrated.

  • Remember HCl is the catalyst.  If you know how many moles of HCl you started with, that’s how many moles you have at the end too.

  • You can work out how many moles of both acids has been neutralised by the NaOH

  • If you know how much of that is HCl then you can work out the moles of CH3COOH

                                    Moles of CH3COOH = Moles of both acids  -  Moles of HCl