Calculations for titration.

Titration calculations question?

• scenario: - 50mL vinegar is diluted to 500mL - 25mL of this diluted vinegar is titrated with 0.1M 26.167mL NaOH solution What is the concentration of acetic acid in this vinegar? ---- I know that no. of NaOH moles = 2.6167x10^-3 moles = no. of moles in 25mL dil. vinegar Which means the no. of moles in 500mL dil. vinegar = (multiply by 20) 0.052334 moles So what do I do now to get back to the original 50mL undiluted vinegar and find the concentration of acetic acid in it?

• Answer:

  NaOH reacts with CH3COOH in 1:1 molar ratio Mol NaOH in 26.167mL of 0.1M solution = 26.167/1000*0.1 = 2.6167*10^-3 mol Therefore there are 2.6167*10^-3 mol in 25mL of titration volume Therefore there are 500/25*(2.6167*10^-3) = 0.052334 mol CH3COOH in 500mL of the diluted solution Therefore there are 0.052334 mol CH3COOH in 50mL of original concentration vinegar Therefore there are 0.052334*1000/50 = 1.05 mol CH3COOH in 1 litre of the original vinegar The original vinegar is 1.05M solution of CH3COOH. If you want concentration in g/L: Molar mass CH3COOH = 60.05g/mol 10.5 mol = 60.05*1.05 = 63.1g/L CH3COOH