ANALYSIS

 

3. Determine the average [CH3COOH] in mol/L

 

I have no idea what this is asking for. The past two questions have made me determine the volume of NaOH at equivalence (22.5mL) and the concentration of CH3COOH (3.75mol/L).

 

All other information: Concentration of NaOH = 1.5mol/L, volume of CH3COOH = 20mL, initial volume of NaOH = 50mL, initial pH of CH3COOH = 2.4, final volume of NaOH trial 1 = 22.8mL, final volume of NaOH trial 2 = 22.2mL, pH of colour change trial 1= 8.7, pH of colour change trial 2 = 8.0.

 

 

Good evening Kimbow,

 

It sounds like you're just determining the concentration of acetic acid. Using the information given, you can find the measured concentration of Acetic Acid and I'm assuming they will not be identical. Find the average (1/2 the sum of the 2 trials) and that should give you your answer. If it wasn't 12:48 am, I'd run through the calculation, but it's pretty straight forward. Let me know if you run into any problems and I'll check back in the morning!

 

-Sonryu