Question

How would your results have been effected if: (provide an explanation for each answer) 

a. each of the flasks you used for the standardization of NaOH had initially turned pink with the addition of phenolphthalein but the color disappeared when you added KHP to each (you did not rinse the flasks and Mart over)?  

b. there was still visible, undissolved KHP in the bottom of each flask when you finished with your titration?  

c. you stopped the titration immediately after the solution first became pink (the color disappeared almost immediately)?  

You were given 25.00 mL of an acetic acid solution of unknown concentration You find that it requires 37.15 mL of a 0.1994 M NaOH solution to exactly neutralize this sample (phenolphthalein was used as an indicator). 

a. What is the molanty of the acetic acid solution? (show work)  

b. What is the percentage of acetic acid in the solution? Assume the density of the solution is 1 g/mL

Answer 1

a) The appearance of pink color on standard flask indicates that the some amount of NaOH is still present in the flask that will result a decreased concentration of NaOH than actually prepared.

b) The presence of undissolved KHP in the flask decreases the amount of KHP in the solution and thus the concentration of KHP will be less, hence it consumes more for titration than actually required.

c) This would give a less volume of acid consumed and hence the concentration of Base will be lesser than original concentration

2) The reaction between Acetic acid and NaOH is given by

CH3COOH + NaOH ----> CH3COONa + H2O

V1= Volume of Acetic acid = 25.00mL

M1 = molarity of Acetic acid = ?

V2 = volume of NaOH = 37.15 mL

M2 = Molarity of NaOH = 0.1994

M1 = M2V2/V1 = 0.1994*37.15/25 = 0.2963 M

a) The molarity of Acetic acid solution = 0.2963 M

b) Percentage of Acetic acid in the solution is

mass acetic acid = 60 g/mol x 0.2963= 17.77g
mass solution = 25.0 mL x 1.0 g/mL = 25.0 g

% by mass = 17.77 x 100 / 25.0= 71.11