You will need to calculate how much of the NaOH added during the titration of a
ID: 479884 • Letter: Y
Question
You will need to calculate how much of the NaOH added during the titration of a sample was needed to:
a. neutralize the HCl that was added to bottle #1 as a volume
b. the portion of the volume added to the bottle #1 sample required to neutralize the acetic acid
c. what the volume required in 'b' equates to in moles of NaOH
d. Stoichiometrically, determine the moles of acetic acid neutralized by the moles NaOH determined in 'c'
e. and FINALLY, the concentration of acetic acid in the 5.00 mL sample you titrated (which is also the EQUILIBRIUM concentration of acetic acid in Bottle #1)
The equilibrium value calculated in 'e' is then used to determine the Change in concentration (x) in the ICE Table.
The volume of 1.000 M HCl added to each bottle was 5.00 mL. If we assume the required mL of NaOH per every mL of HCl (from Step 12) to be 1.041:
a. How much NaOH volume will be needed to titrate the HCl in EVERY 5.00 mL sample of all 3 bottles?
b. If 25.52 mL of NaOH was needed to titrate the entire 5.00 mL sample of Bottle #1, what volume was needed to titrate only the acetic acid?
c. How many moles of NaOH are contained in the volume of NaOH used to titrate the acetic acid assuming the concentration of the NaOH solution was 0.9806 M?
d. How many moles of Acetic Acid was titrated by the number of moles of NaOH calculated in 'c'? HINT: Use stoichiometry from a BALANCED Equation!
Explanation / Answer
a) Since each mL of HCl required 1.041 mL of NaOH, for 5 mL of HCl, the number of mols of NaOH required
= (1.041)*5
= 5.205 mL
b) The volume of NaOH needed to titrate acetic acid = 25.52 mL - 5.205 mL = 20.31 mL
c) The number of mols of NaOH required to titrate NaOH = (20.31 /1000) L * 0.9806 M = 0.0199 moles
d) Since -
CH3COOH (aq) + NaOH (aq) ---------> CH3COONa (aq) + H2O (l)
From the balanced equation 1 mol of acetic acid required 1 mol NaOH and hence 0.0199 moles of acetic acid requires the same amount of NaOH.
Thus number of mols of acetic acid = 0.0199 mols
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.