Need help trying to do a Titration lab, for example: Say you were to start with
ID: 1070614 • Letter: N
Question
Need help trying to do a Titration lab, for example:
Say you were to start with 0.209 grams of an unknown acid. You then transferred that into 20 mL of water. Took that flask of 20 mL of water and 0.209 grams of the unknown solution over to a burette that contains 0.112 M of concentrated NaOH solution. To get to the endpoint of the titration, I dispensed 9.12 mL of the base. I then hovered over the midway point of the titration, and had 5.4pKa and 4.7 mL
then repeated the same steps for the same unknown weak acid, this time I used 0.201 grams, and again transferred it into 20 mL of water which I then titrated with the 0.112 M of concentrated NaOH solution. It took 8.90 mL of base to reach then endpoint and had values of 5.42 pKa and 4.54 mL.
Question #1:
Calculate the moles of OH required (be careful to convert the mL of base to L of base in this calculation)
mol base for trial #1 = __________mol
mol base for trial #2 = __________mol
Question #2
calculate molar mass of weak acid for each trial
g/mol for trial #1 = ___________ g/mol
g/mol for trial #2 = _________g/mol
average for 1 and 2 = ___________g/mol
Third question:
Determine Ka value for weak acid from both trials
Formula = Ka = 10^-pKa
*Use "e" notation such as 1.25e-5.....DO NOT USE 1.25X10^-5
Ka for trial 1 = __________
Ka for trial 2 = ________
Ka for trial 3 = _________
Explanation / Answer
Calculate the moles of OH required:
Trial #1 =>
Concentration of NaOH = 0.112 M
volume of NaOH used = 9.12 ml = 0.00912 L
Moles of base = concentration*volume = 0.112*0.00912 = 0.00102 mol = 1.02 mmol
Trial #2 =>
Concentration of NaOH = 0.112 M
volume of NaOH used = 8.90 ml = 0.00890 L
Moles of base = concentration*volume = 0.112*0.00890 = 0.00010 mol = 1.00 mmol
Question #2
calculate molar mass of weak acid for each trial
g/mol for trial #1 =>
moles of acid = moles of base used = 1.02 mmol
mass of acid used = 0.209 g
molar mass = mass/moles = 0.209/0.00102 = 204.9 g/mol
g/mol for trial #2 =>
moles of acid = moles of base used = 1.00 mmol
mass of acid used = 0.209 g
molar mass = mass/moles = 0.209/0.00100 = 209.0 g/mol
average for 1 and 2 => (204.9+209.0)/2 = 206.95 g/mol
Ka for trial 1 = 10^-pKa = 10^-5.4 = 3.98*10-6
Ka for trial 2 = 10^-5.42 = 3.80*10-6