Question

Questions 6,7,8!

Section D: Standardization of a Solution of NaOH Watch the demonstration video to collect data and observe the titration of NaOH. In the space below, summarize the experiment. (2 points) In a 24 well tray, two drops of approximately 1 M NaOH was added to one well. One drop of the BTB indicator was added to the same well. The solution in the well turned blue. This solution was titrated with 0.1 M HCI. The solution turned yellow after 25 drops of the HCI.

Answer 1

Ans-6:

It is given that ..

M HCl = 0.100

V HCl = 25 drops

M NaOH = to calculate accurately (~ 1 M)

V NaOH = 2 drops

Applying the formula at equivalence point:

MHCl .VHCl = MNaOH.VNaOH

MNaOH = MHCl .VHCl / VNaOH

MNaOH = 0.100 M x 25 drops / 2 drops

MNaOH = 1.25 M

Ans-7:

Since volume is measured in the unit of drops (numbers), and not in mL, therefore the use of calibrated or uncalibrated pipette will not make any difference, because the volume is not measured in mL. Only precaution to be taken, is to count the drops accurately. Also the volume unit appears on both side of the formula, (MHCl .VHCl = MNaOH.VNaOH), hence will cancel out, as long as they are in same unit, e.g. drops, mL, L etc.

Ans-8:

The NaOH cannot be weighed accurately, due to its hygroscopic nature. It is always weighed roughly as per the required molarity, (generally 1M stock is prepared), and the stock solution is then standardized against the known molarity of the HCl.