I need help with this report please. Its an expermient about Acid-Base titration
ID: 971086 • Letter: I
Question
I need help with this report please. Its an expermient about Acid-Base titrations. Questions (Pay att ention to units and significant figures for the calculations). 1. For each of the following circum stances, indicate whether the calculated molarity of NaOH d be lower, higher or unaffected. Explain your answer in each case. a. The inside of the pipet used to transfer the standard HCl solution was wet with water oul blo b. You added 40 mL of water to the titration flask rather than 25 mL c. The buret, wet with water, was not rinsed with NaOH solution before filing the buret with NaOH solution. d. Five (5) drops of phenolphthalein were added to the solution to be titrated rather than three (3) drops.Explanation / Answer
Influence of reaction errors/conditions on molarity
a. If the inside of pipette used to transfer HCl is contaminated with water, the net concentration of HCl would be lower (diluted by excess water). Thus we would be using lower amount of NaOH for reaction and net molarity of NaOH would be lower.
b. If we added 40 ml water insetad of 25 ml to the titration flask. The HCl will be more dilute. The NaOH used would be less. So the molarity of NaOH calculated would also be lower.
c. If the buret is not rinsed with NaOH and had water present. It would dilute the NaOH present. We would be using more of NaOH for the same amount of HCl and thus the molarity of NaOH would be lower.
d. If we added 4 drops of phenophthalein indicator instead of 3 drops. the molarity would still remain the same and unaffected of NaOH.
2. a. It is important to keep NaOH solution covered at all times, since NaOH is hygroscopic in nature and absorbs water from atmosphere which would dilute the solution and lower the molarity.
b. After prolonged exposure, the NaOH solution will have a lower molarity than what we started with.
3. Pink color of solution fades after sometime to colorless. The solution takes up CO2 from atmosphere which makes it more basic and thus the color changes to colorless.
4. Reaction : Ba(OH)2 + 2HNO3 ---> Ba(NO3)2 + 2H2O
moles of HNO3 = 0.2 M x 25 ml = 5 mmol
moles of Ba(OH)2 needed = 5/2 = 2.5 mmol
Volume of Ba(OH)2 needed = 2.5 mmol/0.0293 M = 85.324 ml