1. Purpose: This lab is trying to help us better understand Titration and how it can neutralize bases using acids, and how we can use it to help us when performing and using Titration in labs.
2. Safety: Goggles
3. Prelab Questions:
1. What color is bromothymol blue (BTB) in an acid? Once it is dropped in the BTB will turn into a yellowish color.
2. . What color is BTB in a base? It turns into a bluish color if dropped in. Goes from yellow-green-blue
3. What color is BTB in a neutral solution? If dropped into a neutral solution then it will have greenish and bluish hues
4.How many grams of NaOH is needed to prepare 250.0 mL of 0.50 M NaOH. Show work.: 22.99+16.00+1.008=39.998 g/mol 62.5 moles (39.998/1mol = 249.875 grams
5. Calculate the molarity of 25 mL of HCL that is neutralized by 30.5 mL of 0.50 M NaOH: H30+ (aq) + Cl- (aq) + Na+ (aq) + OH- (aq) --> Na+ (aq) + Cl (aq) + H20 (l))
0.50 M means 0.50 mol/L --> 0.50m mol/mL you added 30.5 mL so:
0.50 *30.5 = 15.25 mmol NaOH.
This means there is also 15.25 mmol HCl in your original solution.
So there is 15.25 mmol / 25 mL --> 15.25/25,0=0, mmol/mL --> 0,61 mol/L
ANSWER: 0.61 M HCl
6. Calculate the molarity of 15 mL of NaOH that is neutralized by 38.8 ML of 0.20 M HCl
0.20 m HCL 0.20 mol/L - 0.20m mol/mL
0.20*38.8 ML = 7.60 mmol HCL
7.60 mmol/15 ml = 0.506 NaOH
7. How many mL of 1.00 M HCl should it take to neutralize 8.0 mL of 2.0 M NaOH?
4 mL HCL.
8. Dilute the HCl down to 0.25 M HCl, you need 50.0 mL of the 0.25 M HCl
Prepare the HCL in 100 mL. Slowly add 2.2 mL of the stock to 0.25 M of deionized water, adjust the final volume of solution 100 mL w/ deionized water.
Data observation: We used a 125 ml beaker and a 250 ml beaker to hold our liquids and we mixed the HCL and the diluted water drop by drop until we reached the green, there were 3 levels, yellow,green and then blue. Once we reached blue we then counted how many drops we used and