A
B
C
D
E
F
1
Solution
pH
Volume (mL)
[H+] (M)
[OHâ] (M)
Volume for pH=7 (L)
2
0.10 M Acetic Acid
3.2
20
3
Baking Soda
8
50
4
0.10 M HCl
1.75
20
5
0.10 M NaOH
12.27
20
6
Ammonium Chloride
6.1
20
7
Bleach
12.7
20
8
Sodium Acetate
7.56
50
9
Vinegar
3.05
20
10
Apple Juice
4.01
20
11
Coke
2.53
20
12
Sodium Chloride
7.75
20
Perform the following calculations in Columns C and D by entering expressions into the appropriate cells in Excel.
Column D - take the measured pH values from Column B and calculate the concentration of H+ in molarity.
Column E - take the calculated concentration of H+ from Column D and calculate the concentration of OHâ in molarity
Column F - calculate the volume you would have to dilute your solution to for it to have a pH of 7 using the formula M1V1 = M2V2. (Hint: for acidic solutions you will need to use [H+] and for basic solutions you will need to use [OHâ])
| A | B | C | D | E | F | |
| 1 | Solution | pH | Volume (mL) | [H+] (M) | [OHâ] (M) | Volume for pH=7 (L) |
| 2 | 0.10 M Acetic Acid | 3.2 | 20 | |||
| 3 | Baking Soda | 8 | 50 | |||
| 4 | 0.10 M HCl | 1.75 | 20 | |||
| 5 | 0.10 M NaOH | 12.27 | 20 | |||
| 6 | Ammonium Chloride | 6.1 | 20 | |||
| 7 | Bleach | 12.7 | 20 | |||
| 8 | Sodium Acetate | 7.56 | 50 | |||
| 9 | Vinegar | 3.05 | 20 | |||
| 10 | Apple Juice | 4.01 | 20 | |||
| 11 | Coke | 2.53 | 20 | |||
| 12 | Sodium Chloride | 7.75 | 20 |
Perform the following calculations in Columns C and D by entering expressions into the appropriate cells in Excel.
Column D - take the measured pH values from Column B and calculate the concentration of H+ in molarity.
Column E - take the calculated concentration of H+ from Column D and calculate the concentration of OHâ in molarity
Column F - calculate the volume you would have to dilute your solution to for it to have a pH of 7 using the formula M1V1 = M2V2. (Hint: for acidic solutions you will need to use [H+] and for basic solutions you will need to use [OHâ])
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Related questions
| 1.) average number of moles of NaOH dispensed |
|---|
| 2.)average concentration of the acid (mol/L) |
|---|
3,) Based on your coarse titration volume, do you expect the acetic acid solution to have a higher or lower concentration than the NaOH solution?
| higher could you please tell me if number 3 is right |
please help with these 2 questions
Part 1: Prepare the NaOH Solution
Lab Results
How many mL of water were used to prepare the NaOH solution?
| 200mL |
Data Analysis
Calculate the molarity of the NaOH solution. The molar mass of NaOH is 39.997 g/mol.
| 0.1M |
Calculate the amount of benzoic acid to be neutralized by 20.00 mL NaOH solution, in both moles and grams. The molar mass of benzoic acid is 122.12 g/mol.
| 1.207 g |
Part 2: Perform a Coarse Titration
Lab Results
Record the following data from your course titration in the table below.
| mass of benzoic acid used (g) | 0.26614g |
|---|---|
| pH of benzoic acid solution before titration | 2.96 |
| volume of NaOH in the burette before titration (mL) | 50.0mL |
| volume of NaOH in the burette after titration (mL) | 28.89mL |
| volume of NaOH dispensed in the titration (mL) | 21.11mL |
Data Analysis
How do you expect your coarse titration volume to compare to your fine titration volumes?
| I expect the values to be a little off from each other since in the first titration we let out a lot at a time and didnt slowly do the titration. should only be off by a few mL. |
Part 3: Perform Fine Titrations
Lab Results
Record the volume of NaOH solution dispensed in the 3 fine titrations.
| 20 mL dispenced ,20.34mL, 20.30mL |
Data Analysis
Calculate the average concentration of the NaOH solution, using the average volume of NaOH solution dispensed in the 3 fine titrations. Report your answer using enough significant figures to distinguish it from the expected concentration of 0.100 M.
| V = (V1 + V2 + V3) / 3 V = (20.03 mL + 20.06 mL + 20.02 mL) / 3 = 20.04 mL M = n / V = 0.00200 moles / 0.02004 L = 0.0998 M Learning objective: Determine the concentration of a solution. 20.21= ----> 0.0999M |
Experiment 2: Use the Standardized NaOH Solution to Determine the Concentration of an Acid
Part 1: Perform a Coarse Titration
Lab Results
What was the pH at the end point of the coarse titration?
| 3.87ph |
Data Analysis
Based on your coarse titration volume, do you expect the acetic acid solution to have a higher or lower concentration than the NaOH solution?
| higher |
Part 2: Perform Fine Titrations
Lab Results
For the 3 fine titrations of the acid of unknown concentration, fill in the following data.
| Titration #1 | Titration #2 | Titration #3 | |
|---|---|---|---|
| volume of acid (mL) | 59.11 | 40.97 | 40.98 |
| volume of NaOH dispensed (mL) | 9.05 | 9.03 | 9.02 |
Data Analysis
Calculate the following quantities and record the data in the table below.
| average volume of NaOH solution dispensed (mL) | 9.03 |
|---|---|
| average number of moles of NaOH dispensed | |
| average concentration of the acid (mol/L) |
Conclusions
Phenolphthalein is pink over the range of pH 8 â 12. Why was it a useful indicator of when the equivalence point was reached?
| This was helpful because once I got to the ph of 7 I knew that I would be getting closer to the end point which would change the color to pink. |
Suppose a student titrated a sample of monoprotic acid of unknown concentration using a previously standardized solution of NaOH.
| volume of 0.125 M NaOH dispensed | 24.68 mL |
|---|---|
| volume of acid solution | 50.00 mL |
Given the data in the figure above, what is the concentration of the unknown acid?
| 0.0617M Lab NOTES: ex1 Take sodium hydroxide from the Materials shelf and add 0.800 g to the beaker. Take water from the Materials shelf and add 200 mL to the beaker. Ph 2.96 inital volume in burett inital volume 50ml 1.) volume43.78mL volumes dispenced =6.22ml 2. volume =38.05 volume dispenced =11.95ml 3.)volume = 32.06 volume dispenced=17.94 4.28.89ml volume dispenced=21.11ml ph=10.95 -------------------------------------- Ex 2 inital ph= 2.96 volume30.33 dispence 20.05 1. volume 39.92ml volume dispenced=10.08 2.volume 29.66 voiume dispence 20.34 final ph10.43 initial valume in burett 50ml -------------------- ex3 final ph9.43 volume 30.28 volume dipenced=20.00 ---------------------------------------------------------------------- 2nd Ex initial ph 3.11 volume 50 1. volume 48.95mL dispenced 1.05mL ph 3.87 2. volume 45.88 dispenced 4.12 ph 4.67 3 volume 42.99 dispenced 7.01 ph 5.29 4 volume 39.93 dispence 10.07 ph 11.45 -------------------- F1 final ph 9.99 volume 59.11 dispenced 9.05 F2 ph 9.59 volume 40.97 dispenced 9.03 f3 ph 9.50 volume 40.98 9.02 |
i need help graphing this information based on data 1 and data 2 based on information below provided
Activity 1: Buffer Solution Preparation
The following eight solutions are prepared in the eight plastic cups labeled with the solution names.
1. Based on the Preparation calculations, prepared 50 mL of a 0.10 M solution of sodium acetate in the cup labeled sodium acetate. 0.050L x 0.10M= 0.005 x 82.03= .41g CH3COONa
2. 100 mL of pure water was added to the cup labeled water
3. 50 mL of a 0.1 M acetic acid solution was added to the cup labeled acetic acid.
4. 10 mL of a 0.1 M NaOH solution (0.10M)(10mL)=(1M)(V2)V2=1mL NaOH
5. 10 mL of a 0.1M HCl solution (.10M)(10mL)=(1M)(V2)V2=1mL HCl
6. 10 mL of a pH 3.7 buffer (9.1 mL of 0.1 M acetic acid and 0.9 mL of sodium acetate)
7. 10 mL of a pH 4.7 buffer (5.0 mL of 0.1 M acetic acid and 5.0 mL of sodium acetate)
8. 10 mL of a pH 5.7 buffer (0.9 mL of 0.1 M acetic acid and 9.1 mL of sodium acetate)
The following amount of Bogenâs Universal Indicator was added to each solution: 10 drops to the acetic acid and the sodium acetate solutions, 20 drops to the 100 mL water, and 2 drops to the NaOH and the HCl solutions.
The 24-well plate was placed on to a sheet of plain white paper as shown below, and one drop of universal indicator was added to each well. The following amount of solutions was transferred the appropriate pipets to each well.
a. 1 mL of 0.1 M acetic acid to all four wells in the first row of the 24-well plate.
b. 1 mL of water to all four wells in the second row of the plate
c. 1 mL of 0.1 M sodium acetate solution to all four wells in the third row
d. 1 mL of the 3.7 pH buffer to all four wells in the fourth row
e. 1 mL of the 4.7 pH buffer to all four wells in the fifth row
f. 1 mL of the 5.7 pH buffers to all four wells in the sixth row
The color of the solution for each row was recorded, and a photo of the plate was taken.
The approximate pH of the solution was determined based on the indicator color chart.
One drop of 0.1 M HCl was added to the second well of each row, 1 drop of 0.1 M NaOH was added to the third well of each row, and 10 drops of 0.1 M NaOH was added to the fourth well of each row.
The color and corresponding pH was recorded.
A photo of the plate showing the colors in each well was taken.
| 0.1M Acetic Acid | water | 0.1M Sodium Acetate | pH 3.7 Buffer | pH 4.7 Buffer | pH 5.7 Buffer | ||||
| Initial Color | yellow | green | pink | orange | yellow | ||||
| Initial pH | 4 | 6 | 7 | 4 | 5 | 6 | |||
| Color after 1 drop HCl | Pink | Pink | |||||||
| pH after 1 drop HCl | 4 | 4 | 7 | 4 | 5 | 6 | |||
| Color after 1 drop NaOH | Pink | blue | blue | blue | |||||
| pH after 1 drop NaOH | 4 | 8 | 8 | 4 | 5 | 8 | |||
| Color after 10 drops NaOH | Pink | purple | purple | purple | |||||
| pH after 10 drops NaOH | 4 | 10 | 10 | 4 | 5 | 10 | |||
Activity 2: Buffering Capacity of a Solution
The 24-well plate was placed on to a sheet of plain white paper at 90°so that it is four rows by six columns
20 mL of pH 4.7 buffer was prepared by doubling the volumes calculate in from activity 1 in the appropriate cup, and 2 mL of the solution was added into all 6 wells in the first row.
5 mL of the pH 4.7 buffer solution was diluted with 15 mL of purified water containing the indicator and poured into a clean cup. 2 mL of this first dilution was added to all the wells in the second row.
5 mL of the first dilution was diluted with 15 mL of purified water containing the indicator and poured into a clean cup. 2 mL of this second dilution was added to all the wells in the third row.
5 mL of the second dilution was diluted with 15 mL of purified water containing the indicator and poured into a clean cup. 2 mL of this third dilution was added to all the wells in the fourth row.
An additional drop of indicator was added to each well to make the color more visible. The color of the first well of each row was recorded.
One drop of 0.1 M NaOH solution was added to the second well in each row, two drops of 0.1 M NaOH solution was added to the third well of each row, three drops to the fourth well, four drops to the fifth well, and five drops to the sixth well. The color and corresponding pH of each well was recorded, and a photo of the plate showing the colors in each well was taken. The sodium acetate molar concentration of each dilution was calculated.
| pH 4.7 Buffer | First Dilution | Second Dilution | Third Dilution | |
| Initial Color | orange | orange | orange | orange |
| Initial pH | 5 | 5 | 5 | 5 |
| Color after 1 drop NaOH | orange | orange | Blue | purple |
| pH after 1 drop NaOH | 5 | 5 | 8 | 10 |
| Color after 2 drops NaOH | orange | yellow | Blue | purple |
| pH after 2 drops NaOH | 5 | 6 | 8 | 10 |
| Color after 3 drops NaOH | orange | green | purple | purple |
| pH after 3 drops NaOH | 5 | 7 | 9 | 10 |
| Color after 4 drops NaOH | orange | purple | purple | purple |
| pH after 4 drops NaOH | 5 | 9 | 9 | 10 |
| Color after 5 drops NaOH | orange | purple | purple | purple |
| pH after 5 drops NaOH | 5 | 9 | 9 | 10 |
| Concentration of sodium acetate | 0.1 M | 2.5x10^-2 M | 6.3x10^ -3 M | 1.6x10^-3 M |