Acid Base and Oxidation Reduction of Titration Lab Report

Lab 10 Acid base Titration lab

Background and definitions:

Comprehensive view:

Example Calculation:

If I used 12.5mL of (unknown concentration) HCl to reach the equivalence point with 10.0mL of 0.10M NaOH, what is the concentration of my unknown HCl?

• 10.0mL NaOH * = 0.010L NaOH used
• 0.10M NaOH =
• * = 0.0010 mol NaOH

When the solution turned pink, we have reached the equivalent point

NaOH and HCl are both mono-protic, therefore

• 0.0010 mol NaOH * = 0.0010mol HCl
• Put the number of moles of HCl over the volume used:
• * = 0.08M HCl

Graphing

Titration curve is a graph plotting the pH of the analyte (HCl) solution (y-axis) versus the total volume of the titrant (NaOH) added. The graph is a sigmoid graph, look something like this: Notice the equivalence point is in the middle of the big jump. Also, the graph continues for a little further after the big jump. This is why during your titration; the instruction was to continue adding more titrant after the change in color occurs (end point)

Pre-Laboratory Questions

**Show ALL calculations for full credit, mind your significant figures

1. Compare and contrast the terms “primary standard” and “secondary standard”.
2. Calculate the mass of oxalic acid dehydrate (C2H2O2•2H2O) needed to prepare500.0mL of 0.2500M oxalic acid solution
3. During a titration, a student used 31.92mL of a NaOH solution to neutralize 30.0mL of 0.2488M oxalic acid solution. (Note: Oxalic acid is diprotic!)
1. Write the balanced equation for the neutralization reaction (see previous question for the formula of oxalic acid)
2. How many moles of oxalic acid were present in the sample?
3. How many moles of NaOH(aq) were needed to neutralize the oxalic acid?
4. What is the molar concentration of the NaOH(aq) ?

Lab and data collection

Steps:

1. In the stock room tap, retrieve by simply clicking on:
1. 1.0M NaOH stock
2. ~ approximate 0.10M HCl stock- your unknown-
1. it is not exactly 0.10M, you will need to calculate the exact concentration from your data average.
3. Phenolphthalein
2. In the glassware tap, retrieve by simply clicking on:
1. Erlenmeyer
1. 250mL
2. Others
1. 50mL Buret (Burrette)
3. Pour 10.0mL of ~0.10M HCl from the stock bottle into the empty Erlenmeyer flask by dragging and dropping the stock bottle on top of the empty one.
1. The program will give you the choice to pour an exact amount once you drag and drop the stock HCl bottle over the empty Erlenmeyer
2. This will be your
4. Pour 0.20 mL of the indicator Phenothalein from the stock bottle into the working Erlenmeyer flask with the 10.0mL ~0.10M HCl (created in the last step). This is done by dragging and dropping the indicator bottle on top of the working Erlenmeyer flask.
1. The program will give you the choice to pour an exact amount once you drag and Phenothalein indicator bottle working Erlenmeyer flask
5. Pour 50.0mL of 1.0M NaOH from the stock bottle into the empty 50.0mL Burette by dragging and dropping the stock bottle on top of the Burette.
1. The program will give you the choice to pour an exact amount once you drag and drop the stock NaOH bottle over the empty Burette
6. Now make sure you single out the working 250mL Erlenmeyer flask (with the 10.0mL ~0.10M HCl and 0.20mL Phenothalein) on the bench with room to work
1. You can elect to remove the other bottles by right click and “remove”
7. Position the Burette over the working Erlenmeyer flask and you are ready to titrate
1. The program will give you the choice to pour an exact amount of NaOH (titrant) from the Burette into the working flask with the 10.0mL HCl (analyte)
8. Use the increment of 0.10mL NaOH
1. Each time you are to add 0.10mL of 1.0M NaOH into your analyte
9. Record the pH and color given in the program in the tables provided below
10. When the color changes to pink, continue to add a few more titrant increments (NaOH)
1. In this activity the end point = equivalent point
2. The reason is for the creation of a complete titration graph
11. Repeat and to have a minimum of 3 trials
12. Fill in and calculate the concentration of your analyte (HCl)
1. You will not receive credit for this lab if you do not show calculations!
13. Graph one of your titration trials!

Data collection and Calculation

Trial 1

 Total volume of NaOH added pH Color

• Titrant = 1.00M NaOH
• Analyte= unknown concentration of HCl
• Indicator= Phenothalein

Trial 2 Trial 3

 Total volume of NaOH added pH Color
 Total volume of NaOH added pH Color

Lab Calculations:

**Show all calculations for full credit

 Trial 1 Trial 2 Trial 3 Initial Buret reading NaOH Final Buret reading NaOH Volume of NaOH added at equivalent point Moles of NaOH at equivalent point Moles of HCl at equivalent point Concentration of HCl Average concentration of HCl unknown

Titration graph:

Post-Laboratory Questions

1. Student A weights 0.747g of KHP (acid) on a laboratory balance. The KHP was titrated with an unknown NaOH solution and the concentration of the NaOH was determined to be 0.113M. Student B diluted 6.0M HCL from the reagent shelf using a volumetric flask to obtain 0.6M HCl. This solution was titrated with the same unknown NaoH solution and found to be 0.104M. Which student should have the more accurate concentration for the unknown NaOH and why?
2. The volume of water added moving a half-drop of titrant from the tip of the buret to the analyte solution during the end of the titration may be ignored in your calculations. Why is this so?
3. Vinegar is an aqueous solution of acetic acid. By law, it must contain 4g acetic acid (CH3COOH, 60.05g/mol) per 100mL solution although it is commonly up to 8% acetic acid. If a 15.00mL aliquot of vinegar required 5.28mL if 0.10M NaOH to reach the end point, is the sample legally vinegar? Show your calculations to support your answer

Lab 11 Oxidation-Reduction Titration

Background and definitions:

Sample calculation:

Calculate oxidation state of Mn in MnO41- :

Oxygen is 2- each, there’s 4 oxygen in MnO41-, therefore total oxidation state of oxygen in this specie is -8

Mn therefore is +7 because the overall charge of MnO41- is -1:

(-8) + (X) = -1

X = +7

• X represent the charge of Mn Pre-Laboratory Questions

1. Explain in terms of electron movement what is meant by the term “oxidation.”
2. Explain in terms of electron movement what is meant by the term “reduction.”
3. Determine the oxidation number for the underlines atomic species. Then indicate if the half-reaction is an oxidation or a reduction. Show your work!!
 Half- Reaction Reagent Oxidation # Product Oxidation # Oxidation Reduction VO21+(aq) + 2H1+(aq) VO2+(aq) + H2O(l) 6H1+(aq) +2KIO3(aq) I2(aq) + 2K+(aq) + H2O(l)
1. List the color change expected at the equivalence point for each of the possible titrations you might perform in parts A, B, and C
1. MnO41-(aq) + Fe2+(aq) Mn2-(aq) + Fe3+(aq)
2. MnO41-(aq) + H2O2(aq) Mn2+(aq) + O(g)
3. ClO1- + I1- + H3O+ I2 + Cl1- + 3H2O (starch present)

I2 + S2O32-I+ S4O62- (starch present)

Lab 11: Only part B titration using MnO4 (titrant) and Fe2+ (analyte) is performed.

Fe2+ = pale green

Fe3+ = yellow- brown

No indicator needed (like in the previous lab) because we can see the change in the color as Fe changes oxidation states.

MnO41- = oxidizing agent – that means Mn, being 7+, easily pulls electrons from someone else, oxidizing them. In the process, because Mn has gained electron(s), it is, therefore, reduced.

This video is the wet lab part: (Same as the first link posted above)

Lab Calculation & Given

**Show ALL calculations to receive full credit.

 Trial 1 Trial 2 Trial 3 Initial volume of titrant 48.00mL 38.10mL 27.00mL Final volume of titrant 38.10mL 27.00mL 36.55mL Volume titrant added Concentrate of analyte

25.00mL of unknown concentration of FeSO4 (analyte) in H2SO4

Simplified reaction mechanism:

MnO41- + 5Fe2+ Mn2+ + 5Fe3+

Average concentration of analyte:

Post-Laboratory Questions

1. Indicate the effect of each of the following procedural mistakes on the reported concentration of the solution. Is the reported concentration too high, too low, or unaffected?
1. The buret was not rinsed with deionized water before being washed with

MnO41-(aq) solution before being filled for the titration.

1. The end point was exceeded by about three drops of titrant.
2. A few drops of MnO41-(aq) solution dribbled on the outside of the buret during filling and fell into the Erlenmeyer.
3. A couple of air bubbles stuck up the sides of the buret.
1. An excess of KI is added to the hypochlorite to first reduce the chloride. Why does the extra I1-(aq) not affect the titration?
2. Write the balanced equation for the titration you performed