Chapter 9: Unit 9. Titration and Stoichiometry of Acid Base Reactions

Titration and Stoichiometry of Acid Base Reactions

Titration is a procedure by which an acid with a base are neutralized. The purpose of such procedure is to determine the concentration of an acid or a base to a precise value (several trials are needed to obtain the averaged precise value).

The process of titration is using an indicator which is an organic compound (a dye) that can change its color by changing the medium from acidic to basic or vice versa. Phenolphthalein is colorless in acidic medium and it has a pink color in basic medium.

In the titration process, on can differentiate between two important points:

• a. End point: The point by which the indicator is changing its color. The signaling of color change is observed by an addition of one drop of the titrant from the buret after the neutralization of the acid with the base is reached.
• b. Equivalent point: the point by which is neutralization is reached and the number of moles of the acid equal number of moles of the base.

Note that: the end point comes always after the equivalent point. One drop of the titrant signals the color change.

The figure below shows the difference between the end point and equivalent point of the titration of HCl with NaOH.

http://pediaa.com/difference–between–equivalence–point–and–endpoint/

A typical setup of the titration is given below:

https://www.khanacademy.org/test–prep/mcat/chemical–processes/titrations–and–solubility–equilibria/a/acid–base–titration–curves

A You Tube video goes over the titration procedure in details:

A simulation of acid base titration is illustrated below:

https://pages.uoregon.edu/tgreenbo/acid_base.html

A video of You Tube explains the simulation above:

Simulation Activity

1.) Select reaction type: Strong acid vs. Strong base

2.) Fill the burette with: Base

3.) Select the acid and base: Acid = HCl ; Base = NaOH

4.) Select the indicator: Phenolphthalein

5.) Push slider up to add a volume of base: Add the NaOH. You can begin by just sliding the slider up, adding about 10–12 mL of NaOH. After that, continue adding the NaOH dropwise, by clicking on the red circle. Continue adding NaOH to the flask until the solution in the flask turns pink. Once the solution has turned pink, you have reached the neutralization point, and your titration is complete.

6.) Print screen and copy and paste the graph into page 2.

7.) Calculate and enter molarity of the base: Fill in the appropriate data in the data list below using the values shown in the simulation. (Be sure to include units!) Then, calculate the molarity of NaOH used. (Hint: Use formula M a V a = M b V b ) Enter the value you get in the box provided in the simulation. Make sure your value is correct!

8.) Repeat steps 1–7 for titration 2 of your choosing.

Submission Instructions

**You will be submitting only Page 2 of this document!!**

Fill in the document in word format (use equation editor where applicable

Save too I:\MMSTC\Homework\Acid Base Titration Lab in your class folder

File Name: Last name_class

Acid–Base Titration Virtual Lab

Titration 1

1.) Write the equation for the neutralization reaction in this titration:

2.) What indicator is used? What color does the indicator change to?

3.) Data:

Molarity of Acid: _____________

Volume of Acid: _____________

Volume of Base: _____________

4.) Calculate the molarity of the base. (Show all work, including the formula & units!)

Molarity of Base: ______________

5. Graph

Titration 2

1.) Write the equation for the neutralization reaction in this titration:

2.) What indicator is used? What color does the indicator change to?

3.) Data:

Molarity of __________: _____________

Volume of ___________: _____________

Volume of unknown: _____________

4.) Calculate the molarity of the unknown. (Show all work, including the formula & units!)

Molarity of unknown: ______________

5. Graph

Another simulation is given by Phet Simulation which illustrates the Acid – Base Reaction (Neutralization):

https://phet.colorado.edu/sims/html/acid–base–solutions/latest/acid–base–solutions_en.htmlhttps://phet.colorado.edu/sims/html/acid-base-solutions/latest/acid-base-solutions_en.html

Simulation Activity:

Acid Bases WebLab PhET Chromebook Version

https://phet.colorado.edu/en/simulation/acid–base–solutions Click <Introduction> to begin.

Part 1: Procedure

1. The lab has 2 tools that allow you to test for pH values: A probe , and pH paper . Use each one by dipping it into the solution to be tested. Try all the given types of solutions and fill in the Data Chart with the pH value 0–14.

2. The circuit with a battery and bulb as shown: is the tool used to test for conduction of a solution. By dipping the wire leads into the solution, the bulb with eitherremain unlit, be dimly lit, be somewhat bright or very bright. Test each solution and record your observation for the bulbs brightness in the chart below.

Part 1: Data	pH Value from Probe	Color & pH Value from pH Paper	Observations from Circuit ToolDescribe the brightness

Part 1: Analysis

1. What pH value range is observed: a. for acids?______________b. for bases?____________

2. Why are some solutions better conductors of electricity?

Part 2 Procedure, Data & Analysis:

Recall: The amount of ionization or dissociation of ions determines the strength of an acid or base. The concentration of [H 3 O + ], hydronium and [OH–], hydroxide ions can be used to calculate pH and pOH as shown on the diagram here:

Note: we use [H3O + ] and [H + ] interchangeably.

• 1. Click on Water Solution, Graph View, Probe Tool. Insert the probe in the water. Notice that the initial concentration of the solution is given before any ionization or dissociation takes place.
• 2. Fill in the missing concentration values for the hydronium and hydroxide ions on the diagram here: Use the concentration value for [H 3 O ⍅ ] to calculate the pH. Show work:3. Use the concentration value for [OH–] to calculate the pOH. Show work:

• 4. Did your answer to #2 match the pH given in the simulation? _________________________ 5. Is the answer to #3 equal to: (14 – pH)? ___________ Show work: _____________________6. Is the solution an acid, a base or neutral based upon the calculated pH?_________________HONORS ONLY: Attach notebook paper to show calculations for the pH and pOH for the other solutions.
• Part 3 Procedure, Analysis, Conclusion: My Solution
• Across the bottom of the screen, click the button. The default setting shows a weak acid with a concentration of 0.010 M. Insert the pH probe to show an initial pH of 4.50. The beaker is shown below:1. Slide the initial concentration bar to the right to increase the number of solute molecules and then slide it to the left.What effect does changing the concentration have on the pH value? (Be specific)

• 2. Return to your default setting and insert the probe. Now slide the strength to the right to make the acid stronger.
  • a. As you increase the strength, describe the change in the number of blue A– ions, orange H 3 O ⍅ ions and the original HA acid:
  • b. As you increase the strength, describe the change in the concentrations of the ions in the solution? Hint: Click <Graph> to see how the concentrations rise and fall.3. Yes or No? Does the pH seem to depend upon the concentration of [H 3 O + ] ions?4. We always assume that strong acids will 100% ionize in water. Click reset and move the slider to strength: strong. Insert the probe. Record pH. Observe the number of ions in the beaker and click <Graph> to observe the concentrations.
  • a. pH Value = ______________
  • b. YES or NO? Does the beaker contain particles that have not ionized and have 0% concentration? If so, what particle seems missing? ____________________. Why is it likely missing?5. Click reset and change to a base. Repeat #1–4 above and describe any different results or simply write, “Same results for bases.”#1:#2:#3:#4:Conclusions: If the answer is no, explainwhy not.6. YES or NO? Can a weak acid be concentrated?7. YES or NO? Can a strong acid be dilute?8. YES or NO? For acids, can increasing the initial concentration increase the pH?9. YES or NO? For Bases, can increasing the initial concentration increase the pH?
• Try different combinations of initial concentration and strength. Dip probe, click <Graph> to record ion concentration. Use concentration to calculate pH and verify probe data.Another Acid Base Simulation is shown below:https://users.wfu.edu/~ylwong/chem/titrationsimulator/index.html

• Simulation ActivityDetermination of acetic acid concentration is the desired simulation to do and will not do the rest of the simulations.. The directions are given within the simulation itself. A snapshot of such directions is given below:Do all steps required and discussed in the simulation for the acetic acid concentration determination.

• Stoichiometry of Acid Base Reaction:A video of You Tube explains the stoichiometry of Acid Base Reaction : https://youtu.be/HJvALCcKYAc

• Examples
• Stoichiometry of Acid Base Reaction:Mole Ratio of Acidto Base is NOT 1:1Example:What volume of 0.800M KOH is needed to titrate 15.0 ml of 0.500M H 2 SO 4 ?H2SO4(aq) + 2 KOH(aq) à K2SO4(aq) + 2 H2O(l)
• Mole Ratio of Acidto Base is 1:2The first step is to calculate the moles of the analyte H2SO 4 : 
• Converting the milliliters to Liters:15.0 mL x [1L / 1000 mL] = 0.0150 L H2SO40.0150 L H 2 SO 4 x [ 0.500 mol H2SO4 / L H2SO4 ] = 0.00750 moles H2SO4The second step is to calculate moles titrant KOH:The mole ratio of KOH : H2SO4 = 2: 1[0.00750 mol H2SO4 ] x [2 mol KOH / 1 mol H 2 SO 4 ] = 0.0150 mol KOHThe third step is to calculate the volume of KOH needed by dividing its number of moles by its molarity:[0.0150 mol KOH] x [L KOH / 0.800 mol KOH] = 0.01875 L = 0.0188 L
• Example: 
• What mass of water can be produced if 156 grams of sulfuric acid solutionH 2 SO 4 react with 250.ml of 1.2 M potassium hydroxide solutionKOH?H2SO4 (aq) + 2 KOH(aq) → K2SO4 (aq) + 2H2O(l)
• Mole Ratio of Acidto Base is 1:2
• Calculations of the water mass based on H2SO4 :First step to calculate number of moles H 2 SO 4 using its number of molar mass:Molar mass of H2SO4 = 2 H + 1 S + 4 O = [2x(1.00 g/mol) + 1x(32.0 g/mol) + 4x(16.0 g/mol)] = 98.0 g/mol[156 g H2SO4 ] x [mol H 2 SO 4 / 98.0 g H2SO 4 ] = 1.59 mol H2SO 4The second step to calculate the mass of water based on H2SO 4 :[1.59 mol H2SO4 ] x [ 2 mol H2O / 1 mol H2SO 4 ] = 3.18 mol H2OUsing the molar mass of H2O = 2 H + 1 O = 2x(1.00 g/mol) + 1x(16.0) = 18.0 g H2O /mol H2OMass of H2O = [3.18 mol H2O] x [18.0 g H 2 O /mol H2O] = 57.2 g H2OThe 57.2 g H2O is the theoretical amount of H 2 O if all the H2SO4 is reacting to completion
• Calculations of the water massbased on KOH: First step to calculate number of moles KOH using its molarity and volume in Liters:Converting the volume from milliliters to Liters:250. mL KOH x [1 L / 1000 mL] = 0.250 L KOH[0.250 L KOH] x [1.2 mol KOH / L KOH] = 0.300 mol KOHThe second step to calculate the mass of water based on KOH:[0.300 mol KOH] x [2 mol H 2 O / 2 mol KOH] = 0.300 mol H2OMass of H2O = [0.300 mol H 2 O] x [18.0 g H 2 O /mol H2O] = 5.40 g H2OThe 5.40 g H2O is the theoretical amount of H 2 O if all the KOH is reacting to completion.The results above indicate that KOH is the limiting reactant because KOH produces less water than the H 2 SO 4 and hence only 5.40 g of water can be made.
• Example:
• Determine the concentration of sulfuric acid required if it takes 28.5 ml of sulfuric acid to titrate a 36.0ml sample of 0.450M KOH.What are we trying to find? That’s right, the concentration or molarity of the sulfuric acid. Recall that molarity = mol solute/L solution. Since we have the volume and the molarity (concentration) of the KOH, we’ll begin with it. 0.0360 L KOH ( 0.450 mol KOH) ( 1 mol H 2 SO 4 ) = 0.00810 mol H 2 SO 4 needed 1 L KOH 2 mol KOH Molarity = moles of solute = 0.00810 mol H 2 SO 4 = 0.284 M or 0.284 mol/L liter of solution 0.0285 L
• Another Example of the acid –base stoichiometry calculations:
• Mole Ratio of Acidto Base is 1:1
• Example:
• Calculate the molarity of an acetic acid solution if 34.57 mLof this solution are needed to neutralize 25.19 mL of 0.1025 M sodium hydroxideCH3 COOH (aq) + NaOH (aq) → Na + (aq) + CH3COOH–(aq) + H2O (l)The first step is to calculate the moles of titrant (in this case, the NaOH):Converting the volume from milliliters to Liters:25.19 mL x [1 L / 1000 mL] = 0.02519 LMultiplying the volume in Liter by molarity of NaOH:0.02519 L x [0.1025 mol / L] = 0.002582 mol NaOHThe second step is to use the balanced chemical equation to calculate the moles of analyte (in this case, the CH 3 COOH) present:CH3COOH (aq) + NaOH (aq) → Na + (aq) + CH3COOH–(aq) + H2O (l)
• Mole Ratio of Acetic Acid to Sodium hydroxide:1: 1
  • 0.2582 gOH x [1 mol CH3COOH / 1 mol NaOH] = 0.002582 mol CH3COOHThe third step is to use the volume of analyte to find the concentration of the analyte.Converting the volume of the analyte from mL into L:
  • 0.2583 34.57 mL x [1L / 1000 mL] = 0.03457 L CH 3 COOHThe concentration of acetic acid: = 0.002582 mol CH3COOH / 0.03457 L CH3COOH = 0.07469 M