### Experiment: Determining the Molarity of Acetic Acid (AA)The following table summarizes the data collected during a titration experiment aimed at determining the molarity of acetic acid using sodium hydroxide (NaOH) as the titrant.| **Quantity** | **Example** | **Trial 1** | **Trial 2** | **Trial 3** ||------------------------------------|-------------|-------------|-------------|-------------|| **M NaOH** (exact concentration) | 0.992 M | 0.953 M | 0.953 M | 0.953 M || **V initial buret reading = Vi** | 0.20 mL | 1.20 mL | 0.52 mL | 0.15 mL || **V final buret reading = Vf** | 12.90 mL | 14.22 mL | 13.71 mL | 13.31 mL || **Vep = VNaOH added = Vf - Vi** | 12.70 mL | 13.02 mL | 13.19 mL | 13.16 mL || **Vep = VNaOH in L** | 0.01270 L | 0.01302 L | 0.01319 L | 0.01316 L || **moles NaOH = MNaOH x VNaOH** | | | | || **moles AA = moles NaOH** | | | | || **V sample = V acid** | 0.0150 L | 0.0150 L | 0.0150 L | 0.0150 L || **Actual molarity of AA** | 0.840 M | 0.827 M | 0.840 M | 0.833 M || **

Since you have posted a question with multiple sub-parts, as per our company guidelines we are…

Quantity Example Trial 1 Trial 2 M NaOH (exact concentration) 0.992 M 0.953 M 0.953 M V initial buret reading = V, 0.20 mL 1.20 mL 0.52 mL %3D V final buret reading = V; 12.90 mL 14.22 mL 13.71 mL %3D Vep = V NAOH added = V, - V, 12.70 mL 13.02 mL 13.19 mL %3D Vep = VNaOH in L 0.01270 L 0.01302 L 0.01319 L %3D moles NaOH = MNAOH X VNAOH moles AA = moles NaOH V sample = V acid 0.0150 L 0.0150 L 0.0150 L (15.00 mL) (15.00 mL) Actual molarity 0.840 M 0.827 M 0.840 M of AA Average molarity for 3 trials => 0.833 M

Quantity Example Trial 1 Trial 2 M NaOH (exact concentration) 0.992 M 0.953 M 0.953 M V initial buret reading = V, 0.20 mL 1.20 mL 0.52 mL %3D V final buret reading = V; 12.90 mL 14.22 mL 13.71 mL %3D Vep = V NAOH added = V, - V, 12.70 mL 13.02 mL 13.19 mL %3D Vep = VNaOH in L 0.01270 L 0.01302 L 0.01319 L %3D moles NaOH = MNAOH X VNAOH moles AA = moles NaOH V sample = V acid 0.0150 L 0.0150 L 0.0150 L (15.00 mL) (15.00 mL) Actual molarity 0.840 M 0.827 M 0.840 M of AA Average molarity for 3 trials => 0.833 M

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

See similar textbooks

Related questions

Q: 3 attermpts left Check my work Be sure to answer all parts. Starting with a 6.701 M stock solution…

A: Given data: Initial concentration of HNO3 = 6.701 M AAt each stage, 25 mL of solution is diluted to…

Q: STARTING AMOUNT 0.1 6.60 If you have 155 mL solution of a 0.762 M FeCl solution, how many grams of…

Q: 5.0 mL of 3.6 M K2CO3 solution is diluted to make a 0.45 M solution, what is the volume of the…

A: Molarity is used for expressing the concentration of solution, which is expressed as mol/L. Given:…

Q: solution stock Concentration (mL) (mL) (HM) 0.100 100.00 Q2a 0.200 100.00 0.400 100.00 4 0.800…

A: Dilution refers to decreasing the concentration of solute in a solution. This is done by adding…

Q: How do you prepare 25 ml of 8.5 mM NaCl from a 20% NaCl stock (convert 20%→g/ml → molar)? What is…

A: The question is based on the concept of molarity.molarity is a method of expressing concentration of…

Q: A solution was prepared by dissolving 1210 mg of K3Fe(CN)6 (329.2 g/mol) in sufficient water to give…

A: This problem is based on general chemistry. Here we will be using general concentration terms in…

Q: How many mL of hydrochloric acid are required to measure to prepare 1 L of a 0.25 M solution if the…

A: To calculate the volume of hydrochloric acid (HCl) required to prepare a 0.25 M solution, you can…

Q: Experinent nome and dewical formaula Cobalt C1) nitkate Sulastance and Clemical frimula mm, 9/ mn…

Q: What would be the mcle fraction of kClinthe Sclution (mclar mass (8.029 12)what is the molarityct…

Q: Craw McGraw-Hill Education Campus Saved Ex. 1b - Mass/volume percent concentration - calculate using…

A: Volume of solution = 0.350 L Mass of KCl = 13.0 g We have to find m/v percent of solution.

Q: Solute S has a distribution constant of 3.7 between water (phase 1) and hexane (phase 2). Calculate…

A: According to the mole concept, in terms of mass, the amount of substance in moles is equal to the…

Q: I asked thsi question but confused on how the person got there molar mass wt . so I will try again .…

Q: (d) the grams of Ca3(PO4)2 that can be obtained from 123 mL of 0.750 M Ca(NO3)2 g Caз(РОд)2 the…

A: The coefficients existing with each species reacting as well as formed in particular balanced…

Q: The volume of water needed to dissolve 0.0694 grams of lead fluoride is ______ L. Assume no volume…

A: Here we have to find the volume of water needed to dissolve 0.0694 grams of lead fluoride. Given,…

Q: The K, value for NaCl is approximately 37.5 M²/L3 at 25'C. Determine the absolute and percent errors…

A: To calculate the absolute error and the percent error in the experimental values. The experimental…

Q: Full Screen Accessibility The Ksp of the phosphate fertilizer CaHPO4 2H20 is 2.7 x 10-7 at 25°C. The…

A: Given: Ksp=2.7 x 10-7 CaHPO4.2H2O --> Ca2+ + HPO42- + 2 H2O s…

Q: Table 4. Calculations for standardization of sodium hyroxide Trial 1 Trial 2 Trial 3 [1] Tared mass…

A: We have to find the molarity of NaOH using moles of KHP

Q: Explain why one compound (Urea/cholesterol) dissolved in water while the other one did not using…

A: Interaction of two or more compounds with each other can be explained on the basis of like dissolve…

Q: Question 8 The addition of a nonvolative solute to a solvent will create a solution with a freezing…

A: The addition of nonvolatile solute to a volatile liquid solvent than the solution vapour pressure…

Q: Determine the volume of 0.150 M NaOH solution required to neutralize 130 mL of a 0.115 M HCl…

Q: K QRSTU v Q2b: Serial Dilution (M,V=M,V2) Dilute aliquots of the previously made solution solution…

Q: Given the following data Dissolved Oxygen Titration Data Trial 1 Preparation of titrant 0.4455 g…

Q: What NaCl concentration results when 299 mL of a 0.780 M NaCl solution is mixed with 587 mL of a…

A: In this question, we will determine the resulting concentration of NaCl when two different…

Q: An aqueous solution of ethanol, CH;CH,OH, has a concentration of 6.99 mol/L and has a density of…

A: Answer :- a) Mass percent of CH3CH2OH = 34.0% b) Mole fraction of CH3CH2OH = 0.167…

Q: Question 1 of 10 Subm How many moles of Na,SO. are in a 0.140 m aqueous solution that was prepared…

A: Given: The mass of water, (w) = 2.30 Kg. The molality of the solution, (m) = 0.140 m.

Q: lo what volume should you dilute 41.0 mL of a 5.95 M KI solution so that 24.5 mL of the diluted…

Q: From: http://ths.sps.lane.edu/chemweb/unit6/problems/molarity/ 0.450 moles of NaCl are dissolved in…

A: Here we have been asked two questions related to molarity. Moarity can be defined as the number of…

Q: 7) If you are told a line has a slope = 12 g/mL and it came from a plot using a formula of y = (B/4)…

Q: What is the mass in grams of NaCN in 120.0 mL of a 2.40 x 10 M solution? STARTING AMOUNT ADD FACTOR…

A: Given: Volume of solution = 120 ml = 0.120 L Molarity of solution= 2.40 × 10-5 M

Q: How do you find weight of Na2C2O4 in unknown?

Q: Macmillan Learning How many grams of solute are present in 575 mL of 0.560 M KBr? mass of solute: 60

Concept explainers

Kinetic-Theory of Gases

Kinetic theory of gases is the model or method which was developed in the 19th century by Maxwell and Boltzmann. This is possible as the interatomic force which is of short-range forces that are important for solids and liquids can be neglected for gases. This theory or gas law gives a molecular interpretation of the pressure and temperature of a gas and is consistent with gas laws and Avogadro’s hypothesis. This theory is very helpful to correctly explain the specific heat capacities of many gases, kinetic energy and it also explains the transport qualities such as viscosity, thermal conductivity, and diffusion with molecular parameters. As stated in the kinetic-molecular theory, the temperature of a substance or gas particle is related to the average kinetic energy of the particles of that substance. When a substance is heated, some of the absorbed energy is stored within the particles, while some of the energy increases the motion of the particles, called average kinetic energy.

Question

100%

### Experiment: Determining the Molarity of Acetic Acid (AA)

The following table summarizes the data collected during a titration experiment aimed at determining the molarity of acetic acid using sodium hydroxide (NaOH) as the titrant.

| **Quantity** | **Example** | **Trial 1** | **Trial 2** | **Trial 3** |
|------------------------------------|-------------|-------------|-------------|-------------|
| **M NaOH** (exact concentration) | 0.992 M | 0.953 M | 0.953 M | 0.953 M |
| **V initial buret reading = V<sub>i</sub>** | 0.20 mL | 1.20 mL | 0.52 mL | 0.15 mL |
| **V final buret reading = V<sub>f</sub>** | 12.90 mL | 14.22 mL | 13.71 mL | 13.31 mL |
| **V<sub>ep</sub> = V<sub>NaOH</sub> added = V<sub>f</sub> - V<sub>i</sub>** | 12.70 mL | 13.02 mL | 13.19 mL | 13.16 mL |
| **V<sub>ep</sub> = V<sub>NaOH</sub> in L** | 0.01270 L | 0.01302 L | 0.01319 L | 0.01316 L |
| **moles NaOH = M<sub>NaOH</sub> x V<sub>NaOH</sub>** | | | | |
| **moles AA = moles NaOH** | | | | |
| **V sample = V acid** | 0.0150 L | 0.0150 L | 0.0150 L | 0.0150 L |
| **Actual molarity of AA** | 0.840 M | 0.827 M | 0.840 M | 0.833 M |
| **

Expert Solution

This question has been solved!

Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.

This is a popular solution!

SEE SOLUTION Check out a sample Q&A here

Step 1

Step 2

Step 3

Step 4

Step 5

Trending now

This is a popular solution!

Step by step

Solved in 5 steps

SEE SOLUTION Check out a sample Q&A here

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.

Similar questions

EX 2.Given-> Volume of Na+ = 500 ml Molarity of Na+= 0.0100M Molar mass of Na2CO3 = 105.99 gm/mole Millimole of Na+ = molarity × volume Number of millimole = 0.0100 × 500 = 5 millimole Na2CO3 ---> 2Na+ + CO32- Millimole of Na2CO3 = millimole of Na+/2 Millimole of Na2CO3 =5/2 = 2.5 millimole Mole of Na2CO3 = 2.5 × 10-3mole (1 mole = 10^3 millimole) Weight of Na2CO3 required = mole × molar mass = 2.5 × 10-3 × 105.99 =0.26 gm Hence, 0.26 gm Na2CO3 must dissolve in 500 ml of water.
Please fill in blanks as shown on problem!!
3:36 1 ul LTE O Send a chat What volume in of a 0.32 M Mg(NO3)2 solution contains 45 L g of Mg(NO3)2 ? STARTING AMOUNT ADD FACTOR ANSWER RESET *( ) 210.34 6.022 × 10" 45 0.30 140 0.32 1 10 148.33 0.95 86.32 0.097
Answer it asap i will upvote you
i have hard time geting it stright even after I do the experiment can u help me with the one pls Material ChemicalsSealed NO2 tubes, KSCN solution (0.002 M), acidified Fe(NO3)3 solution (0.2 M), KSCN crystals, Fe(NO3)3 crystals, KOH (concentrated (8 M)), distilled waterApparatus:Bunsen burner, ring stand, wire gauze, matches, thermometer, dropper, test tubes, glass stirring rod, beaker tongs Procedure Part A: Fill a 600 mL beaker to the 3/4 mark with water. Bring to a boil using a Bunsen burner. In another 600 mL beaker, fill to the 1/2 mark with water; add ice in order to lower the temperature to about 0°C. Note the room temperature. Observe the NO2 thoroughly at room temperature. Place the sealed NO2 tube in the cold bath. Note your observations. Remove the test tube from the cold bath. Let the test tube stand until it reaches room temperature. Note your observations. Place the test tube in the hot bath. Note your observations. Remove the test tube from the hot bath. Let the test…
A student combined 37.2 mL of 1.69 M KMnO4 solution with 16.60 mL of a 0.891 M KMnO4 solution. Calculate the concentration of the final solution. Tap here r pull up for additional resources Q A 2 W S # 3 E D $ 4 R % 5 F LL Question 5 of 17 T < 60 G Y & 7 H C +00 8 J 1 1 4 7 +/- K 5 C
Sample Quiz 10 a berkite kat A 1.132 g sample of KHP (HKC8H4O4) required 28.05 mL of NaOH solution to reach endpoint. Calculate the molarity of the NaOH. 7. Notify: sa Chemistry 60 lab Apga od safety lothing, All cals you are all Page 120
Find out the molarity of an old KOH solutionYou measure 1.456 g of KHP (MW=204.22 g/mole) dissolve it in 50.0 of H20 add an indicator and begin to titrate with the old KOH KHP (aq) +KOH (aq) =K2P (aq) +H2O (l) the initial and volume burette readings are 1.35 mL & 24.3 mL then what is the concentration of KOH ?
using the given values for trial 1, how would you find the mass of pure liquid H2O2?
I need help with this question please! Use dimensional analysis please!?
SOLVE STEP BY STEPDO NOT USE CHATGPTMAKE THE RESOLUTION QUICKLY PLEASE 18. How many mL of hydrochloric acid are required to measure to prepare 1 L of a 0.25 M solution if the density of HCl is 1.17 g/mL and its purity is 36%? (PA of H is 1,008 amu, Cl is 35,451 amu)
What volume of 0.150 M Li2S solution is required to completely react with 130ml of 0.150 M Co(NO3)2?