Titration:3. In the titration of 80.0 mL of 0.150 M ethylamine, CH₂CH₂NH₂, with 0.100 M HCl, find thepH at each of the following points in the titration. (K = 4.7 x 10-4)Titration pointInitially14 completiona. Initially, before and HCL has been added.b. At 1/4 completion.c. At the halfway point in the titration.d. At the endpoint.excess.e. 10.0 mL of HCI beyond the endpoint.HalfwayEndpointExcessPH

Answered: Image /qna-images/answer/5cd9dd0f-bd85-45aa-9020-48cc86c8b6cc.jpg

Titration: 3. In the titration of 80.0 mL of 0.150 M ethylamine, CH3CH₂NH₂, with 0.100 M HCl, find the pH at each of the following points in the titration. (K = 4.7 x 10-4) Titration point Initially 14 completion Halfway Endpoint Excess a. Initially, before and HCl has been added. b. At 1/4 completion. c. At the halfway point in the titration. d. At the endpoint. Kexcess. e. 10.0 mL of HCI beyond the endpoint. PH

Titration: 3. In the titration of 80.0 mL of 0.150 M ethylamine, CH3CH₂NH₂, with 0.100 M HCl, find the pH at each of the following points in the titration. (K = 4.7 x 10-4) Titration point Initially 14 completion Halfway Endpoint Excess a. Initially, before and HCl has been added. b. At 1/4 completion. c. At the halfway point in the titration. d. At the endpoint. Kexcess. e. 10.0 mL of HCI beyond the endpoint. PH

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: <CHE154 S20 Ch18 Sec4-6 Chapter 18 Reading Question 4 Part A When 25.00 mL of 0.0500 M HBr is…

A: Given, Volume of HBr = 25.00 mL = 0.025 L (1 mL = 0.001 L) Molarity of HBr = 0.0500 M = 0.0500…

Q: Use the References to access important values if needed for this question. How many grams of solid…

Q: Titration Curves. Solve the ff. Problems 1. A 25mL aliquot of 0.1M NaOH is titrated with 0.10M HCl.…

Q: A 25.0mL sample of 0.125M pyridine is titrated with 0.100M hydrochloric acid. Calculate the pH at…

A: From given datas With Kb of pyridine, at different volume of added acid pH is calculated…

Q: 10- 8- 6- OH 2- H30¹ **** 10 strong acid strong base B A ... 20 30 40 Volume titrant (mL) Aside from…

A: Answer:Since, on adding the volume of titrant, pH of the solution is decreasing, therefore it is…

Q: 8. A buffer consists of a mixture of a weak acid and its conjugate base but the pH 4.00 buffer…

A: above molecule contain both acid as well as salt . so it can act as buffer. if we add hCl to the…

Q: D)

A: Since,you have posted a question with multiple sub-parts,we will solve the first three sub-parts for…

Q: 3. Weak base titration: You are titrating a 31.25 mL solution of 0.100 M trimethylamine (pK=4.20)…

A: According to the Bartleby guidelines for the multipart questions I can solve only first three…

Q: A 25.0mL samnple of a solution of an unknown compound is titrated witha 0.115M NaOH solution. The…

A: For the both at Equivalence point and at Pka its pH is same Around = 8.5

Q: 4. Explain why pH = pKa when the "half-way point" (half-way to the equivalence point) is…

A: -> Half equivalence point represents the point at which exactly half of the acid in the buffer…

Q: Not Secure- gmc.mrooms3.net glish - United States (en_us) When titrating a weak base with HCl at…

A: In a weak base-strong acid titration, the acid and base will react to form an acidic solution.

Q: 1. 10.0 mL of 0.40 M weak acid is being neutralized with 0.10 M NAOH. The results of this titration…

A: a. For the titration of a weak acid with a strong base, the pH at the equivalence point is greater…

Q: Which is the best indicator for the equivalence point during the titration of K2O solution with HCl.…

A: K2O forms KOH in its solution. KOH is a strong base and HCl is a strong acid. The best indicator for…

Q: Determine the pH during the titration of 33.2 mL of 0.220 M ammonia (NH3, Kb = 1.8 10-5) by 0.220 M…

A: we have to calculate the pH at the given points for the titration of 33.2 mL of 0.220 M ammonia by…

Q: B)

A: The milliliters of NaOH are needed to reach the equivalence point is to be calculated. The…

Q: 1. Consider the titration of 50.0 mL of 0.100 M HA (a weak acid with K.-4.0 x 10) with 0.100 M KOH.…

Q: Questions 1. Consider the titration of 25.00 mL of 0.200 M fantastic acid (Ka = 7.0x10-6). The…

Q: 25.0 mL of 0.25 M HCN (for HCN; Ka = 6.2 x 10-10) sample is titrated with 0.50 M NaOH. Calculate the…

A: The question is based on the concept of the pH of the solution.It is defined as a negative logarithm…

Q: Which is the best indicator for the equivalence point during the titration of KOH solution with…

A: Equivalence point in titration is the point at which the number of moles of the standard titrant…

Q: Calculate the pH when 2.0 M HCl is used to titrate 50 mL of 1.0 M NH3 (Kb = 1.8 x 10-5 ). a.) before…

A: This is problem of titration of weak base and strong base which goes through formation of basic…

Q: Homework: Construct a titration curve for 10 mL 0.01 M NaNH₂ titrated by 0.01 M HCl in NH3 solvent.…

A: As per the given question i was provided the solution i hope it will helps you.Explanation:Step…

Q: C)

Q: G)

A: The volume of 1 mL is equal to 10-3 L. The number of moles of NaOH in 75 mL of 0.0500 M is…

Q: 11. A 35.0 mL sample of 0.100 M acetic acid is titrated with a 0.150 M NaOH solution. The Ka of…

A: pH is an indicator to check the hydrogen ion’s amount concentration is available in a solution. The…

Q: A 30.0-mL sample of 0.20 M HCOOH was titrated with 0.20 M NaOH. The following data were collected…

A: Answer:Buffer solution is a type of solution that resists the change in its pH on adding small…

Q: pH

Q: The following acid–base indicators are available to follow the titration. Which of them would be…

Concept explainers

Solutions

Solutions can be considered as homogeneous mixtures which contain two or more than two chemical substances. The chemical substance which is the greater part is defined as solvent. And the chemical substance that is dissolved in a solvent is referred to as solute. Solution chemistry plays an important role in chemistry since it supports many commercial applications of solutions.

Question

Titration:
3. In the titration of 80.0 mL of 0.150 M ethylamine, CH₂CH₂NH₂, with 0.100 M HCl, find the
pH at each of the following points in the titration. (K = 4.7 x 10-4)
Titration point
Initially
14 completion
a. Initially, before and HCL has been added.
b. At 1/4 completion.
c. At the halfway point in the titration.
d. At the endpoint.
excess.
e. 10.0 mL of HCI beyond the endpoint.
Halfway
Endpoint
Excess
PH

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: Calculate volume required to reach equivalence point and halfway pont

VIEW

Step 2: Before addition of acid

VIEW

Step 3: completion of 1/4 point

VIEW

Step 4: halfway point

VIEW

Step 5: equivalence point

VIEW

Step 6: 10.0 mL beyond the end point

VIEW

Solution

VIEW

Trending now

This is a popular solution!

Step by step

Solved in 7 steps with 5 images

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

1. How to pick the best indicator for any titration? a. The one that is pink, if the solution is acidic b. The one that is blue if the solution is basic c. The one with pKa of the solution = pH (predicted at the equivalence point) d. It's a trial and error procedure e. None of the above
E)
Based on that experimental titration curve of the IU unknown sample, can you sketch out the titration curve if both the starting weak acid solution and the NaOH solution are diluted 10 times? What about diluting another 10 times on top of that? Particularly, you should pay attention to the size of the pH jumps around each equivalence point, as well as the relative positions of the flat regions before or after the pH jumps. For the sake of simplicity, no need to consider activity in this exercise.
Consider the titration of a 20.0 mL sample of 0.105 M HC2H3O2 with 0.125 M NaOH. Determine eachof the following. (Ka HC2H3O2 = 1.8 x 10-5) d) the pH at one-half of the equivalence point e) the pH at the equivalence point f) the pH after adding 21.8 mL of base beyond the equivalence point
5 Please answer all questions. Thanks
4. You are titrating 50.0 ml of 0.100 M Na:CO, with 0.400 M AGNOS. Calculate the pAg* values.. a) 1.0 ml. before the equivalence point b) at the equivalence point c) 1.0 mL after the equivalence point
Consider titration of 25 mL of .IM CH₂ CO₂H with 1 M NaOH. Calculate the pH after the following volumes of NaOH: adding a) O mL b) 12.5 mL c) 25mL d) 37.5 mL
Read sec. 14.7 2, then calculate the volume at equivalence and the pH at each of the following volumes of titrant for the titration of a 75.00 mL of a 0.049 mol L1 solution of nitrous acid (pKa = 3.37) titrated with 0.198 mol L' sodium hydroxide. %3D Ve = 18.6 mL VOH pH VOH 7.05 pH : VOH = = 12.1 pH %3D VOH = Ve %3D pH %3D Von = 27.8 pH = VoH = 55.7 pH Submit Question Jump to Answer
Calculate the pH during the titration of 20.00 mL of0.1 M NaOH with 0.100 M HCl after theaddition of the following volumes of reagent:a. 0.00 mLb. 10.00 mLc. 15.00 mLd. 20.00 mLe. 24.5 mLf. 25.00 mLg. 30 mLh. 35 mL
Answer using this data for the remaining questions on this page: For 1.0L of an acetic acid solution: Ka = 1.8 x 10-5 and [HA] = 0.250M 2. Calculate the pH of the solution. 3. What is the pka? 4. Calculate the pH of the solution if you make it a buffer by adding enough solid sodium acetate to make the concentration of acetate ion equal to 0.35M. The volume of the solution does not change. 5. What is the new concentration of [HA] if 2.0mL of 5.00M of HCI is added to 1.000L of the solution above? 6. What is the new pH?