jr pecs SiaIn titrating 10 ml of 0.1 N HA (pKa=4),the pH after the addition of 7 ml 0.1NNaOH isO 2.323.253.614.50O 4.36In titrating 10 ml of 0.1 M basic amino acid (pKa1=.10).the pH after the addition of 15 ml 0.1N NaOH will bepKa2 = 4, pKa3 D

Buffer is the solution of mixture of weak acid/weak base with its salt of strong base/strong acid

Answered: In titrating 10 ml of 0.1N HA…

Science

Chemistry

In titrating 10 ml of 0.1N HA (pKa=4),the pH after the addition NaOH is O 2.32 3.25 O 3.61 4.50 4.36

Appl Of Ms Excel In Analytical Chemistry

2nd Edition

ISBN:9781285686691

Author:Crouch

Publisher:Crouch

Chapter7: Neutralization Titrations And Graphical Representations

Section: Chapter Questions

Problem 8P

See similar textbooks

Related questions

Q: Find the pH of the equivalence point(s) and the volume (mL) of 0.447 M HNO3 needed to reach the…

A: The equilibrium constant for a reaction is expressed in terms of the ratio of the concentration of…

Q: Determine the pH at the equivalence point in the titration of 50.0 mL of 0.300 M CH;COOH with 0.300…

Q: A 50.00-mL aliquot of 0.0700 M NaCN is titrated with 0.140 M HCI. The reaction is: CN' (aq) + H3O*…

A: The question is based on the concept of titrations. we are titrating a strong hydrochloric acid…

Q: Which of the following mixtures of acids exhibit(s) two end points when titrated with 0.10 M NAOH? O…

A: Option b is correct.

Q: A flask contains 50 mL of a buffer that is 0.1 M in acetic acid and 0.20 M in acetate ion. What will…

Q: In titrating 10 ml of 0.1 M neutral amino acid (pKal= 2, pKa2 = 8),the pH after the addition of 5 ml…

Q: A 50.0-mL sample of 0.10 MHNO 2 (Ka 4.0 x 10-4) is titrated with 0.18 M NaOH. The pH after 11.1 mL…

A: Given -> Volume of HNO2 = 50.0 ml Molarity of HNO2 = 0.10 M Molarity of NaOH = 0.18 M Volume of…

Q: How many mL of 1.25 M H3PO4 are required to reach the equivalence point when titrated with 45.0 mL…

A: The balanced reaction taking place is given as, => 2 H3PO4 + 3 Ca(OH)2 --------> 6 HOH +…

Q: Assuming the Ka is 2.37 x 10-5 for acetic acid, calculate the pH at the half equivalence point for a…

A: Acetic acid is a weak acid while NaOH is a strong base. So the titration of acidic acid with NaOH is…

Q: Calculate the pH of a titration of 50.00 mL of 0.100 M Phenylacetic acid, Ka = 4.9 x 10-5, with…

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

Q: 125 mL of a 0.098 M aniline (pKb = 9.40) solution is titrated with 0.217 M HCI. What is the pH after…

A: Aniline is a weak monoprotic base. Hence we can assume aniline as An Since 0 mL of HCl is added.…

Q: It required 38.72 mL of 0.068M NaOH to reach the equivalence point in the titration of 25 mL of…

Q: 25 mL of 0.10 M acetic acid is titrated with 0.10 M NaOH. What is the pH after 30 ml of NaOH have…

Q: Consider the titration of 100.0 ml of 0.100 M HOCI by 0.100 M KOH at 25°C. K, for HOCI = 3.5x10-8.…

A: Since you have posted a question having multiple subparts, we will solve only the first three…

Q: Calculate the pH during the titration of 40.0 mL of 0.100 M propanoic acid (1.3 x 10-5) after each…

Q: A flask contains 50 mL of a buffer that is 0.10M in acetic acid and 0.20 M in acetate ion. What will…

Q: You titrate 0.1Lof 0.2 M HCI with 0.1 M NaOH. The half equivalence point will be reached when added,…

A: .

Q: Calculate the pH of a titration of 50.00 mL of 0.100 M Phenylacetic acid, Ka = 4.9 x 10-5, with…

A: The relation for ionic product of water is as follows, Kw = Ka × Kb The…

Q: 1. Calculate the volume, in mL, of a 0.50 M NAOH solution that will be required to titrate 1.2 of…

Q: If you add 5.0 mL of 0.20 M HCI to 25 mL lactic acid/lactate buffer that is 0.10 M lactic acid and…

A: The pH of a buffer solution is determined by Henderson equation…

Q: A 60.0 ml solution of a weak acid, 0.300 M HA is titrated with a 0.200M NaOH solution (Ka, HA = 7.90…

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

Q: The pH at the equivalence point for the titration of a weak acid with a strong base will always be O…

A: Answer: pH defines the acidity of the solution. Mathematically, it is the negative logarithm of H+…

Q: Determine the pH during the titration of 25.8 mL of 0.487 M benzoic acid (K, = %3D 6.3×10-5) by…

Q: HA is a weak acid with a Ka = 4.5 × 10-6 and a 25.0 mL sample of 0.10 M HA is titrated with 0.10 M…

Q: You titrate 0.1 L of 0.2 M HCI with 0.1 M NaOH. The half equivalence point will be reached when…

Q: Aditi titrated 80.00 mL of 0.50 M methylamine, CH3NH2, with 0.40 M HCI. CH3NH₂ Kb = 4.4 x 10-4 What…

Q: c) A student titrated 50.0 mL of a 0.10 M solution of a certain weak acid with NaOH. The results are…

A: Identify a pH value between 2.5 and 7.5 at which the concentration of the weak acid being titrated…

Q: ated Find the pH at the end point in the titration of 0.2 M potassium acid maleate, KHM. with 0.2 M…

A: The reaction between KHM and NaOH: KHM+NaOH→NaKM+H2O

Q: Calculate the pH in the titration of 50.00 mL of 0.060 M acetic acid (CH3COOH) with a 0.120M sodium…

A: (A)pH when 0mL of NaOH solution is added:

Q: 3. (а) When 40 mL of benzoic acid (C,H5COOH) and 35 mL of potassium hydroxide were titrated, the pH…

Q: A sample of 0,444 M methylamine (CH3NH2, K 4.29 x 10-) is titrated with 0.444M HCL What is the pH at…

A: Whenver an acid and base reacts with each other than they form salt and water and the process is…

Q: A 20 mL of 0.10 M HCI was titrated with 0.06 M NaOH. What is the pH at the equivalence point? O 4.23…

A: The pH at equivalence point can be calculated as follows

Q: Calculate the pH of a titration of 50.00 mL of 0.100 M Phenylacetic acid, Ka = 4.9 x 10-5, with…

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

Q: 4) Calculate the pH at the points in the titration of 25.00 mL of 0.10 M NH3 when (a) no acid…

A: Ammonia is a weak acid, when it titrated with strong acid like hydrochloric acid, the pH of the…

Q: A 20.0 mL sample of 0.150 M ethylamine is titrated with 0.0981 M HCl.What is the pH after the…

A: The volume of 1 L is equal to 1000 mL. The number of mole (n) of ethylamine is calculated as shown…

Q: in the titration of 50.0 mL 0.0200 M weak acid HA using 0.0150 M NaOH, what is the solution pH at…

A: Number of moles = Molarity x volume (L) Molarity of HA = 0.0200 M Volume of HA = 50.0 mL = 0.050 L…

Q: Use the titration curve to calculate the pKa of HA when 0.00 mL of NaOH has been added.

A: In an acid -base titration , pKa of acid can be determined from the pH vs Volume of base plot . This…

Q: What is the pH

Q: the pH of the solution at the half equivalence point

A: In titrating 0.20 M acetic acid, CH3COOH, with 0.20 M NaOH, the pH of the solution at the half…

Q: Consider the titration of a 25.0-mL sample of 0.115 M CH3COOH with 0.100 M КОН. Ką CH3COOH =…

A: At equivalence point no. Of moles of acid is equal to the no. Of moles of base. Therefore,…

Q: Derive the titration curve of the following solution: a. 20.00 mL, 0.25 M CaCO, with 0.20 M AgNOs…

A: Since you have asked multiple questions, we will solve the first question for you. If you want any…

Q: #6 Calculate the pH during the titration of 40.00 mL of 0.1000 M HCl with 0.1000 M NaOH solution…

A: Given, Volume of HCl = 40.00 mL = 0.040 L (1mL = 0.001 L) Molarity of HCl = 0.1000 M = 0.1000…

Q: HA is a weak acid with a Ka = 6.5 × 10-6 and a 45.0 mL sample of 0.10 M HA is titrated with 0.10 M…

A: HA is a weak acid with a Ka = 6.5 × 10-6 and a 45.0 mL sample of 0.10 M HA is titrated with 0.10 M…

Q: 26 mL of 0.11 M acetic acid solution is titrated with 0.10 M NaOH. What is the pH after 12 mL of…

A: Volume of CH3COOH = 26 mL Molarity of CH3COOH = 0.11 M Volume of NaOH = 12 mL Molarity of NaOH =…

Q: Calculate the pH during the titration of 40.00 mL of 0.1000 M HCl with 0.1000 M NaOH solution after…

A: pH of the solution after addition of 26.00mL of base to 40.00mL of 0.1000M HCl with 0.1000M NaOH…

Q: 57.2 of 0,45M HC2H3O2 is titrated with 0.20M NAOH. What is the pH at the equivalence point? Ka of…

A: For the neutralization: N(acid) x V(acid) = N(base) x V(base) N = normality, V = volume Normality =…

Q: You are titrating a weak acid and determine that the half-equivalence point is 7.35 mL. However, the…

A: Weak acid- an acid dissociate slowly into its ions in an aqueous solution to give H+ ions e.g.…

Q: 4) Calculate the pH at the points in the titration of 25.00 mL of 0.10 M NH3 when (a) no acid…

A: 4) (a) The reaction can be given as follows: NH3(aq)+H2O(l)↔NH4+(aq)+OH-(aq) The moles of NH3 can be…

Q: 16. Calculate the pH at the equivalence point in the titration of 25 mL of 0.10 M formic acid with a…

Q: Determine the pH at the equivalence point in the titration of 50.0 mL of 0.300 M CH3COOH with 0.300…

Question

jr pecs Sia
In titrating 10 ml of 0.1 N HA (pKa=4),the pH after the addition of 7 ml 0.1N
NaOH is
O 2.32
3.25
3.61
4.50
O 4.36
In titrating 10 ml of 0.1 M basic amino acid (pKa1=.
10).the pH after the addition of 15 ml 0.1N NaOH will be
pKa2 = 4, pKa3 D

Expert Solution

This question has been solved!

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

SEE SOLUTION Check out a sample Q&A here

Step 1

VIEW

Step 2

VIEW

Step by step

Solved in 2 steps with 1 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

Calculate the pH at the following points in a titration of 40 mL (0.040 L) of 0.100 M barbituric acid (Ka=9.8105) with 0.100 M KOH. (a) no KOH added. (b) 20 mL of KOH solution added. (c) 39 mL of KOH solution added. (d) 40 mL of KOH solution added. (e) 41 mL of KOH solution added
Consider the titration of 100.0 mL of 0.200 M acetic acid (Ka = 1.8 105) by 0.100 M KOH. Calculate the pH of the resulting solution after the following volumes of KOH have been added. a. 0.0 mL b. 50.0 mL c. 100.0 mL d. 150.0 mL e. 200.0 mL f. 250.0 mL
Lactic acid is a common by-product of cellular respiration and is often said to cause the burn associated with strenuous activity. A 25.0-mL sample of 0.100 M lactic acid (HC3H5O3, pKa = 3.86) is titrated with 0.100 M NaOH solution. Calculate the pH after the addition of 0.0 mL, 4.0 mL, 8.0 mL, 12.5 mL, 20.0 mL, 24.0 mL, 24.5 mL, 24.9 mL, 25.0 mL, 25.1 mL, 26.0 mL, 28.0 mL, and 30.0 mL of the NaOH. Plot the results of your calculations as pH versus milliliters of NaOH added.
In a titration, 25.00 mL of 0.1000 M triethyl amine ((CH3CH2)3N) is titrated with 0.0500 M hydrochloric acid (HCl). The following data was obtained: mL HCl added pH 0.00 11.86 25.00 10.71 50.00 6.79 What is pKa for the triethylammonium ion? 3.29 10.71 11.86 6.79
Find the pH during the titration of 20.00 mL of 0.1000 M butanoic acid, CH₃CH₂CH₂COOH (Kₐ=1.54X10⁻⁵), with0.1000 M NaOH solution after the following additions of titrant:(a) 0 mL(b) 10.00 mL(c) 15.00 mL(d) 19.00 mL(e) 19.95 mL(f ) 20.00 mL(g) 20.05 mL(h) 25.00 mL
Consider the titration of 30.0 mL of 0.0700 M C2H5NH2 (a weak base; Kb = 0.000640) with 0.100 M HBrO4. Calculate the pH after the following volumes of titrant have been added: (a) 0.0 mL pH = (b) 5.3 mL pH = (c) 10.5 mL pH = (d) 15.8 mL pH = (e) 21.0 mL pH = (f) 27.3 mL pH =
Consider the titration of 50.0 mL of 0.1000 M diprotic acid H2A (Ka1 = 1.63 × 10-5, Ka2 = 6.09 × 10-10) titrated with 0.2000 M NaOH. Determine the pH of the solution after the addition of 50.00 mL of KOH.
Consider the titration of 50.00 ml of a solution of 0.2574 M of the fully protonated form of the diprotic amino acid glycine (pKa1 = 2.34, pKa2 = 9.60) with 0.3786 M KOH. What is the pH after the addition of 33.99 mL of the KOH solution?
Derive the titration curve when titrating 100.0mL of 0.100M weak base aniline (pka = 4.601) with 1.000M HClO4 using the following volumes of acid, 0.00mL, 2.00mL, 5.00mL, 8.00mL, 10.00mL, 13.00mL? (find the pH values and make a graph of pH vs VHClO4
For the titration of 50.00 mL of 0.300 M RbC7H5O2 (WB) with 0.200 M HNO3 (SA). Ka (HC7H5O2) = 6.3 x 10-5 Express pH value to two decimal places. Determine the pH after the addition 12.50 mL of HNO3.
Titration is performed on 30 mL of 0.04 M Hydrazine (NH2NH2) (Kb: 2.10-6 ) with 0.02 M HCl in the following volumes . Find the pH values.)(ka:1,4x10-5) (volumes:30 mL ; 45 mL ; 90 mL ; 120 mL; DN)
A 100.0 mL100.0 mL solution containing 0.951 g0.951 g of maleic acid (MW=116.072 g/mol)(MW=116.072 g/mol) is titrated with 0.364 M KOH.0.364 M KOH. Calculate the pH of the solution after the addition of 45.0 mL45.0 mL of the KOHKOH solution. Maleic acid has p?apKa values of 1.92 and 6.27. pH=pH=