Chemistry: The Molecular Science
Chemistry: The Molecular Science
5th Edition
ISBN: 9781285199047
Author: John W. Moore, Conrad L. Stanitski
Publisher: Cengage Learning
bartleby

Concept explainers

bartleby

Videos

Question
Book Icon
Chapter 15, Problem 43QRT

(a)

Interpretation Introduction

Interpretation:

The pH of 30mL 0.100M benzoic acid solution when titrated with 10mL 0.100M NaOH has to be calculated.

Concept Introduction:

The pH is defined as the negative logarithm to the hydrogen ion concentration and it tells the nature of the solution. The physical constant which is used to determine the strength of an acid is known as acid dissociation constant. It is denoted by Ka. The greater Ka value, stronger is the acid.

(a)

Expert Solution
Check Mark

Answer to Problem 43QRT

The pH of 30mL 0.100M benzoic acid solution when titrated with 10mL 0.100M NaOH is 3.62_.

Explanation of Solution

The given molarity of benzoic acid is 0.100M.

The given molarity of titrant, NaOH is 0.100M.

The given volume of benzoic acid is 30mL.

The given volume of NaOHis 10mL.

The given value of Ka of benzoic acid is 1.2×104.

The total volume of the solution after titration is 30mL+10mL=40mL.

The titration reaction of benzoic acid and NaOH is given below.

    C6H5COOH(aq)+ NaOH(aq) C6H5COONa(aq) +H2O(l)

The conversion of mL into L is shown below.

    1ml=103L

So, the volume of benzoic acid is calculated below.

    30mL=0.030L

The volume of NaOHis calculated below.

    10mL=0.010L

The total volume of the solution is calculated below.

    40mL=0.040L

The mole, n, of the given substance is calculated as below.

n=Volume×molarity (1)

Substitute the values of volume and molarity of benzoic acid in equation (1).

    n=Volume×molarity=0.030L×0.1mol/L=3×103mol

Substitute the values of volume and molarity of NaOH in equation (1).

    n=Volume×molarity=0.010L×0.1mol/L=1×103mol

So, the moles of benzoic acid is 3×103mol and the moles 10mL NaOH solution is 1×103mol. So, 1×103mol of NaOH neutralizes 1×103mol of benzoic acid solution after titration. Thus, this reaction leads to the formation of 1×103mol of sodium benzoate ions and 2×103mol benzoic acid exists un-neutralized inside the solution after the complete titration.

The molarity of un-neutralized 2×103mol benzoic acid and 1×103mol of sodium benzoate ions is calculated by the formula given below.

    Molarity=NumberofmolesTotalvolume                                                                          ……(2)

Substitute the values of number of moles of benzoic acid and total volume of the solution in equation (2).

    Molarity=2×103mol0.040L=0.05M

Substitute the values of number of moles of sodium benzoate ions and total volume of the solution in equation (2).

    Molarity=1×103mol0.040L=0.025M

The Henderson-Hasselbalch equation can be represented as follows.

    pH =pKa+log[conjugatebase][conjugateacid]pH =logKa+log[conjugatebase][conjugateacid]                                                          ……(3)

Where,

  • [conjugatebase] is the concentration of conjugate base that is of sodium benzoate in the solution.
  • [conjugateacid] is the concentration of conjugate acid that is of un-neutralized benzoic acid in the solution.

Substitute the value of Ka of benzoic acid, concentration of conjugate base and conjugate acid in equation (3).

pH =log(1.2×104)+log[0.025][0.0500]pH =3.92+(0.3010)=3.62_

Therefore the pH of 30mL 0.100M benzoic acid solution when titrated with 10mL 0.100M NaOH  is 3.62_.

(b)

Interpretation Introduction

Interpretation:

The pH of 30mL 0.100M benzoic acid solution when titrated with 30mL 0.100M NaOH has to be calculated.

Concept Introduction:

Refer to part (a).

(b)

Expert Solution
Check Mark

Answer to Problem 43QRT

The pH of 30mL 0.100M benzoic acid solution when titrated with 30mL 0.100M NaOH is 8.31_.

Explanation of Solution

The given molarity of benzoic acid is 0.100M.

The given molarity of titrant, NaOH is 0.100M.

The given volume of benzoic acid is 30mL.

The given volume of NaOHis 30mL.

The given value of Ka of benzoic acid is 1.2×104.

The total volume of the solution after titration is 30mL+30mL=60mL.

The titration reaction of benzoic acid and NaOH is given below.

    C6H5COOH(aq)+ NaOH(aq) C6H5COONa(aq) +H2O(l)

The conversion of mL into L is shown below.

    1ml=103L

So, the volume of benzoic acid and NaOH is calculated below.

    30mL=0.030L

The total volume of the solution is calculated below.

    60mL=0.060L

The mole, n, of the given substance is calculated as below.

n=Volume×molarity (1)

Substitute the values of volume and molarity of benzoic acid in equation (1).

    n=Volume×molarity=0.030L×0.1mol/L=3×103mol

Substitute the values of volume and molarity of NaOH in equation (1).

    n=Volume×molarity=0.030L×0.1mol/L=3×103mol

So, the moles of benzoic acid is 3×103mol and the moles 30mL NaOH solution is 3×103mol. So, 3×103mol of NaOH neutralizes 3×103mol of benzoic acid solution and gives the equivalence point. Thus, this reaction leads to the formation of 3×103mol of sodium benzoate ions after the complete titration.

The molarity of 3×103mol of sodium benzoate ions is calculated by the formula given below.

    Molarity=NumberofmolesTotalvolume                                                                          ……(2)

Substitute the values of number of moles of sodium benzoate ions and total volume of the solution in equation (2).

    Molarity=3×103mol0.060L=0.050M

Thus, the equivalence point of the reaction of benzoic acid and NaOH is shown below.

    C6H5COOH(aq)+ NaOH(aq) C6H5COONa(aq) +H2O(l)

The value of Kw is 1.0×1014.

The expression for calculating the value of Kb of acetate ion in the solution is given below.

    Kb=KwKa                                                                                  ……(4)

Where,

  • Kb is the equilibrium constant of base.
  • Kw is the equilibrium constant of water.
  • Ka is the equilibrium constant of acid.

Substitute the value of Ka of benzoic acid and Kw is the equilibrium constant of water in equation (4).

    Kb=1.0×10141.2×104=8.3×1011

The concentration of C6H5COOH and OH is supposed to be xand the concentration of sodium benzoate, C6H5COO is supposed to 0.0500x.

The expression for calculating the equilibrium constant, Kb is given below.

    Kb=[C6H5COOH][OH][C6H5COO]                                                                      ……(5)

Substitute the respective values of concentration and value of Kb equation (5).

    8.3×1011=[x][x][0.0500x]8.3×1011=x2[0.0500x]

The value of Kb is small, so, 0.0500x=x. Simplify the above equation as follows.

    8.3×1011=x2[0.0500]8.3×1011×0.0500=x2x=2.0×106

The equation to calculate the value ofpOH is given below.

    pOH=log[OH]                                                                                             ……(6)

Substitute 2.0×106 for [OH] in equation (6).

    pOH=log[OH]=log(2.0×106)=5.69

The relation between pHand pOH is given below.

    pH+pOH=14                                                                                            ……(7)

Substitute 5.69 for Chemistry: The Molecular Science, Chapter 15, Problem 43QRT , additional homework tip  1 in equation (7).

    pH+5.69=14pH=145.69=8.31_

Therefore, the pH of 30mL 0.100M benzoic acid solution when titrated with 30mL 0.100M NaOH is 8.31_

(c)

Interpretation Introduction

Interpretation:

The pH of 30mL 0.100M benzoic acid solution when titrated with 40mL 0.100M NaOH has to be calculated.

Concept Introduction:

Refer to part (a).

(c)

Expert Solution
Check Mark

Answer to Problem 43QRT

The pH of 30mL 0.100M benzoic acid solution when titrated with 40mL 0.100M NaOH is 12.15_.

Explanation of Solution

The given molarity of benzoic acid is 0.100M.

The given molarity of titrant, NaOH is 0.100M.

The given volume of benzoic acid is 30mL.

The given volume of NaOHis 40mL.

The given value of Ka of benzoic acid is 1.2×104.

The total volume of the solution after titration is 30mL+40mL=70mL.

The titration reaction of benzoic acid and NaOH is given below.

    C6H5COOH(aq)+ NaOH(aq) C6H5COONa(aq) +H2O(l)

The conversion of mL into L is shown below.

    1ml=103L

So, the volume of benzoic acid is calculated below.

    9781305498129

The total volume of the solution is calculated below.

    70mL=0.070L

So, the concentration of hydroxyl ion is more than the required concentration for equivalence point. The excess volume of OH ions is 40mL30mL=10mL.

So, the volume of excess OH ions in liters is calculated below.

    10mL=0.010L

The mole, n, of the given substance is calculated as below.

n=Volume×molarity (1)

Substitute the values of volume and molarity of excess OHn equation (1).

    n=Volume×molarity=0.010L×0.1mol/L=1×103mol

The molarity of 1×103mol of excess OHis calculated by the formula given below.

    Molarity=NumberofmolesTotalvolume                                                                          ……(2)

Substitute the values of number of moles of excess OHand total volume of the solution in equation (2).

    Molarity=1×103mol0.070L=0.0140M

The equation to calculate the value ofpOH is given below.

    pOH=log[OH]                                                                                             ……(6)

Substitute 0.0140M for [OH] in equation (6).

    pOH=log[OH]=log(0.0140M)=1.85

The relation between pHand pOH is given below.

    pH+pOH=14                                                                                            ……(7)

Substitute 5.69 for Chemistry: The Molecular Science, Chapter 15, Problem 43QRT , additional homework tip  2 in equation (7).

    pH+1.85=14pH=141.85=12.15_

Therefore, the pH of 30mL 0.100M benzoic acid solution when titrated with 30mL 0.100M NaOH is 12.15_.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!
Students have asked these similar questions
Calculate the pH at the following points in a titration of 40.0 mL of 0.100 M barbituric acid(Ka = 9.8 × 10−5) with 0.100 M KOH. (a) no KOH added (b) 20.0 mL of KOH solution added (c)39.0 mL of KOH solution added (d) 40.0 mL of KOH solution added (e) 41.0 mL of KOHSketch an appropriate pH titration curve indicating the buffer region, equivalence point,and excess base region. Why is the pH at the equivalence point not 7.00?
You are asked to prepare a pH = 3.00 buffer solution startingfrom 1.25 L of a 1.00 M solution of hydrofluoric acid(HF) and any amount you need of sodium fluoride (NaF).(a) What is the pH of the hydrofluoric acid solution priorto adding sodium fluoride? (b) How many grams of sodiumfluoride should be added to prepare the buffer solution?Neglect the small volume change that occurs when the sodiumfluoride is added.
Consider the titration of 36.0 mL of 0.117 M ammonia with 0.0752 M HCl. (See the Acid-Base Table.) (a) How many mL of HCl are required to reach the equivalence point?5.60   mL(b) What is the pH at the equivalence point?  5.30(c) What is the pH of the solution after the addition of 15.7 mL of acid?  (d) What is the pH of the solution after the addition of 82.9 mL of acid?  only need part c and d please

Chapter 15 Solutions

Chemistry: The Molecular Science

Ch. 15.2 - Draw the titration curve for the titration of 50.0...Ch. 15.2 - Use the Ka expression and value for acetic acid to...Ch. 15.2 - Explain why the curve for the titration of acetic...Ch. 15.4 - Write the Ksp expression for each of these...Ch. 15.4 - The Ksp of AgBr at 100 C is 5 1010. Calculate the...Ch. 15.4 - A saturated solution of silver oxalate. Ag2C2O4....Ch. 15.4 - Prob. 15.9CECh. 15.5 - Consider 0.0010-M solutions of these sparingly...Ch. 15.5 - Prob. 15.11PSPCh. 15.5 - Calculate the solubility of PbCl2 in (a) pure...Ch. 15.5 - Prob. 15.13PSPCh. 15.6 - (a) Determine whether AgCl precipitates from a...Ch. 15.6 - Prob. 15.15PSPCh. 15 - Prob. 1SPCh. 15 - Choose a weak-acid/weak-base conjugate pair from...Ch. 15 - Prob. 4SPCh. 15 - Define the term buffer capacity.Ch. 15 - What is the difference between the end point and...Ch. 15 - What are the characteristics of a good acid-base...Ch. 15 - A strong acid is titrated with a strong base, such...Ch. 15 - Repeat the description for Question 4, but use a...Ch. 15 - Use Le Chatelier’s principle to explain why PbCl2...Ch. 15 - Describe what a complex ion is and give an...Ch. 15 - Define the term “amphoteric”. Ch. 15 - Distinguish between the ion product (Q) expression...Ch. 15 - Describe at least two ways that the solubility of...Ch. 15 - Briefly describe how a buffer solution can control...Ch. 15 - Identify each pair that could form a buffer. (a)...Ch. 15 - Identify each pair that could form a buffer. (a)...Ch. 15 - Many natural processes can be studied in the...Ch. 15 - Which of these combinations is the best to buffer...Ch. 15 - Without doing calculations, determine the pH of a...Ch. 15 - Without doing calculations, determine the pH of a...Ch. 15 - Select from Table 15.1 a conjugate acid-base pair...Ch. 15 - Select from Table 15.1 a conjugate acid-base pair...Ch. 15 - Calculate the mass of sodium acetate, NaCH3COO,...Ch. 15 - Calculate the mass in grams of ammonium chloride,...Ch. 15 - A buffer solution can be made from benzoic acid,...Ch. 15 - A buffer solution is prepared from 5.15 g NH4NO3...Ch. 15 - You dissolve 0.425 g NaOH in 2.00 L of a solution...Ch. 15 - A buffer solution is prepared by adding 0.125 mol...Ch. 15 - If added to 1 L of 0.20-M acetic acid, CH3COOH,...Ch. 15 - If added to 1 L of 0.20-M NaOH, which of these...Ch. 15 - Calculate the pH change when 10.0 mL of 0.100-M...Ch. 15 - Prob. 29QRTCh. 15 - Prob. 30QRTCh. 15 - Prob. 31QRTCh. 15 - The titration curves for two acids with the same...Ch. 15 - Explain why it is that the weaker the acid being...Ch. 15 - Prob. 34QRTCh. 15 - Consider all acid-base indicators discussed in...Ch. 15 - Which of the acid-base indicators discussed in...Ch. 15 - It required 22.6 mL of 0.0140-M Ba(OH)2 solution...Ch. 15 - It took 12.4 mL of 0.205-M H2SO4 solution to...Ch. 15 - Vitamin C is a monoprotic acid. To analyze a...Ch. 15 - An acid-base titration was used to find the...Ch. 15 - Calculate the volume of 0.150-M HCl required to...Ch. 15 - Calculate the volume of 0.225-M NaOH required to...Ch. 15 - Prob. 43QRTCh. 15 - Prob. 44QRTCh. 15 - Prob. 45QRTCh. 15 - Explain why rain with a pH of 6.7 is not...Ch. 15 - Identify two oxides that are key producers of acid...Ch. 15 - Prob. 48QRTCh. 15 - Prob. 49QRTCh. 15 - Prob. 50QRTCh. 15 - Prob. 51QRTCh. 15 - A saturated solution of silver arsenate, Ag3AsO4,...Ch. 15 - Prob. 53QRTCh. 15 - Prob. 54QRTCh. 15 - Prob. 55QRTCh. 15 - Prob. 56QRTCh. 15 - Prob. 57QRTCh. 15 - Prob. 58QRTCh. 15 - Prob. 59QRTCh. 15 - Prob. 60QRTCh. 15 - Prob. 61QRTCh. 15 - Prob. 62QRTCh. 15 - Prob. 63QRTCh. 15 - Prob. 64QRTCh. 15 - Predict what effect each would have on this...Ch. 15 - Prob. 66QRTCh. 15 - Prob. 67QRTCh. 15 - The solubility of Mg(OH)2 in water is...Ch. 15 - Prob. 69QRTCh. 15 - Prob. 70QRTCh. 15 - Prob. 71QRTCh. 15 - Prob. 72QRTCh. 15 - Write the chemical equation for the formation of...Ch. 15 - Prob. 74QRTCh. 15 - Prob. 75QRTCh. 15 - Prob. 76QRTCh. 15 - Prob. 77QRTCh. 15 - Prob. 78QRTCh. 15 - Prob. 79QRTCh. 15 - Prob. 80QRTCh. 15 - Prob. 81QRTCh. 15 - Solid sodium fluoride is slowly added to an...Ch. 15 - Prob. 83QRTCh. 15 - Prob. 84QRTCh. 15 - A buffer solution was prepared by adding 4.95 g...Ch. 15 - Prob. 86QRTCh. 15 - Prob. 87QRTCh. 15 - Prob. 88QRTCh. 15 - Prob. 89QRTCh. 15 - Which of these buffers involving a weak acid HA...Ch. 15 - Prob. 91QRTCh. 15 - Prob. 92QRTCh. 15 - When 40.00 mL of a weak monoprotic acid solution...Ch. 15 - Each of the solutions in the table has the same...Ch. 15 - Prob. 95QRTCh. 15 - Prob. 97QRTCh. 15 - The average normal concentration of Ca2+ in urine...Ch. 15 - Explain why even though an aqueous acetic acid...Ch. 15 - Prob. 100QRTCh. 15 - Prob. 101QRTCh. 15 - Prob. 102QRTCh. 15 - Prob. 103QRTCh. 15 - Prob. 104QRTCh. 15 - Apatite, Ca5(PO4)3OH, is the mineral in teeth. On...Ch. 15 - Calculate the maximum concentration of Mg2+...Ch. 15 - Prob. 107QRTCh. 15 - Prob. 108QRTCh. 15 - The grid has six lettered boxes, each of which...Ch. 15 - Consider the nanoscale-level representations for...Ch. 15 - Consider the nanoscale-level representations for...Ch. 15 - Prob. 112QRTCh. 15 - Prob. 113QRTCh. 15 - Prob. 114QRTCh. 15 - Prob. 115QRTCh. 15 - You want to prepare a pH 4.50 buffer using sodium...Ch. 15 - Prob. 117QRTCh. 15 - Prob. 118QRTCh. 15 - Prob. 119QRTCh. 15 - Prob. 120QRTCh. 15 - Prob. 121QRTCh. 15 - Prob. 122QRTCh. 15 - You are given four different aqueous solutions and...Ch. 15 - Prob. 124QRTCh. 15 - Prob. 126QRTCh. 15 - Prob. 15.ACPCh. 15 - Prob. 15.BCP
Knowledge Booster
Background pattern image
Chemistry
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
SEE MORE QUESTIONS
Recommended textbooks for you
Text book image
General Chemistry - Standalone book (MindTap Cour...
Chemistry
ISBN:9781305580343
Author:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell
Publisher:Cengage Learning
Text book image
Chemistry & Chemical Reactivity
Chemistry
ISBN:9781337399074
Author:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
Publisher:Cengage Learning
Text book image
Chemistry & Chemical Reactivity
Chemistry
ISBN:9781133949640
Author:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
Publisher:Cengage Learning
Text book image
Chemistry: Principles and Reactions
Chemistry
ISBN:9781305079373
Author:William L. Masterton, Cecile N. Hurley
Publisher:Cengage Learning
Text book image
General, Organic, and Biological Chemistry
Chemistry
ISBN:9781285853918
Author:H. Stephen Stoker
Publisher:Cengage Learning
Text book image
World of Chemistry, 3rd edition
Chemistry
ISBN:9781133109655
Author:Steven S. Zumdahl, Susan L. Zumdahl, Donald J. DeCoste
Publisher:Brooks / Cole / Cengage Learning
General Chemistry | Acids & Bases; Author: Ninja Nerd;https://www.youtube.com/watch?v=AOr_5tbgfQ0;License: Standard YouTube License, CC-BY