EBK GENERAL CHEMISTRY: THE ESSENTIAL CO
EBK GENERAL CHEMISTRY: THE ESSENTIAL CO
7th Edition
ISBN: 8220106637203
Author: Chang
Publisher: YUZU
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Chapter 17, Problem 17.21QP

(a)

Interpretation Introduction

Interpretation:

The pH at the equivalence point of titration HClVsNH3 has to be calculated.

Concept Information:

  • pH is the logarithm of the reciprocal of the concentration  of H3O+ in a solution.
  •  pH is used to determine the acidity or basicity of an aqueous solution.
  • pH=-log[H3O+]

Acid ionization constantKa

The equilibrium expression for the reaction HA(aq)H+(aq)+A-(aq) is given below.

Ka=[H+][A-][HA]

Where Ka acid ionization is constant, [H+]  is concentration of hydrogen ion, [A-]  is concentration of acid anion, [HA] is concentration of the acid

(a)

Expert Solution
Check Mark

Answer to Problem 17.21QP

The pH at the equivalence point of titration HClVsNH3 is 5.28

Explanation of Solution

Concentration of HCl is 0.10M

Concentration of NH3 is 0.10M

Find the concentration of sodium formate

HCl(aq) + NH3(aq)  NH4Cl(aq)Initial concentration(M): 0.010 00Change in concentration (M):-0.10 -0.10+0.10Finalconcentration (M): 000.10Both the concentrations ofare same and to reach equivalence point, equal amount of solution should be needed. Therefore, the volume of solution is twofolded.The concentration of NH4Cl is=0.10 mol2L=0.050M

The concentration of ammonium chloride can be calculated from the final concentration of ammonium chloride in equilibrium table.  The solution gets doubled and the strength of ammonium chloride is divided by two.

Calculate the pH of the titration

NH4+(aq) + H2O(l)  NH3(aq) + H3O+(aq)Initial concentration(M): 0.050 00Change in concentration (M):-x +x+xEquilibriumconcentration (M): 0.050-xxxKb value for NH3 is 1.8 ×10-5Ka=KwKb=1.0 ×10-141.8 ×10-5=5.6×10-10Kb=[NH3][H3O+][NH4+]5.56×10-10=(x2)(0.050-x)x2=(5.56×10-10)×(0.050)x2=2.78×1011Takingsqurerootbothsidex=5.27×10-6

xisverysmallandneglectit,x = [H3O+] = 5.27 ×10-6MpH=-log[H3O+]=-log(5.27 ×10-6)pH=5.28

The concentration of H3O+ can be calculated from the final concentration of H3O+ in equilibrium table using base dissociation constant.  The pH is determined by taking negative logarithm of H3O+ concentration.

(b)

Interpretation Introduction

Interpretation:

The pH at the equivalence point of titration CH3COOHVsNaOH has to be calculated.

Concept Information:

  • pH is the logarithm of the reciprocal of the concentration  of H3O+ in a solution.
  •  pH is used to determine the acidity or basicity of an aqueous solution.
  • pH=-log[H3O+]

Base ionization constant Kb

The equilibrium expression for the ionization of weak base B will be,

B(aq)+H2O(l)HB+(aq)+OH-(aq)

Kb=[HB+][OH-][B]

Where Kb is base ionization constant, [OH] is concentration of hydroxide ion, [HB+] is concentration of conjugate acid, [B] is concentration of the base

Relationship between Kaand Kb

Ka×Kb=Kw

(b)

Expert Solution
Check Mark

Answer to Problem 17.21QP

The pH at the equivalence point of titration CH3COOHVsNaOH is 8.22

Explanation of Solution

Concentration of CH3COOH is 0.10M

Concentration of NaOH is 0.10M

Find the concentration of sodium formate

CH3COOH(aq) + NaOH(aq)  CH3COO-Na+(aq)+H2OInitial concentration(M): 0.010 00Change in concentration (M):-0.10 -0.10+0.10Finalconcentration (M): 000.10Both the concentrations ofare same and to reach equivalence point, equal amount of solution should be needed. Therefore, the volume of solution is twofolded.The concentration of CH3COO-Na+is=0.10 mol2L=0.050M

The concentration of sodium acetate can be calculated from the final concentration of sodium acetate in equilibrium table.  The solution gets doubled and the strength of sodium acetate is divided by two.

Calculate the pH of the titration

CH3COO-Na+(aq) + H2O(l)  CH3COOH(aq) + NaOH(aq)Initial concentration(M): 0.050 00Change in concentration (M):-x +x+xEquilibriumconcentration (M): 0.050-xxxKa value for NaOH is 13.8Kb=KwKa=1.0 ×10-1413.8=7.24×10-16Kb=[CH3COOH][NaOH][CH3COONa]7.24×10-16=(x2)(0.050-x)x2=(7.24×10-16)×(0.050)x2=3.62×1017Takingsqurerootbothsidex=6.01×10-9

xisverysmallandneglectit,x = [NaOH] = 6.01×10-9MpH=-log[NaOH]=-log(6.01×10-9)pH=8.22

Therefore given solution pH  value is 8.22

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Chapter 17 Solutions

EBK GENERAL CHEMISTRY: THE ESSENTIAL CO

Ch. 17.5 - Prob. 2PECh. 17.5 - Prob. 3PECh. 17.6 - Prob. 1PECh. 17.6 - Prob. 1RCCh. 17.7 - Prob. 1PECh. 17.7 - Prob. 1RCCh. 17 - Prob. 17.1QPCh. 17 - Prob. 17.2QPCh. 17 - Prob. 17.3QPCh. 17 - 17.4 The pKbs for the bases X−, Y−, and Z− are...Ch. 17 - 17.5 Specify which of these systems can be...Ch. 17 - 17.6 Specify which of these systems can be...Ch. 17 - 17.7 The pH of a bicarbonate–carbonic acid buffer...Ch. 17 - Prob. 17.8QPCh. 17 - 17.9 Calculate the pH of the buffer system 0.15 M...Ch. 17 - 17.10 What is the pH of the buffer 0.10 M...Ch. 17 - 17.11 The pH of a sodium acetate–acetic acid...Ch. 17 - 17.12 The pH of blood plasma is 7.40. Assuming the...Ch. 17 - Prob. 17.13QPCh. 17 - Prob. 17.14QPCh. 17 - 17.16 A student wishes to prepare a buffer...Ch. 17 - Prob. 17.17QPCh. 17 - Prob. 17.18QPCh. 17 - Prob. 17.19QPCh. 17 - 17.20 A 5.00-g quantity of a diprotic acid is...Ch. 17 - Prob. 17.21QPCh. 17 - Prob. 17.22QPCh. 17 - 17.23 The diagrams shown here represent solutions...Ch. 17 - 16.38 The diagrams shown here represent solutions...Ch. 17 - 17.25 Explain how an acid-base indicator works in...Ch. 17 - 17.26 What are the criteria for choosing an...Ch. 17 - 17.27 The amount of indicator used in an acid-base...Ch. 17 - 17.28 A student carried out an acid-base titration...Ch. 17 - 17.29 Referring to Table 17.1, specify which...Ch. 17 - 17.30 The ionization constant Ka of an indicator...Ch. 17 - 17.31 Define solubility, molar solubility, and...Ch. 17 - 17.32 Why do we usually not quote the Ksp values...Ch. 17 - 17.33 Write balanced equations and solubility...Ch. 17 - Prob. 17.34QPCh. 17 - Prob. 17.35QPCh. 17 - 17.36 Silver chloride has a larger Ksp than silver...Ch. 17 - Prob. 17.38QPCh. 17 - 17.39 The molar solubility of MnCO3 is 4.2 × 10−6...Ch. 17 - Prob. 17.40QPCh. 17 - Prob. 17.41QPCh. 17 - 17.42 Using data from Table 17.2, calculate the...Ch. 17 - 17.43 What is the pH of a saturated zinc hydroxide...Ch. 17 - 17.44 The pH of a saturated solution of a metal...Ch. 17 - Prob. 17.45QPCh. 17 - 17.46 A volume of 75 mL of 0.060 M NaF is mixed...Ch. 17 - 17.47 How does a common ion affect solubility? Use...Ch. 17 - Prob. 17.48QPCh. 17 - Prob. 17.49QPCh. 17 - Prob. 17.50QPCh. 17 - Prob. 17.51QPCh. 17 - 17.52 Calculate the molar solubility of BaSO4 (a)...Ch. 17 - Prob. 17.55QPCh. 17 - Prob. 17.56QPCh. 17 - 17.57 If 2.50 g of CuSO4 are dissolved in 9.0 ×...Ch. 17 - 17.58 Calculate the concentrations of Cd2+, , and...Ch. 17 - Prob. 17.59QPCh. 17 - Prob. 17.60QPCh. 17 - Prob. 17.61QPCh. 17 - Prob. 17.62QPCh. 17 - Prob. 17.63QPCh. 17 - 16.88 In a group 1 analysis, a student adds HCl...Ch. 17 - 17.65 Both KCl and NH4Cl are white solids. Suggest...Ch. 17 - 17.66 Describe a simple test that would enable you...Ch. 17 - Prob. 17.67QPCh. 17 - Prob. 17.68QPCh. 17 - Prob. 17.69QPCh. 17 - 17.70 The pKa of the indicator methyl orange is...Ch. 17 - Prob. 17.71QPCh. 17 - Prob. 17.72QPCh. 17 - 17.73 The two curves shown here represent the...Ch. 17 - 17.74 The two curves shown here represent the...Ch. 17 - Prob. 17.75QPCh. 17 - 17.76 A solution is made by mixing exactly 500 mL...Ch. 17 - Prob. 17.77QPCh. 17 - Prob. 17.78QPCh. 17 - 17.79 For which of these reactions is the...Ch. 17 - Prob. 17.80QPCh. 17 - Prob. 17.81QPCh. 17 - Prob. 17.82QPCh. 17 - Prob. 17.83QPCh. 17 - 17.84 Find the approximate pH range suitable for...Ch. 17 - Prob. 17.85QPCh. 17 - 17.86 Which of these substances will be more...Ch. 17 - Prob. 17.87QPCh. 17 - Prob. 17.88QPCh. 17 - Prob. 17.89QPCh. 17 - Prob. 17.90QPCh. 17 - Prob. 17.91QPCh. 17 - 17.92 When a KI solution was added to a solution...Ch. 17 - Prob. 17.93QPCh. 17 - Prob. 17.94QPCh. 17 - Prob. 17.95QPCh. 17 - 17.96 Solid NaI is slowly added to a solution that...Ch. 17 - Prob. 17.97QPCh. 17 - 17.98 (a) Assuming complete dissociation and no...Ch. 17 - 17.99 Acid-base reactions usually go to...Ch. 17 - 17.100 Calculate x, the number of molecules of...Ch. 17 - Prob. 17.101QPCh. 17 - 17.102 What reagents would you employ to separate...Ch. 17 - 17.103 CaSO4 (Ksp = 2.4 × 10−5) has a larger Ksp...Ch. 17 - 17.104 How many milliliters of 1.0 M NaOH must be...Ch. 17 - Prob. 17.105QPCh. 17 - Prob. 17.106QPCh. 17 - Prob. 17.107QPCh. 17 - Prob. 17.108QPCh. 17 - Prob. 17.109QPCh. 17 - Prob. 17.111SPCh. 17 - Prob. 17.112SPCh. 17 - Prob. 17.113SPCh. 17 - Prob. 17.114SPCh. 17 - Prob. 17.115SPCh. 17 - Prob. 17.116SPCh. 17 - 17.117 The titration curve shown here represents...Ch. 17 - Prob. 17.118SP
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