Chemistry: The Molecular Nature of Matter and Change - Standalone book
Chemistry: The Molecular Nature of Matter and Change - Standalone book
7th Edition
ISBN: 9780073511177
Author: Martin Silberberg Dr., Patricia Amateis Professor
Publisher: McGraw-Hill Education
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Chapter 19, Problem 19.59P

(a)

Interpretation Introduction

Interpretation:

pH of the calculation has to be calculated for 23.4mL of 0.0390MHNO2.

Concept introduction:

pH definition:

The concentration of hydrogen ion is measured using pH scale.  The acidity of aqueous solution is expressed by pH scale.

pH of a solution is defined as the negative base-10 logarithm of the hydrogen or hydronium ion concentration.

  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

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

(a)

Expert Solution
Check Mark

Answer to Problem 19.59P

The pH of a solution made by mixing 23.4mL of 0.0390MHNO2 is 7.76

Explanation of Solution

Given,

The balance equation is given below,

K+(aq) +OH(aq)+HNO2(aq)K+(aq) +NO2(aq)+H2O(l)

Potassium ions on the product side are written as separate spices because they have no effect on pH of the solution.

Next calculate the volume of KOH,

Volume of KOH in mL,

KOH=(0.0390molHNO2L)(103L1mL)(23.4mL)(1molKOH1molKOH)(L0.0588molKOH)(1mL103L)

Determination the moles of HNO2present,

MolesofHNO2=(0.0390molHNO21mL)(103L1mL)(23.4mL)=0.0009126molHNO2

The equivalent point, 0.0009126molKOH will be added so, the moles acid = moles base.

The KOH will react with an equal amount of the acid 0molHNO2 will remain and 0.0009126molNO2 will be formed.

ICE Table of equilibrium,

 HNO2(aq) +KOH(aq)H2O(l) +NO2(aq)+K+(aq)Initial (M)0.0009126mol0.0009126mol   0Change(M)0.0009126mol0.0009126mol+0.0009126molFinal  00+0.0009126mol

Determination the liters of solution present at the equivalent point,

Volume

   =[(23.4+15.5204)mL](103L/1mL)=0.0389204L

Concentration of NO2 at equivalent point,

Molarity,

  =(0.0009126molNO2)(0.0389204L)=0.023447858M

Next calculate Kbfor NO2

The equilibrium value for HNO2=7.1×104

So,

  Kb=KW/Ka=(1.0×1014)(7.1×104)=1.40845×1011 

Using a reaction table for the equilibrium reaction of NO2

   NO2 +H2OHNO2 +OHInitial (M):0.023447858M  00Change(M):x+x+xEquilibrium: 0.023447858Mx+x+x

Determination the hydroxide ion concentration from the Kb and then determine the pH from the pOH.

  Kb=1.40845×1011=[HNO2][OH][NO2]=[x][x][0.023447858x]=x2[0.023447858][OH]=x=5.7468×107M

Calculation for pH and pOH

  pOH=log(5.7468×107)=6.240574pH=14.00pOH=14.006.240574=7.759426pH=7.76

(b)

Interpretation Introduction

Interpretation:

pH of the calculation has to be calculated for 17.3mL of 0.130MH2CO3 with two equivalent point.

Concept introduction:

pH definition:

The concentration of hydrogen ion is measured using pH scale.  The acidity of aqueous solution is expressed by pH scale.

The pH of a solution is defined as the negative base-10 logarithm of the hydrogen or hydronium ion concentration.

  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

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 19.59P

The pH of a solution made by mixing 17.3mL of 0.130MH2CO3 is 11.35

Explanation of Solution

The balance equation is given below,

KOH(aq)+H2CO3(aq)K+(aq)+HCO3(aq)+H2O(l)KOH(aq)+HCO3(aq)K+(aq)+CO32(aq)+H2O(l)

Potassium ions on the product side are written as separate spices because they have no effect on the pH of the solution.

Next calculate the volume of KOH,

Volume of KOH in mL,

KOH=(0.130molH2CO3L)(103L1mL)(17.3mL)(1molKOH1molH2CO3)(L0.0588molKOH)(1mL103L)=38.248299=38.2mLofKOH

The solution requires an equal volume to reach second equivalence point 76.4mL

Determination the moles of HCO3produced,

MolesofH2CO3=(0.130molH2CO3L)(103L1mL)(17.3mL)(1molHCO31molH2CO3)=0.002249molHCO3-

An equal number of moles of CO32 will be present at the second equivalent point.

Determination the liters of solution present at the first equivalent point,

Volume

  =[(17.3+38.2482299)mL](103L/1mL)=0.055548L

Determination the liters of solution present at the second equivalent point,

Volume

  =[(17.3+38.2482299+38.2482299)mL](103L/1mL)=0.0937966L

Concentration of HCO3 at equivalent point,

Molarity,

  =(0.022249molHCO3)(0.055548L)=0.0404875M

Concentration of CO32 at equivalent point,

Molarity,

  =(0.022249molCO32)(0.0937966L)=0.023977M

Next calculate Kbfor HCO3

The equilibrium value for Ka=4.5×107HCO3

  Kb=KW/Ka=(1.0×1014)(4.5×107)=2.222×108 

The equilibrium value for Ka=4.7×1011CO32

  Kb=KW/Ka=(1.0×1014)(4.7×1011)=2.1276596×104

Determination the hydroxide ion concentration from the Kb and then determine the pH from pOH.

Calculation for first equilibrium point:

  Kb=2.222×108=[H2CO3][OH][HCO3]=[x][x][0.0404875x]=x2[0.0404875][OH]=x=2.999387×105M

Calculation for pH and pOH

  pOH=log(2.999387×105)=4.522967495pH=14.00pOH=14.004.522967495=9.4770pH=9.48

Calculation for second equilibrium point:

  Kb=2.222×108=[HCO3][OH][CO32]=[x][x][0.023977x]=x2[0.023977][OH]=x=2.2584677×103M

Calculation for pH and pOH

  pOH=log(2.2586477×103)=2.6461515pH=14.00pOH=14.002.6461515=11.3538pH=11.35

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Chapter 19 Solutions

Chemistry: The Molecular Nature of Matter and Change - Standalone book

Ch. 19.3 - Prob. 19.6AFPCh. 19.3 - Prob. 19.6BFPCh. 19.3 - Prob. 19.7AFPCh. 19.3 - Prob. 19.7BFPCh. 19.3 - Prob. 19.8AFPCh. 19.3 - Prob. 19.8BFPCh. 19.3 - Prob. 19.9AFPCh. 19.3 - Prob. 19.9BFPCh. 19.3 - Prob. 19.10AFPCh. 19.3 - Prob. 19.10BFPCh. 19.3 - Prob. 19.11AFPCh. 19.3 - Prob. 19.11BFPCh. 19.3 - Prob. 19.12AFPCh. 19.3 - Prob. 19.12BFPCh. 19.3 - An environmental technician collects a sample of...Ch. 19.3 - A lake that has a surface area of 10.0 acres (1...Ch. 19.4 - Cyanide ion is toxic because it forms stable...Ch. 19.4 - Prob. 19.13BFPCh. 19.4 - Prob. 19.14AFPCh. 19.4 - Calculate the solubility of PbCl2 in 0.75 M NaOH....Ch. 19 - Prob. 19.1PCh. 19 - Prob. 19.2PCh. 19 - Prob. 19.3PCh. 19 - The scenes below depict solutions of the same...Ch. 19 - Prob. 19.5PCh. 19 - Prob. 19.6PCh. 19 - Prob. 19.7PCh. 19 - Prob. 19.8PCh. 19 - Prob. 19.9PCh. 19 - Does the pH increase or decrease with each of the...Ch. 19 - What are the [H3O+] and the pH of a propanoic...Ch. 19 - What are the [H3O+] and the pH of a benzoic...Ch. 19 - Prob. 19.13PCh. 19 - Prob. 19.14PCh. 19 - Prob. 19.15PCh. 19 - Find the pH of a buffer that consists of 0.95 M...Ch. 19 - Prob. 19.17PCh. 19 - Prob. 19.18PCh. 19 - Prob. 19.19PCh. 19 - Find the pH of a buffer that consists of 0.50 M...Ch. 19 - A buffer consists of 0.22 M KHCO3 and 0.37 M...Ch. 19 - A buffer consists of 0.50 M NaH2PO4 and 0.40 M...Ch. 19 - What is the component concentration ratio,...Ch. 19 - Prob. 19.24PCh. 19 - Prob. 19.25PCh. 19 - Prob. 19.26PCh. 19 - Prob. 19.27PCh. 19 - A buffer that contains 0.40 M of a base, B, and...Ch. 19 - A buffer that contains 0.110 M HY and 0.220 M Y−...Ch. 19 - A buffer that contains 1.05 M B and 0.750 M BH+...Ch. 19 - A buffer is prepared by mixing 204 mL of 0.452 M...Ch. 19 - A buffer is prepared by mixing 50.0 mL of 0.050 M...Ch. 19 - Prob. 19.33PCh. 19 - Prob. 19.34PCh. 19 - Prob. 19.35PCh. 19 - Choose specific acid-base conjugate pairs to make...Ch. 19 - An industrial chemist studying bleaching and...Ch. 19 - Oxoanions of phosphorus are buffer components in...Ch. 19 - Prob. 19.39PCh. 19 - Prob. 19.40PCh. 19 - Prob. 19.41PCh. 19 - Prob. 19.42PCh. 19 - The scenes below depict the relative...Ch. 19 - Prob. 19.44PCh. 19 - What species are in the buffer region of a weak...Ch. 19 - Prob. 19.46PCh. 19 - Prob. 19.47PCh. 19 - Prob. 19.48PCh. 19 - Prob. 19.49PCh. 19 - Prob. 19.50PCh. 19 - Prob. 19.51PCh. 19 - Prob. 19.52PCh. 19 - Use figure 19.9 to find an indicator for these...Ch. 19 - Prob. 19.54PCh. 19 - Prob. 19.55PCh. 19 - Prob. 19.56PCh. 19 - Prob. 19.57PCh. 19 - Prob. 19.58PCh. 19 - Prob. 19.59PCh. 19 - Prob. 19.60PCh. 19 - Prob. 19.61PCh. 19 - Prob. 19.62PCh. 19 - Prob. 19.63PCh. 19 - Prob. 19.64PCh. 19 - Prob. 19.65PCh. 19 - Write the ion-product expressions for (a) silver...Ch. 19 - Prob. 19.67PCh. 19 - Prob. 19.68PCh. 19 - Prob. 19.69PCh. 19 - The solubility of silver carbonate is 0.032 M at...Ch. 19 - Prob. 19.71PCh. 19 - Prob. 19.72PCh. 19 - The solubility of calcium sulfate at 30°C is 0.209...Ch. 19 - Prob. 19.74PCh. 19 - Prob. 19.75PCh. 19 - Prob. 19.76PCh. 19 - Calculate the molar solubility of Ag2SO4 in (a)...Ch. 19 - Prob. 19.78PCh. 19 - Prob. 19.79PCh. 19 - Prob. 19.80PCh. 19 - Prob. 19.81PCh. 19 - Prob. 19.82PCh. 19 - Write equations to show whether the solubility of...Ch. 19 - Prob. 19.84PCh. 19 - Prob. 19.85PCh. 19 - Prob. 19.86PCh. 19 - Does any solid PbCl2 form when 3.5 mg of NaCl is...Ch. 19 - Prob. 19.88PCh. 19 - Prob. 19.89PCh. 19 - Prob. 19.90PCh. 19 - A 50.0-mL volume of 0.50 M Fe(NO3)3 is mixed with...Ch. 19 - Prob. 19.92PCh. 19 - Prob. 19.93PCh. 19 - Prob. 19.94PCh. 19 - Write a balanced equation for the reaction of in...Ch. 19 - Prob. 19.96PCh. 19 - Prob. 19.97PCh. 19 - Prob. 19.98PCh. 19 - What is [Ag+] when 25.0 mL each of 0.044 M AgNO3...Ch. 19 - Prob. 19.100PCh. 19 - Prob. 19.101PCh. 19 - Prob. 19.102PCh. 19 - When 0.84 g of ZnCl2 is dissolved in 245 mL of...Ch. 19 - When 2.4 g of Co(NO3)2 is dissolved in 0.350 L of...Ch. 19 - Prob. 19.105PCh. 19 - A microbiologist is preparing a medium on which to...Ch. 19 - As an FDA physiologist, you need 0.700 L of formic...Ch. 19 - Tris(hydroxymethyl)aminomethane [(HOCH2)3CNH2],...Ch. 19 - Water flowing through pipes of carbon steel must...Ch. 19 - Gout is caused by an error in metabolism that...Ch. 19 - In the process of cave formation (Section 19.3),...Ch. 19 - Phosphate systems form essential buffers in...Ch. 19 - The solubility of KCl is 3.7 M at 20°C. Two...Ch. 19 - It is possible to detect NH3 gas over 10−2 M NH3....Ch. 19 - Manganese(II) sulfide is one of the compounds...Ch. 19 - The normal pH of blood is 7.40 ± 0.05 and is...Ch. 19 - A bioengineer preparing cells for cloning bathes a...Ch. 19 - Sketch a qualitative curve for the titration of...Ch. 19 - A solution contains 0.10 M ZnCl2 and 0.020 M...Ch. 19 - Prob. 19.120PCh. 19 - The scene at right depicts a saturated solution of...Ch. 19 - Prob. 19.122PCh. 19 - The acid-base indicator ethyl orange turns from...Ch. 19 - Prob. 19.124PCh. 19 - Prob. 19.125PCh. 19 - Prob. 19.126PCh. 19 - Prob. 19.127PCh. 19 - Prob. 19.128PCh. 19 - Prob. 19.129PCh. 19 - Calcium ion present in water supplies is easily...Ch. 19 - Calculate the molar solubility of Hg2C2O4 (Ksp =...Ch. 19 - Environmental engineers use alkalinity as a...Ch. 19 - Human blood contains one buffer system based on...Ch. 19 - A 0.050 M H2S solution contains 0.15 M NiCl2 and...Ch. 19 - Quantitative analysis of Cl− ion is often...Ch. 19 - An ecobotanist separates the components of a...Ch. 19 - Some kidney stones form by the precipitation of...Ch. 19 - Prob. 19.138PCh. 19 - Prob. 19.139PCh. 19 - Because of the toxicity of mercury compounds,...Ch. 19 - A 35.0-mL solution of 0.075 M CaCl2 is mixed with...Ch. 19 - Rainwater is slightly acidic due to dissolved CO2....Ch. 19 - Prob. 19.143PCh. 19 - Ethylenediaminetetraacetic acid (abbreviated...Ch. 19 - Buffers that are based on...Ch. 19 - NaCl is purified by adding HCl to a saturated...Ch. 19 - Scenes A to D represent tiny portions of 0.10 M...Ch. 19 - Prob. 19.148PCh. 19 - Prob. 19.149PCh. 19 - Prob. 19.150P
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