Chemistry: The Molecular Nature of Matter and Change
Chemistry: The Molecular Nature of Matter and Change
8th Edition
ISBN: 9781259631757
Author: Martin Silberberg Dr., Patricia Amateis Professor
Publisher: McGraw-Hill Education
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Chapter 18, Problem 18.167P

(a)

Interpretation Introduction

Interpretation:

[H3O+][OH-], pH, and pOH for 0.240 M acetic acid has to be calculated.

Concept introduction:

An equilibrium constant (K) is the ratio of concentration of products and reactants raised to appropriate stoichiometric coefficient at equlibrium.

For the general base B,

  B(aq)+H2O(l)BH+(aq)+OH-(aq)

The relative strength of an acid and base in water can be also expressed quantitatively with an equilibrium constant as follows:

  Kb=[BH+][OH-][B]                                                                                                 (1)

An equilibrium constant (K) with subscript b indicate that it is an equilibrium constant of an base in water.

  Base - dissociation constants can be expressed as pKb values,pKb = -log Kb  and10 - pKb = K

Percent dissociation can be calculated by using following formula,

  Percent dissociated =  dissociationinitial×100

The Kb value is calculating by using following formula,

  Kw = Ka × Kb

The Relationships Among pH, pOH, and pKw

  Kw = [H3O+][OH-] = 1.0×1014

  pH = -log(H3O+)pOH = -log(OH)

  pKw = pH + pOH = 14.00 

(a)

Expert Solution
Check Mark

Explanation of Solution

Given,

Acetic acid is an acid therefore it can donate the proton to the water.

The balance equation is given below,

  CH3COOH (aq) +H2O(l)CH3COO(aq) + H3O+(aq)

0.240 M acetic acid Solution.

Therefore,

ICE table:

CH3COOH (aq) +H2O(l)CH3COO(aq) + H3O+(aq)

Initial concentration0.240 M-00
Change-x + x+ x
At equilibrium0.240-x xx

The initial concentration is 0.240 M acetic acid.

  CH3COOH (aq) +H2O(l)CH3COO(aq) + H3O+(aq)

The value of Ka is calculated by using following formula,

    Ka =[CH3COO][H3O+][CH3COOH ]

Given,

  Ka =[CH3COO][H3O+][CH3COOH ]=1.8×105

Let consider,

  0.24x=0.240[CH3COO][H3O+][CH3COOH ]=1.8×105,x2(0.240)=1.8×105x2=4.32×106x=2.07×103M=[H3O+]

Concentration of [H3O+]=2.07×103M

X is small when compared to 0.240M, therefore,

  Percent dissociated =  dissociationinitial×1002.07×1030.240M×100=0.86

The dissociation value is very less, and it is valid.

  Kw=[OH-][H3O+][OH-]=Kw[H3O+][OH-]=1.0×10-142.07×103[OH-]=4.83×10-12

Therefore,

  pH = -log(H3O+)pH=-log(2.07×103)pH=2.68pOH = -log(OH)pOH = -log(4.83×10-12)pOH =11.31

(b)

Interpretation Introduction

Interpretation:

[H3O+][OH-], pH, and pOH for 0.240 M ammonia has to be calculated.

Concept introduction:

An equilibrium constant (K) is the ratio of concentration of products and reactants raised to appropriate stoichiometric coefficient at equlibrium.

For the general base B,

  B(aq)+H2O(l)BH+(aq)+OH-(aq)

The relative strength of an acid and base in water can be also expressed quantitatively with an equilibrium constant as follows:

  Kb=[BH+][OH-][B]                                                                                                 (1)

An equilibrium constant (K) with subscript b indicate that it is an equilibrium constant of an base in water.

  Base - dissociation constants can be expressed as pKb values,pKb = -log Kb  and10 - pKb = K

Percent dissociation can be calculated by using following formula,

  Percent dissociated =  dissociationinitial×100

The Kb value is calculating by using following formula,

  Kw = Ka × Kb

The Relationships Among pH, pOH, and pKw

  Kw = [H3O+][OH-] = 1.0×1014

  pH = -log(H3O+)pOH = -log(OH)

  pKw = pH + pOH = 14.00 

(b)

Expert Solution
Check Mark

Explanation of Solution

Given,

Ammonia is a base therefore it can accept the proton from the water.

The balance equation is given below,

NH3 (aq) +H2O (l)NH4+(aq) + HO(aq)

0.240 M ammonia Solution.

Therefore,

ICE table:

NH3 (aq) +H2O (l)NH4+(aq) + HO(aq)

Initial concentration0.240 M-00
Change-x + x+ x
At equilibrium0.240-x xx

The initial concentration is 0.240 M acetic acid.

    NH3 (aq) +H2O (l)NH4+(aq) + HO(aq)

The value of Kb is calculated by using following formula,

    Kb =[NH4+][HO][NH3]

Given,

  Kb =[NH4+][HO][NH3]=1.8×105

Let consider,

  0.24x=0.240[NH4+][HO][NH3]=1.8×105,x2(0.240)=1.8×105x2=4.32×106x=2.07×103M=[HO]

Concentration of [H3O+]=2.07×103M

X is small when compared to 0.240M, therefore,

  Percent dissociated =  dissociationinitial×1002.07×1030.240M×100=0.86

The dissociation value is very less, and it is valid.

  Kw=[OH-][H3O+][H3O+]=Kw[OH-][H3O+]=1.0×10-142.07×103[H3O+]=4.83×10-12

Therefore,

  pH = -log(H3O+)pH=-log(4.83×10-12)pH11.31pOH = -log(OH)pOH = -log(2.07×103)pOH =2.68

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Chapter 18 Solutions

Chemistry: The Molecular Nature of Matter and Change

Ch. 18.3 - The left-hand scene in the margin represents the...Ch. 18.3 - The right-hand scene depicts an aqueous solution...Ch. 18.4 - The conjugate acid of ammonia is the weak acid ....Ch. 18.4 - Prob. 18.7BFPCh. 18.4 - Cyanic acid (HOCN) is an extremely acrid, unstable...Ch. 18.4 - Prob. 18.8BFPCh. 18.4 - Prob. 18.9AFPCh. 18.4 - Prob. 18.9BFPCh. 18.4 - Prob. 18.10AFPCh. 18.4 - Prob. 18.10BFPCh. 18.6 - Pyridine (C5H5N, see the space-filling model)...Ch. 18.6 - Prob. 18.11BFPCh. 18.6 - Prob. 18.12AFPCh. 18.6 - Prob. 18.12BFPCh. 18.7 - Write equations to predict whether solutions of...Ch. 18.7 - Write equations to predict whether solutions of...Ch. 18.7 - Determine whether solutions of the following salts...Ch. 18.7 - Prob. 18.14BFPCh. 18.9 - Prob. 18.15AFPCh. 18.9 - Prob. 18.15BFPCh. 18 - Prob. 18.1PCh. 18 - Prob. 18.2PCh. 18 - Prob. 18.3PCh. 18 - What do “strong” and “weak” mean for acids and...Ch. 18 - Prob. 18.5PCh. 18 - Prob. 18.6PCh. 18 - Prob. 18.7PCh. 18 - Which of the following are Arrhenius...Ch. 18 - Prob. 18.9PCh. 18 - Prob. 18.10PCh. 18 - Prob. 18.11PCh. 18 - Prob. 18.12PCh. 18 - Use Appendix C to rank the following in order of...Ch. 18 - Prob. 18.14PCh. 18 - Classify each as a strong or weak acid or...Ch. 18 - Prob. 18.16PCh. 18 - Prob. 18.17PCh. 18 - Prob. 18.18PCh. 18 - Prob. 18.19PCh. 18 - Prob. 18.20PCh. 18 - Prob. 18.21PCh. 18 - Which solution has the higher pH? Explain. A 0.1 M...Ch. 18 - (a) What is the pH of 0.0111 M NaOH? Is the...Ch. 18 - (a) What is the pH of 0.0333 M HNO3? Is the...Ch. 18 - Prob. 18.25PCh. 18 - (a) What is the pH of 7.52×10−4 M CsOH? Is the...Ch. 18 - Prob. 18.27PCh. 18 - Prob. 18.28PCh. 18 - Prob. 18.29PCh. 18 - Prob. 18.30PCh. 18 - Prob. 18.31PCh. 18 - Prob. 18.32PCh. 18 - Prob. 18.33PCh. 18 - Prob. 18.34PCh. 18 - The two molecular scenes shown depict the relative...Ch. 18 - Prob. 18.36PCh. 18 - Prob. 18.37PCh. 18 - Prob. 18.38PCh. 18 - A Brønstcd-Lowry acid-base reaction proceeds in...Ch. 18 - Prob. 18.40PCh. 18 - Prob. 18.41PCh. 18 - Prob. 18.42PCh. 18 - Give the formula of the conjugate...Ch. 18 - Give the formula of the conjugate base: Ch. 18 - Give the formula of the conjugate...Ch. 18 - Prob. 18.46PCh. 18 - Prob. 18.47PCh. 18 - In each equation, label the acids, bases, and...Ch. 18 - Prob. 18.49PCh. 18 - Prob. 18.50PCh. 18 - Prob. 18.51PCh. 18 - Prob. 18.52PCh. 18 - Prob. 18.53PCh. 18 - The following aqueous species constitute two...Ch. 18 - Prob. 18.55PCh. 18 - Use Figure 18.8 to determine whether Kc > 1...Ch. 18 - Prob. 18.57PCh. 18 - Prob. 18.58PCh. 18 - Prob. 18.59PCh. 18 - Prob. 18.60PCh. 18 - Prob. 18.61PCh. 18 - Prob. 18.62PCh. 18 - Prob. 18.63PCh. 18 - Prob. 18.64PCh. 18 - Prob. 18.65PCh. 18 - Prob. 18.66PCh. 18 - Prob. 18.67PCh. 18 - Prob. 18.68PCh. 18 - Hypochlorous acid, HClO, has a pKa of 7.54. What...Ch. 18 - Prob. 18.70PCh. 18 - Prob. 18.71PCh. 18 - Prob. 18.72PCh. 18 - Prob. 18.73PCh. 18 - Prob. 18.74PCh. 18 - Prob. 18.75PCh. 18 - Prob. 18.76PCh. 18 - Prob. 18.77PCh. 18 - Prob. 18.78PCh. 18 - Prob. 18.79PCh. 18 - Prob. 18.80PCh. 18 - Prob. 18.81PCh. 18 - Prob. 18.82PCh. 18 - Formic acid, HCOOH, the simplest carboxylic acid,...Ch. 18 - Across a period, how does the electronegativity of...Ch. 18 - How does the atomic size of a nonmetal affect the...Ch. 18 - Prob. 18.86PCh. 18 - Prob. 18.87PCh. 18 - Prob. 18.88PCh. 18 - Prob. 18.89PCh. 18 - Choose the stronger acid in each of the following...Ch. 18 - Prob. 18.91PCh. 18 - Prob. 18.92PCh. 18 - Prob. 18.93PCh. 18 - Use Appendix C to choose the solution with the...Ch. 18 - Prob. 18.95PCh. 18 - Prob. 18.96PCh. 18 - Prob. 18.97PCh. 18 - Prob. 18.98PCh. 18 - Prob. 18.99PCh. 18 - Prob. 18.100PCh. 18 - Prob. 18.101PCh. 18 - Prob. 18.102PCh. 18 - Prob. 18.103PCh. 18 - Prob. 18.104PCh. 18 - Prob. 18.105PCh. 18 - Prob. 18.106PCh. 18 - Prob. 18.107PCh. 18 - Prob. 18.108PCh. 18 - What is the pKb of ? What is the pKa of the...Ch. 18 - Prob. 18.110PCh. 18 - Prob. 18.111PCh. 18 - Prob. 18.112PCh. 18 - Prob. 18.113PCh. 18 - Prob. 18.114PCh. 18 - Prob. 18.115PCh. 18 - Prob. 18.116PCh. 18 - Prob. 18.117PCh. 18 - Prob. 18.118PCh. 18 - Prob. 18.119PCh. 18 - Prob. 18.120PCh. 18 - Prob. 18.121PCh. 18 - Prob. 18.122PCh. 18 - Prob. 18.123PCh. 18 - Prob. 18.124PCh. 18 - Explain with equations and calculations, when...Ch. 18 - Prob. 18.126PCh. 18 - Prob. 18.127PCh. 18 - Rank the following salts in order of increasing pH...Ch. 18 - Rank the following salts in order of decreasing pH...Ch. 18 - Prob. 18.130PCh. 18 - Prob. 18.131PCh. 18 - Prob. 18.132PCh. 18 - Prob. 18.133PCh. 18 - Prob. 18.134PCh. 18 - Prob. 18.135PCh. 18 - Prob. 18.136PCh. 18 - Prob. 18.137PCh. 18 - Prob. 18.138PCh. 18 - Prob. 18.139PCh. 18 - Which are Lewis acids and which are Lewis...Ch. 18 - Prob. 18.141PCh. 18 - Prob. 18.142PCh. 18 - Prob. 18.143PCh. 18 - Prob. 18.144PCh. 18 - Classify the following as Arrhenius,...Ch. 18 - Chloral (Cl3C—CH=O) forms a monohydrate, chloral...Ch. 18 - Prob. 18.147PCh. 18 - Prob. 18.148PCh. 18 - Prob. 18.149PCh. 18 - Prob. 18.150PCh. 18 - Prob. 18.151PCh. 18 - Prob. 18.152PCh. 18 - Prob. 18.153PCh. 18 - Prob. 18.154PCh. 18 - The strength of an acid or base is related to its...Ch. 18 - Prob. 18.156PCh. 18 - Three beakers contain 100. mL of 0.10 M HCl,...Ch. 18 - Prob. 18.158PCh. 18 - Prob. 18.159PCh. 18 - Prob. 18.160PCh. 18 - Prob. 18.161PCh. 18 - Prob. 18.162PCh. 18 - What is the pH of a vinegar with 5.0% (w/v) acetic...Ch. 18 - Prob. 18.164PCh. 18 - Prob. 18.165PCh. 18 - Prob. 18.166PCh. 18 - Prob. 18.167PCh. 18 - Prob. 18.168PCh. 18 - Prob. 18.169PCh. 18 - Prob. 18.170PCh. 18 - Prob. 18.171PCh. 18 - Prob. 18.172PCh. 18 - Prob. 18.173PCh. 18 - Prob. 18.174PCh. 18 - Prob. 18.175PCh. 18 - Prob. 18.176PCh. 18 - Prob. 18.177PCh. 18 - Prob. 18.178PCh. 18 - Prob. 18.179PCh. 18 - Prob. 18.180PCh. 18 - Prob. 18.181PCh. 18 - Prob. 18.182PCh. 18 - Prob. 18.183PCh. 18 - Prob. 18.184PCh. 18 - Drinking water is often disinfected with Cl2,...Ch. 18 - Prob. 18.186P
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