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 18.71P

(a)

Interpretation Introduction

Interpretation:

The [H3O+], pH, [OH-], and pOH has to be calculated for 0.735 M solution.

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 acid HA,

  HA(aq)+H2O(l)H3O+(aq)+A(aq)

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

  Ka=[H3O+][A][HA]                                                                                                   (1)

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

  Acid - dissociation constants can be expressed as pKa values,pKa = -log Ka  and10 - pKa = K

Percent dissociation can be calculated by using following formula,

  Percent dissociated =  dissociationinitial×100

The Ka value is calculating by using following formula,

  Kw = Ka × Kb

(a)

Expert Solution
Check Mark

Explanation of Solution

12.5 % acid dissociated in a 0.735 M solution.

The given compound is acid therefore it can donate the proton to the water.

The balance equation is given below,

  HA (aq) +H2O(l)H3O(aq) +   A (aq)

Percent dissociation can be calculated by using following formula,

  Percent dissociated =  dissociationinitial×10012.5 %=  x0.735×100x =  0.092M

The dissociation of acid is 0.092M

Therefore,

Construct ICE table:

  HA (aq) +H2O(l)H3O(aq) +   A (aq)

Initial concentration0.735 M-00
Change -x + x+ x
     
At equilibrium0.735-x xx

The initial concentration is 0.735 M HA.

  HA (aq) +H2O(l)H3O(aq) +   A (aq)The value Kais calculating by using following formula, Ka =[A][H3O][HA]

  [Dissociated acid] = x =[A-] =[H3O+]=0.092M

[H3O+] of the solution is 0.092M

Therefore,

  pH=-log[H3O+]pH=-log(0.092)pH=1.04

pH of the solution is 1.04

The Kb value is calculating by using following formula,

  Kw = Ka × Kb=[H3O+][OH][OH] = Kw[H3O+][OH] = 1×10140.092[OH] = 1.09×1013

Therefore,

  pOH=-log[OH]pOH=-log(1.09×1013)pOH=12.96

pOH of the solution is 12.96

(b)

Interpretation Introduction

Interpretation:

The Ka has to be calculated for 0.20 M solution.

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 acid HA,

  HA(aq)+H2O(l)H3O+(aq)+A(aq)

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

  Ka=[H3O+][A][HA] (1)

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

  Acid - dissociation constants can be expressed as pKa values,pKa = -log Ka  and10 - pKa = K

Percent dissociation can be calculated by using following formula,

  Percent dissociated =  dissociationinitial×100

The Ka value is calculating by using following formula,

  Kw = Ka × Kb

(b)

Expert Solution
Check Mark

Explanation of Solution

12.5 % acid dissociated in a 0.735 M solution.

The given compound is acid therefore it can donate the proton to the water.

The balance equation is given below,

  HA (aq) +H2O(l)H3O(aq) +   A (aq)

Percent dissociation can be calculated by using following formula,

  Percent dissociated =  dissociationinitial×10012.5 %=  x0.735×100x =  0.092M

The dissociation of acid is 0.092M

Therefore,

Construct ICE table:

  HA (aq) +H2O(l)H3O(aq) +   A (aq)

Initial concentration0.735 M-00
Change -x + x+ x
     
At equilibrium0.735-x xx

The initial concentration is 0.735 M HA.

  HA (aq) +H2O(l)H3O(aq) +   A (aq)The value Kais calculating by using following formula, Ka =[A][H3O][HA]

  [Dissociated acid] = x =[A-] =[H3O+]=0.092M[HA] = 0.735 M -0.092 M[HA] = 0.643 M

Therefore,

  Ka =[A][H3O][HA]Ka =[0.092][0.092][0.643]Ka =1.32×102

The Ka of the acid is 1.32×102.

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Chapter 18 Solutions

Chemistry: The Molecular Nature of Matter and Change

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