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 18.76P

(a)

Interpretation Introduction

Interpretation:

The pH of the solution has to be calculated for 0.175 M HY (Ka = 1.50×104).

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

Given,

Solution has 0.175 M HY (Ka = 1.50×104).

 HY is weak acid therefore it can donate the proton to the water.

The balance equation is given below,

  HY (aq) +H2O(l)H3O(aq) +   Y-(aq)

 0.175 M HY (Ka = 1.50×104) Solution.

Therefore,

Construct ICE table:

  HY (aq) +H2O(l)H3O(aq) +   Y-(aq)

Initial concentration0.175 M-00
Change -x + x+ x
At equilibrium0.175-x xx

The initial concentration is 0.175 M HZ.

  HY (aq) +H2O(l)H3O(aq) +   Y-(aq)The value Kais calculating by using following formula, Ka =[ Y-][H3O][HY]given,Ka =[ Y-][H3O][HY]= 1.50×104,

Let consider,

  Ka =[ Y-][H3O][HY]= 1.50×104

Let consider,

  Ka =[ Y-][H3O][HY]= 1.50×104x2(0.175x)=1.50×104

Assume, x is negligible so 0.175  x  0.175

  x2(0.175x)=1.50×104x2=2.62×105x=5.12×103

Percent dissociation can be calculated by using following formula,

  Percent dissociated =  dissociationinitial×100Percent dissociated =  5.12×1030.175×100Percent dissociated =  2.9%

Percentage of error is valid,

Therefore,

  pH=-log(H3O+)pH=-log(5.12×103)pH=2.3

 pH of 0.175 M HY is 2.3.

(b)

Interpretation Introduction

Interpretation:

The pOH of the solution has to be calculated for 0.175 M HX (Ka = 2.00×102).

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

Given,

Solution has 0.175 M HX (Ka = 2.00×102).

 HX is weak acid therefore it can donate the proton to the water.

The balance equation is given below,

  HX (aq) +H2O(l)H3O(aq) +   X-(aq)

 0.175 M HX (Ka = 2.00×102) Solution.

Therefore,

Construct ICE table:

  HX (aq) +H2O(l)H3O(aq) +   X-(aq)

Initial concentration0.175 M-00
Change -x + x+ x
At equilibrium0.175-x xx

The initial concentration is 0.045 M HZ.

  HX (aq) +H2O(l)H3O(aq) +   X-(aq)The value Kais calculating by using following formula, Ka =[ X-][H3O][HX]given,Ka =[ X-][H3O][HX]= 2.00×102,

Let consider,

  Ka =[ X-][H3O][HX]= 2.00×102

Let consider,

  Ka =[ X-][H3O][HX]= 2.00×102x2(0.175x)=2.00×102

Assume, x is negligible so 0.045  x  0.045

  x2(0.175x)=2.00×102x2=3.5×103x=0.059

Percent dissociation can be calculated by using following formula,

  Percent dissociated =  dissociationinitial×100Percent dissociated =  0.0590.175×100Percent dissociated =  34%

The dissociation of HX is not negligible in comparison to the initial concentration. Therefore, the equilibrium expression can be solved using the quadratic formula.

The quadratic formula is given below,

  x2(0.175x)=2.00×102x2=(0.175x)×2.00×102x2+2.00×102x3.5×103=0

  a=1,b=2.0×102,c=3.5×103x=b±b24ac2ax=(2.0×102)±(2.0×102)24(1)(3.5×103)2(1)x=5.0×102M=  [H3O+]

The Kb value is calculating by using following formula,

  Kw = Ka × KbKb = KwKaKb = 1×10145.0×102Kb = 2.0×1013

Therefore,

  pOH=-log(OH)pOH=-log(2.0×1013)pOH=12.69

 pOH of 0.175 M HX is 12.69.

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Chapter 18 Solutions

Chemistry: The Molecular Nature of Matter and Change

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