pH of the solution has to be calculated for 0 .65 M potassium formate (HCOOK) . 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 ) +H 2 O ( 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: K b = [ BH + ] [ OH - ] [ B ] 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 pK b values, pK b = -log K b and 10 - pK b = K b Percent dissociation can be calculated by using following formula, Percent dissociated = dissociation initial ×100 The K b value is calculating by using following formula, K w = K a × K b

Question

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Answer 1

Question

Definition Definition Number that is expressed before molecules, ions, and atoms such that it balances out the number of components present on either section of the equation in a chemical reaction. Stoichiometric coefficients can be a fraction or a whole number and are useful in determining the mole ratio among the reactants and products. In any equalized chemical equation, the number of components on either side of the equation will be the same.

Chapter 18, Problem 18.113P

(a)

Interpretation Introduction

Interpretation:

$pH$ of the solution has to be calculated for $0.65 M potassium formate (HCOOK)$ .

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:

$Kb=[BH+][OH-][B]$

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

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

Percent dissociation can be calculated by using following formula,

$Percent dissociated = dissociationinitial×100$

The $Kb$ value is calculating by using following formula,

$Kw = Ka × Kb$

(a)

Expert Solution

Explanation of Solution

Given,

Potassium formate ion is a base therefore it can accept the proton from the water. Potassium formate dissociates into potassium ions ( $K+$ ) and formate ions ( $HCOO–$ ). potassium ions ( $K+$ ) is the conjugate acid of $KOH$ , so $K+$ does not react with water. formate ions ( $HCOO–$ ) is the conjugate base of $HCOOH$ , so it does react with a base-dissociation reaction.

The balance equation is given below,

$HCOO-(aq) + H2O(l) ⇌ HCOOH(aq) + OH-(aq)$

$0.100 M sodium phenolate$ Solution.

Therefore,

ICE table:

$HCOO-(aq) + H2O(l) ⇌ HCOOH(aq) + OH-(aq)$

Initial concentration

0.100 M

-

0

Change

-x

+ x

At equilibrium

0.

0-x

x

The initial concentration is $0.65 M potassium formate (HCOOK)$ .

$HCOO-(aq) + H2O(l) ⇌ HCOOH(aq) + OH-(aq) The value Kb of HCOO− is calculating by using following formula, Kw = Ka × KbKb = Kw KaKb = 1×10−14 1.8×10−4Kb = 5.55×10−11Kb =[HCOOH][OH−][HCOO-]given,Kb =[HCOOH][OH−][HCOO-]= 5.55×10−11,let consider,0.65−x= 0.65[HCOOH][OH−][HCOO-]= 5.55×10−11,x2(0.65)= 5.55×10−11x2 = 3.6×10−11x = 6.00×10−6 M = [OH−]$

$x is small when compared to 0.65 M, therefore, Percent dissociated = dissociationinitial×1006.00×10−6 M 0.65 M×100= 9.24×10−4The dissociation value is very less, and it is validKw=[OH-][H3O+][H3O+]=1.0×10-146.00×10−6 [H3O+]=1.66×10-9Therefore,PH = - log (H3O+)PH = - log (1.66×10-9)PH = 8.77$

$pH$ of $0.65 M potassium formate (HCOOK)$ is 8.77.

(b)

Interpretation Introduction

Interpretation:

$pH$ of the solution has to be calculated for $0.85 M NH4Br$ .

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:

$Kb=[BH+][OH-][B]$

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

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

Percent dissociation can be calculated by using following formula,

$Percent dissociated = dissociationinitial×100$

The $Kb$ value is calculating by using following formula,

$Kw = Ka × Kb$

(b)

Expert Solution

Explanation of Solution

Given,

Ammonium bromide dissociates into ammonium ion ( $NH4+$ ) and bromide ions ( $Br–$ ). Bromide ion ( $Br–$ ) is the conjugate base of a strong acid so it will not influence the pH of the solution. Ammonium ion is the conjugate acid of a weak base, so an acid-dissociation reaction determines the pH of the solution.

The balance equation is given below,

$NH4+(aq) + H2O(l) ⇌ NH3(aq) + H3O+(aq)$

$0.85 M NH4Br$ Solution.

Therefore,

ICE table:

$NH4+(aq) + H2O(l) ⇌ NH3(aq) + H3O+(aq)$

Initial concentration	0.85 M	-	0	0
Change	-x		+ x	+ x
At equilibrium	0.85-x		x	x

The initial concentration is $0.15 M methylammonium bromide (CH3NH3Br)$ .

$NH4+(aq) + H2O(l) ⇌ NH3(aq) + H3O+(aq) The value Ka of NH4+ is calculating by using following formula, Kw = Ka × KbKa = Kw KbKa = 1×10−14 1.76×10−5Ka = 5.68×10−10Ka =[ CH3NH2][H3O+][NH4+]given,Ka =[ CH3NH2][H3O+][NH4+]= 5.68×10−10,let consider,0.850−x= 0.850[ CH3NH2][H3O+][NH4+]= 5.68×10−10,x2(0.850)= 5.68×10−10x2 = 4.82×10−10x = 2.19×10−5 M$

$x is small when compared to 0.850 M, therefore,Percent dissociated = dissociationinitial×100 2.19×10−5 M 0.850 M×100= 2.58×10−3The dissociation value is very less, and it is validTherefore,PH = - log (H3O+)PH = - log (2.19×10−5)PH =4.66$

$pH$ of $0.85 M NH4Br$ is 4.66.

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Answer 2

Question

Definition Definition Number that is expressed before molecules, ions, and atoms such that it balances out the number of components present on either section of the equation in a chemical reaction. Stoichiometric coefficients can be a fraction or a whole number and are useful in determining the mole ratio among the reactants and products. In any equalized chemical equation, the number of components on either side of the equation will be the same.

Chapter 18, Problem 18.113P

(a)

Interpretation Introduction

Interpretation:

$pH$ of the solution has to be calculated for $0.65 M potassium formate (HCOOK)$ .

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:

$Kb=[BH+][OH-][B]$

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

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

Percent dissociation can be calculated by using following formula,

$Percent dissociated = dissociationinitial×100$

The $Kb$ value is calculating by using following formula,

$Kw = Ka × Kb$

(a)

Expert Solution

Explanation of Solution

Given,

Potassium formate ion is a base therefore it can accept the proton from the water. Potassium formate dissociates into potassium ions ( $K+$ ) and formate ions ( $HCOO–$ ). potassium ions ( $K+$ ) is the conjugate acid of $KOH$ , so $K+$ does not react with water. formate ions ( $HCOO–$ ) is the conjugate base of $HCOOH$ , so it does react with a base-dissociation reaction.

The balance equation is given below,

$HCOO-(aq) + H2O(l) ⇌ HCOOH(aq) + OH-(aq)$

$0.100 M sodium phenolate$ Solution.

Therefore,

ICE table:

$HCOO-(aq) + H2O(l) ⇌ HCOOH(aq) + OH-(aq)$

Initial concentration

0.100 M

-

0

Change

-x

+ x

At equilibrium

0.

0-x

x

The initial concentration is $0.65 M potassium formate (HCOOK)$ .

$HCOO-(aq) + H2O(l) ⇌ HCOOH(aq) + OH-(aq) The value Kb of HCOO− is calculating by using following formula, Kw = Ka × KbKb = Kw KaKb = 1×10−14 1.8×10−4Kb = 5.55×10−11Kb =[HCOOH][OH−][HCOO-]given,Kb =[HCOOH][OH−][HCOO-]= 5.55×10−11,let consider,0.65−x= 0.65[HCOOH][OH−][HCOO-]= 5.55×10−11,x2(0.65)= 5.55×10−11x2 = 3.6×10−11x = 6.00×10−6 M = [OH−]$

$x is small when compared to 0.65 M, therefore, Percent dissociated = dissociationinitial×1006.00×10−6 M 0.65 M×100= 9.24×10−4The dissociation value is very less, and it is validKw=[OH-][H3O+][H3O+]=1.0×10-146.00×10−6 [H3O+]=1.66×10-9Therefore,PH = - log (H3O+)PH = - log (1.66×10-9)PH = 8.77$

$pH$ of $0.65 M potassium formate (HCOOK)$ is 8.77.

(b)

Interpretation Introduction

Interpretation:

$pH$ of the solution has to be calculated for $0.85 M NH4Br$ .

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:

$Kb=[BH+][OH-][B]$

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

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

Percent dissociation can be calculated by using following formula,

$Percent dissociated = dissociationinitial×100$

The $Kb$ value is calculating by using following formula,

$Kw = Ka × Kb$

(b)

Expert Solution

Explanation of Solution

Given,

Ammonium bromide dissociates into ammonium ion ( $NH4+$ ) and bromide ions ( $Br–$ ). Bromide ion ( $Br–$ ) is the conjugate base of a strong acid so it will not influence the pH of the solution. Ammonium ion is the conjugate acid of a weak base, so an acid-dissociation reaction determines the pH of the solution.

The balance equation is given below,

$NH4+(aq) + H2O(l) ⇌ NH3(aq) + H3O+(aq)$

$0.85 M NH4Br$ Solution.

Therefore,

ICE table:

$NH4+(aq) + H2O(l) ⇌ NH3(aq) + H3O+(aq)$

Initial concentration	0.85 M	-	0	0
Change	-x		+ x	+ x
At equilibrium	0.85-x		x	x

The initial concentration is $0.15 M methylammonium bromide (CH3NH3Br)$ .

$NH4+(aq) + H2O(l) ⇌ NH3(aq) + H3O+(aq) The value Ka of NH4+ is calculating by using following formula, Kw = Ka × KbKa = Kw KbKa = 1×10−14 1.76×10−5Ka = 5.68×10−10Ka =[ CH3NH2][H3O+][NH4+]given,Ka =[ CH3NH2][H3O+][NH4+]= 5.68×10−10,let consider,0.850−x= 0.850[ CH3NH2][H3O+][NH4+]= 5.68×10−10,x2(0.850)= 5.68×10−10x2 = 4.82×10−10x = 2.19×10−5 M$

$x is small when compared to 0.850 M, therefore,Percent dissociated = dissociationinitial×100 2.19×10−5 M 0.850 M×100= 2.58×10−3The dissociation value is very less, and it is validTherefore,PH = - log (H3O+)PH = - log (2.19×10−5)PH =4.66$

$pH$ of $0.85 M NH4Br$ is 4.66.

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Answer 3

Question

Definition Definition Number that is expressed before molecules, ions, and atoms such that it balances out the number of components present on either section of the equation in a chemical reaction. Stoichiometric coefficients can be a fraction or a whole number and are useful in determining the mole ratio among the reactants and products. In any equalized chemical equation, the number of components on either side of the equation will be the same.

Chapter 18, Problem 18.113P

(a)

Interpretation Introduction

Interpretation:

$pH$ of the solution has to be calculated for $0.65 M potassium formate (HCOOK)$ .

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:

$Kb=[BH+][OH-][B]$

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

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

Percent dissociation can be calculated by using following formula,

$Percent dissociated = dissociationinitial×100$

The $Kb$ value is calculating by using following formula,

$Kw = Ka × Kb$

(a)

Expert Solution

Explanation of Solution

Given,

Potassium formate ion is a base therefore it can accept the proton from the water. Potassium formate dissociates into potassium ions ( $K+$ ) and formate ions ( $HCOO–$ ). potassium ions ( $K+$ ) is the conjugate acid of $KOH$ , so $K+$ does not react with water. formate ions ( $HCOO–$ ) is the conjugate base of $HCOOH$ , so it does react with a base-dissociation reaction.

The balance equation is given below,

$HCOO-(aq) + H2O(l) ⇌ HCOOH(aq) + OH-(aq)$

$0.100 M sodium phenolate$ Solution.

Therefore,

ICE table:

$HCOO-(aq) + H2O(l) ⇌ HCOOH(aq) + OH-(aq)$

Initial concentration

0.100 M

-

0

Change

-x

+ x

At equilibrium

0.

0-x

x

The initial concentration is $0.65 M potassium formate (HCOOK)$ .

$HCOO-(aq) + H2O(l) ⇌ HCOOH(aq) + OH-(aq) The value Kb of HCOO− is calculating by using following formula, Kw = Ka × KbKb = Kw KaKb = 1×10−14 1.8×10−4Kb = 5.55×10−11Kb =[HCOOH][OH−][HCOO-]given,Kb =[HCOOH][OH−][HCOO-]= 5.55×10−11,let consider,0.65−x= 0.65[HCOOH][OH−][HCOO-]= 5.55×10−11,x2(0.65)= 5.55×10−11x2 = 3.6×10−11x = 6.00×10−6 M = [OH−]$

$x is small when compared to 0.65 M, therefore, Percent dissociated = dissociationinitial×1006.00×10−6 M 0.65 M×100= 9.24×10−4The dissociation value is very less, and it is validKw=[OH-][H3O+][H3O+]=1.0×10-146.00×10−6 [H3O+]=1.66×10-9Therefore,PH = - log (H3O+)PH = - log (1.66×10-9)PH = 8.77$

$pH$ of $0.65 M potassium formate (HCOOK)$ is 8.77.

(b)

Interpretation Introduction

Interpretation:

$pH$ of the solution has to be calculated for $0.85 M NH4Br$ .

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:

$Kb=[BH+][OH-][B]$

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

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

Percent dissociation can be calculated by using following formula,

$Percent dissociated = dissociationinitial×100$

The $Kb$ value is calculating by using following formula,

$Kw = Ka × Kb$

(b)

Expert Solution

Explanation of Solution

Given,

Ammonium bromide dissociates into ammonium ion ( $NH4+$ ) and bromide ions ( $Br–$ ). Bromide ion ( $Br–$ ) is the conjugate base of a strong acid so it will not influence the pH of the solution. Ammonium ion is the conjugate acid of a weak base, so an acid-dissociation reaction determines the pH of the solution.

The balance equation is given below,

$NH4+(aq) + H2O(l) ⇌ NH3(aq) + H3O+(aq)$

$0.85 M NH4Br$ Solution.

Therefore,

ICE table:

$NH4+(aq) + H2O(l) ⇌ NH3(aq) + H3O+(aq)$

Initial concentration	0.85 M	-	0	0
Change	-x		+ x	+ x
At equilibrium	0.85-x		x	x

The initial concentration is $0.15 M methylammonium bromide (CH3NH3Br)$ .

$NH4+(aq) + H2O(l) ⇌ NH3(aq) + H3O+(aq) The value Ka of NH4+ is calculating by using following formula, Kw = Ka × KbKa = Kw KbKa = 1×10−14 1.76×10−5Ka = 5.68×10−10Ka =[ CH3NH2][H3O+][NH4+]given,Ka =[ CH3NH2][H3O+][NH4+]= 5.68×10−10,let consider,0.850−x= 0.850[ CH3NH2][H3O+][NH4+]= 5.68×10−10,x2(0.850)= 5.68×10−10x2 = 4.82×10−10x = 2.19×10−5 M$

$x is small when compared to 0.850 M, therefore,Percent dissociated = dissociationinitial×100 2.19×10−5 M 0.850 M×100= 2.58×10−3The dissociation value is very less, and it is validTherefore,PH = - log (H3O+)PH = - log (2.19×10−5)PH =4.66$

$pH$ of $0.85 M NH4Br$ is 4.66.

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Answer 4

Question

Definition Definition Number that is expressed before molecules, ions, and atoms such that it balances out the number of components present on either section of the equation in a chemical reaction. Stoichiometric coefficients can be a fraction or a whole number and are useful in determining the mole ratio among the reactants and products. In any equalized chemical equation, the number of components on either side of the equation will be the same.

Chapter 18, Problem 18.113P

(a)

Interpretation Introduction

Interpretation:

$pH$ of the solution has to be calculated for $0.65 M potassium formate (HCOOK)$ .

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:

$Kb=[BH+][OH-][B]$

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

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

Percent dissociation can be calculated by using following formula,

$Percent dissociated = dissociationinitial×100$

The $Kb$ value is calculating by using following formula,

$Kw = Ka × Kb$

(a)

Expert Solution

Explanation of Solution

Given,

Potassium formate ion is a base therefore it can accept the proton from the water. Potassium formate dissociates into potassium ions ( $K+$ ) and formate ions ( $HCOO–$ ). potassium ions ( $K+$ ) is the conjugate acid of $KOH$ , so $K+$ does not react with water. formate ions ( $HCOO–$ ) is the conjugate base of $HCOOH$ , so it does react with a base-dissociation reaction.

The balance equation is given below,

$HCOO-(aq) + H2O(l) ⇌ HCOOH(aq) + OH-(aq)$

$0.100 M sodium phenolate$ Solution.

Therefore,

ICE table:

$HCOO-(aq) + H2O(l) ⇌ HCOOH(aq) + OH-(aq)$

Initial concentration

0.100 M

-

0

Change

-x

+ x

At equilibrium

0.

0-x

x

The initial concentration is $0.65 M potassium formate (HCOOK)$ .

$HCOO-(aq) + H2O(l) ⇌ HCOOH(aq) + OH-(aq) The value Kb of HCOO− is calculating by using following formula, Kw = Ka × KbKb = Kw KaKb = 1×10−14 1.8×10−4Kb = 5.55×10−11Kb =[HCOOH][OH−][HCOO-]given,Kb =[HCOOH][OH−][HCOO-]= 5.55×10−11,let consider,0.65−x= 0.65[HCOOH][OH−][HCOO-]= 5.55×10−11,x2(0.65)= 5.55×10−11x2 = 3.6×10−11x = 6.00×10−6 M = [OH−]$

$x is small when compared to 0.65 M, therefore, Percent dissociated = dissociationinitial×1006.00×10−6 M 0.65 M×100= 9.24×10−4The dissociation value is very less, and it is validKw=[OH-][H3O+][H3O+]=1.0×10-146.00×10−6 [H3O+]=1.66×10-9Therefore,PH = - log (H3O+)PH = - log (1.66×10-9)PH = 8.77$

$pH$ of $0.65 M potassium formate (HCOOK)$ is 8.77.

(b)

Interpretation Introduction

Interpretation:

$pH$ of the solution has to be calculated for $0.85 M NH4Br$ .

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:

$Kb=[BH+][OH-][B]$

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

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

Percent dissociation can be calculated by using following formula,

$Percent dissociated = dissociationinitial×100$

The $Kb$ value is calculating by using following formula,

$Kw = Ka × Kb$

(b)

Expert Solution

Explanation of Solution

Given,

Ammonium bromide dissociates into ammonium ion ( $NH4+$ ) and bromide ions ( $Br–$ ). Bromide ion ( $Br–$ ) is the conjugate base of a strong acid so it will not influence the pH of the solution. Ammonium ion is the conjugate acid of a weak base, so an acid-dissociation reaction determines the pH of the solution.

The balance equation is given below,

$NH4+(aq) + H2O(l) ⇌ NH3(aq) + H3O+(aq)$

$0.85 M NH4Br$ Solution.

Therefore,

ICE table:

$NH4+(aq) + H2O(l) ⇌ NH3(aq) + H3O+(aq)$

Initial concentration	0.85 M	-	0	0
Change	-x		+ x	+ x
At equilibrium	0.85-x		x	x

The initial concentration is $0.15 M methylammonium bromide (CH3NH3Br)$ .

$NH4+(aq) + H2O(l) ⇌ NH3(aq) + H3O+(aq) The value Ka of NH4+ is calculating by using following formula, Kw = Ka × KbKa = Kw KbKa = 1×10−14 1.76×10−5Ka = 5.68×10−10Ka =[ CH3NH2][H3O+][NH4+]given,Ka =[ CH3NH2][H3O+][NH4+]= 5.68×10−10,let consider,0.850−x= 0.850[ CH3NH2][H3O+][NH4+]= 5.68×10−10,x2(0.850)= 5.68×10−10x2 = 4.82×10−10x = 2.19×10−5 M$

$x is small when compared to 0.850 M, therefore,Percent dissociated = dissociationinitial×100 2.19×10−5 M 0.850 M×100= 2.58×10−3The dissociation value is very less, and it is validTherefore,PH = - log (H3O+)PH = - log (2.19×10−5)PH =4.66$

$pH$ of $0.85 M NH4Br$ is 4.66.

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Answer 5

Chapter 18, Problem 18.113P

(a)

Interpretation Introduction

Interpretation:

$pH$ of the solution has to be calculated for $0.65 M potassium formate (HCOOK)$ .

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:

$Kb=[BH+][OH-][B]$

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

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

Percent dissociation can be calculated by using following formula,

$Percent dissociated = dissociationinitial×100$

The $Kb$ value is calculating by using following formula,

$Kw = Ka × Kb$

(a)

Expert Solution

Explanation of Solution

Given,

Potassium formate ion is a base therefore it can accept the proton from the water. Potassium formate dissociates into potassium ions ( $K+$ ) and formate ions ( $HCOO–$ ). potassium ions ( $K+$ ) is the conjugate acid of $KOH$ , so $K+$ does not react with water. formate ions ( $HCOO–$ ) is the conjugate base of $HCOOH$ , so it does react with a base-dissociation reaction.

The balance equation is given below,

$HCOO-(aq) + H2O(l) ⇌ HCOOH(aq) + OH-(aq)$

$0.100 M sodium phenolate$ Solution.

Therefore,

ICE table:

$HCOO-(aq) + H2O(l) ⇌ HCOOH(aq) + OH-(aq)$

Initial concentration

0.100 M

-

0

Change

-x

+ x

At equilibrium

0.

0-x

x

The initial concentration is $0.65 M potassium formate (HCOOK)$ .

$HCOO-(aq) + H2O(l) ⇌ HCOOH(aq) + OH-(aq) The value Kb of HCOO− is calculating by using following formula, Kw = Ka × KbKb = Kw KaKb = 1×10−14 1.8×10−4Kb = 5.55×10−11Kb =[HCOOH][OH−][HCOO-]given,Kb =[HCOOH][OH−][HCOO-]= 5.55×10−11,let consider,0.65−x= 0.65[HCOOH][OH−][HCOO-]= 5.55×10−11,x2(0.65)= 5.55×10−11x2 = 3.6×10−11x = 6.00×10−6 M = [OH−]$

$x is small when compared to 0.65 M, therefore, Percent dissociated = dissociationinitial×1006.00×10−6 M 0.65 M×100= 9.24×10−4The dissociation value is very less, and it is validKw=[OH-][H3O+][H3O+]=1.0×10-146.00×10−6 [H3O+]=1.66×10-9Therefore,PH = - log (H3O+)PH = - log (1.66×10-9)PH = 8.77$

$pH$ of $0.65 M potassium formate (HCOOK)$ is 8.77.

(b)

Interpretation Introduction

Interpretation:

$pH$ of the solution has to be calculated for $0.85 M NH4Br$ .

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:

$Kb=[BH+][OH-][B]$

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

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

Percent dissociation can be calculated by using following formula,

$Percent dissociated = dissociationinitial×100$

The $Kb$ value is calculating by using following formula,

$Kw = Ka × Kb$

(b)

Expert Solution

Explanation of Solution

Given,

Ammonium bromide dissociates into ammonium ion ( $NH4+$ ) and bromide ions ( $Br–$ ). Bromide ion ( $Br–$ ) is the conjugate base of a strong acid so it will not influence the pH of the solution. Ammonium ion is the conjugate acid of a weak base, so an acid-dissociation reaction determines the pH of the solution.

The balance equation is given below,

$NH4+(aq) + H2O(l) ⇌ NH3(aq) + H3O+(aq)$

$0.85 M NH4Br$ Solution.

Therefore,

ICE table:

$NH4+(aq) + H2O(l) ⇌ NH3(aq) + H3O+(aq)$

Initial concentration	0.85 M	-	0	0
Change	-x		+ x	+ x
At equilibrium	0.85-x		x	x

The initial concentration is $0.15 M methylammonium bromide (CH3NH3Br)$ .

$NH4+(aq) + H2O(l) ⇌ NH3(aq) + H3O+(aq) The value Ka of NH4+ is calculating by using following formula, Kw = Ka × KbKa = Kw KbKa = 1×10−14 1.76×10−5Ka = 5.68×10−10Ka =[ CH3NH2][H3O+][NH4+]given,Ka =[ CH3NH2][H3O+][NH4+]= 5.68×10−10,let consider,0.850−x= 0.850[ CH3NH2][H3O+][NH4+]= 5.68×10−10,x2(0.850)= 5.68×10−10x2 = 4.82×10−10x = 2.19×10−5 M$

$x is small when compared to 0.850 M, therefore,Percent dissociated = dissociationinitial×100 2.19×10−5 M 0.850 M×100= 2.58×10−3The dissociation value is very less, and it is validTherefore,PH = - log (H3O+)PH = - log (2.19×10−5)PH =4.66$

$pH$ of $0.85 M NH4Br$ is 4.66.

Answer 6

Chapter 18, Problem 18.113P

(a)

Interpretation Introduction

Interpretation:

$pH$ of the solution has to be calculated for $0.65 M potassium formate (HCOOK)$ .

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:

$Kb=[BH+][OH-][B]$

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

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

Percent dissociation can be calculated by using following formula,

$Percent dissociated = dissociationinitial×100$

The $Kb$ value is calculating by using following formula,

$Kw = Ka × Kb$

(a)

Expert Solution

Explanation of Solution

Given,

Potassium formate ion is a base therefore it can accept the proton from the water. Potassium formate dissociates into potassium ions ( $K+$ ) and formate ions ( $HCOO–$ ). potassium ions ( $K+$ ) is the conjugate acid of $KOH$ , so $K+$ does not react with water. formate ions ( $HCOO–$ ) is the conjugate base of $HCOOH$ , so it does react with a base-dissociation reaction.

The balance equation is given below,

$HCOO-(aq) + H2O(l) ⇌ HCOOH(aq) + OH-(aq)$

$0.100 M sodium phenolate$ Solution.

Therefore,

ICE table:

$HCOO-(aq) + H2O(l) ⇌ HCOOH(aq) + OH-(aq)$

Initial concentration

0.100 M

-

0

Change

-x

+ x

At equilibrium

0.

0-x

x

The initial concentration is $0.65 M potassium formate (HCOOK)$ .

$HCOO-(aq) + H2O(l) ⇌ HCOOH(aq) + OH-(aq) The value Kb of HCOO− is calculating by using following formula, Kw = Ka × KbKb = Kw KaKb = 1×10−14 1.8×10−4Kb = 5.55×10−11Kb =[HCOOH][OH−][HCOO-]given,Kb =[HCOOH][OH−][HCOO-]= 5.55×10−11,let consider,0.65−x= 0.65[HCOOH][OH−][HCOO-]= 5.55×10−11,x2(0.65)= 5.55×10−11x2 = 3.6×10−11x = 6.00×10−6 M = [OH−]$

$x is small when compared to 0.65 M, therefore, Percent dissociated = dissociationinitial×1006.00×10−6 M 0.65 M×100= 9.24×10−4The dissociation value is very less, and it is validKw=[OH-][H3O+][H3O+]=1.0×10-146.00×10−6 [H3O+]=1.66×10-9Therefore,PH = - log (H3O+)PH = - log (1.66×10-9)PH = 8.77$

$pH$ of $0.65 M potassium formate (HCOOK)$ is 8.77.

Answer 7

Chapter 18, Problem 18.113P

(a)

Interpretation Introduction

Interpretation:

$pH$ of the solution has to be calculated for $0.65 M potassium formate (HCOOK)$ .

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:

$Kb=[BH+][OH-][B]$

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

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

Percent dissociation can be calculated by using following formula,

$Percent dissociated = dissociationinitial×100$

The $Kb$ value is calculating by using following formula,

$Kw = Ka × Kb$

Answer 8

(a)

Expert Solution

Explanation of Solution

Given,

Potassium formate ion is a base therefore it can accept the proton from the water. Potassium formate dissociates into potassium ions ( $K+$ ) and formate ions ( $HCOO–$ ). potassium ions ( $K+$ ) is the conjugate acid of $KOH$ , so $K+$ does not react with water. formate ions ( $HCOO–$ ) is the conjugate base of $HCOOH$ , so it does react with a base-dissociation reaction.

The balance equation is given below,

$HCOO-(aq) + H2O(l) ⇌ HCOOH(aq) + OH-(aq)$

$0.100 M sodium phenolate$ Solution.

Therefore,

ICE table:

$HCOO-(aq) + H2O(l) ⇌ HCOOH(aq) + OH-(aq)$

Initial concentration

0.100 M

-

0

Change

-x

+ x

At equilibrium

0.

0-x

x

The initial concentration is $0.65 M potassium formate (HCOOK)$ .

$HCOO-(aq) + H2O(l) ⇌ HCOOH(aq) + OH-(aq) The value Kb of HCOO− is calculating by using following formula, Kw = Ka × KbKb = Kw KaKb = 1×10−14 1.8×10−4Kb = 5.55×10−11Kb =[HCOOH][OH−][HCOO-]given,Kb =[HCOOH][OH−][HCOO-]= 5.55×10−11,let consider,0.65−x= 0.65[HCOOH][OH−][HCOO-]= 5.55×10−11,x2(0.65)= 5.55×10−11x2 = 3.6×10−11x = 6.00×10−6 M = [OH−]$

$x is small when compared to 0.65 M, therefore, Percent dissociated = dissociationinitial×1006.00×10−6 M 0.65 M×100= 9.24×10−4The dissociation value is very less, and it is validKw=[OH-][H3O+][H3O+]=1.0×10-146.00×10−6 [H3O+]=1.66×10-9Therefore,PH = - log (H3O+)PH = - log (1.66×10-9)PH = 8.77$

$pH$ of $0.65 M potassium formate (HCOOK)$ is 8.77.

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

(b)

Interpretation Introduction

Interpretation:

$pH$ of the solution has to be calculated for $0.85 M NH4Br$ .

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:

$Kb=[BH+][OH-][B]$

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

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

Percent dissociation can be calculated by using following formula,

$Percent dissociated = dissociationinitial×100$

The $Kb$ value is calculating by using following formula,

$Kw = Ka × Kb$

(b)

Expert Solution

Explanation of Solution

Given,

Ammonium bromide dissociates into ammonium ion ( $NH4+$ ) and bromide ions ( $Br–$ ). Bromide ion ( $Br–$ ) is the conjugate base of a strong acid so it will not influence the pH of the solution. Ammonium ion is the conjugate acid of a weak base, so an acid-dissociation reaction determines the pH of the solution.

The balance equation is given below,

$NH4+(aq) + H2O(l) ⇌ NH3(aq) + H3O+(aq)$

$0.85 M NH4Br$ Solution.

Therefore,

ICE table:

$NH4+(aq) + H2O(l) ⇌ NH3(aq) + H3O+(aq)$

Initial concentration	0.85 M	-	0	0
Change	-x		+ x	+ x
At equilibrium	0.85-x		x	x

The initial concentration is $0.15 M methylammonium bromide (CH3NH3Br)$ .

$NH4+(aq) + H2O(l) ⇌ NH3(aq) + H3O+(aq) The value Ka of NH4+ is calculating by using following formula, Kw = Ka × KbKa = Kw KbKa = 1×10−14 1.76×10−5Ka = 5.68×10−10Ka =[ CH3NH2][H3O+][NH4+]given,Ka =[ CH3NH2][H3O+][NH4+]= 5.68×10−10,let consider,0.850−x= 0.850[ CH3NH2][H3O+][NH4+]= 5.68×10−10,x2(0.850)= 5.68×10−10x2 = 4.82×10−10x = 2.19×10−5 M$

$x is small when compared to 0.850 M, therefore,Percent dissociated = dissociationinitial×100 2.19×10−5 M 0.850 M×100= 2.58×10−3The dissociation value is very less, and it is validTherefore,PH = - log (H3O+)PH = - log (2.19×10−5)PH =4.66$

$pH$ of $0.85 M NH4Br$ is 4.66.

Answer 13

(b)

Interpretation Introduction

Interpretation:

$pH$ of the solution has to be calculated for $0.85 M NH4Br$ .

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:

$Kb=[BH+][OH-][B]$

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

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

Percent dissociation can be calculated by using following formula,

$Percent dissociated = dissociationinitial×100$

The $Kb$ value is calculating by using following formula,

$Kw = Ka × Kb$

Answer 14

(b)

Expert Solution

Explanation of Solution

Given,

Ammonium bromide dissociates into ammonium ion ( $NH4+$ ) and bromide ions ( $Br–$ ). Bromide ion ( $Br–$ ) is the conjugate base of a strong acid so it will not influence the pH of the solution. Ammonium ion is the conjugate acid of a weak base, so an acid-dissociation reaction determines the pH of the solution.

The balance equation is given below,

$NH4+(aq) + H2O(l) ⇌ NH3(aq) + H3O+(aq)$

$0.85 M NH4Br$ Solution.

Therefore,

ICE table:

$NH4+(aq) + H2O(l) ⇌ NH3(aq) + H3O+(aq)$

Initial concentration	0.85 M	-	0	0
Change	-x		+ x	+ x
At equilibrium	0.85-x		x	x

The initial concentration is $0.15 M methylammonium bromide (CH3NH3Br)$ .

$NH4+(aq) + H2O(l) ⇌ NH3(aq) + H3O+(aq) The value Ka of NH4+ is calculating by using following formula, Kw = Ka × KbKa = Kw KbKa = 1×10−14 1.76×10−5Ka = 5.68×10−10Ka =[ CH3NH2][H3O+][NH4+]given,Ka =[ CH3NH2][H3O+][NH4+]= 5.68×10−10,let consider,0.850−x= 0.850[ CH3NH2][H3O+][NH4+]= 5.68×10−10,x2(0.850)= 5.68×10−10x2 = 4.82×10−10x = 2.19×10−5 M$

$x is small when compared to 0.850 M, therefore,Percent dissociated = dissociationinitial×100 2.19×10−5 M 0.850 M×100= 2.58×10−3The dissociation value is very less, and it is validTherefore,PH = - log (H3O+)PH = - log (2.19×10−5)PH =4.66$