Chemical Principles: The Quest for Insight
Chemical Principles: The Quest for Insight
7th Edition
ISBN: 9781464183959
Author: Peter Atkins, Loretta Jones, Leroy Laverman
Publisher: W. H. Freeman
Question
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Chapter 6, Problem 6D.17E

(a)

Interpretation Introduction

Interpretation:

The pH of 0.63 M aqueous NaCH3CO2 has to be calculated.

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+]

On rearranging, the concentration of hydrogen ion [H+] can be calculated using pH as follows,

  [H+]=10-pH

(a)

Expert Solution
Check Mark

Answer to Problem 6D.17E

The pH of 0.63 M aqueous NaCH3CO2 is 9.27.

Explanation of Solution

The proton transfer equilibrium reaction for  the base of CH3COO- is given below.

  CH3COO(aq)+H2O(l)CH3COOH(aq)+OH(aq)

The equilibrium expression for the above reaction is given below.

  Kb=[CH3COOH][OH][CH3COO]

 CH3COOCH3COOHOH-
Initial concentration0.6300
Change in concentration-x+x+x
Equilibrium concentration0.63-xxx

The Kb of CH3COO value is calculated using given formula,

  Kw=Ka×KbKb = KwKa=1.0×10141.8×105=5.56×1010

The equilibrium concentration values obtained in the above table is substituted in the above equation and is given below.

  Kb=x×x0.63-x

The Kb value is 5.56×1010 and it is substituted in above equation and then the value of x is calculated.

  5.56×1010=x20.63-x

The above equation, assume that the x present in 0.63-x is very small than 0.63 then it can be negligible and as follows,

  5.56×10-10=x20.10x2=0.63×(5.56×10-10)x=0.63×(5.56×10-10)=1.87×10-5(x=[OH-])

The pOH of the solution is calculated using the given equation.

  pOH=-log[OH       =-log(1.87×105)=4.73

The general equilibrium expression to find out the pH of the solution is given below,

  pH+pOH = 14          pH = 14-pOH                = 14-4.73                                = 9.27

Therefore, the calculated pH value of 0.63 M aqueous NaCH3CO2 is 9.27.

(b)

Interpretation Introduction

Interpretation:

The pH of 0.65 M aqueous potassium cyanide has to be calculated.

Concept introduction:

Refer to part (a).

(b)

Expert Solution
Check Mark

Answer to Problem 6D.17E

The pH of 0.65 M aqueous potassium cyanide is 11.56.

Explanation of Solution

The proton transfer equilibrium reaction for  the base of CN- is given below.

  CN(aq)+H2O(l)HCN(aq)+OH(aq)

The equilibrium expression for the above reaction is given below.

  Kb=[HCN][OH][CN]

 CN-HCNOH-
Initial concentration0.6500
Change in concentration-x+x+x
Equilibrium concentration0.65-xxx

The Kb of CN value is calculated using given formula,

  Kw=Ka×KbKb = KwKa=1.0×10144.9×1010=2.04×105

The equilibrium concentration values obtained in the above table is substituted in the above equation and is given below.

  Kb=x×x0.65-x

The Kb value is 2.04×105 and it is substituted in above equation and then the value of x is calculated.

  2.04×105=x20.65-x

The above equation, assume that the x present in 0.65-x is very small than 0.65 then it can be negligible and as follows,

  2.04×10-5=x20.65x2=0.65×(2.04×10-5)x=0.65×(2.04×10-5)=3.64×10-3(x=[OH-])

The pOH of the solution is calculated using the given equation.

  pOH=-log[OH       =-log(3.64×103)=2.44

The general equilibrium expression to find out the pH of the solution is given below,

  pH+pOH = 14          pH = 14-pOH                = 14-2.44                                = 11.56

Therefore, the calculated pH value of 0.65 M potassium cyanide is 11.56.

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Chapter 6 Solutions

Chemical Principles: The Quest for Insight

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