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General Chemistry - Standalone boo...

11th Edition
Steven D. Gammon + 7 others
ISBN: 9781305580343

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BuyFindarrow_forward

General Chemistry - Standalone boo...

11th Edition
Steven D. Gammon + 7 others
ISBN: 9781305580343
Textbook Problem

Which of the following aqueous solutions has the highest pH and which has the lowest? a 0.1 M NH3; b 0.1 M NH4Br; c 0.1 M NaF; d 0.1 M NaCl.

Interpretation Introduction

Interpretation:

Among the given aqueous solutions, the solution having highest pH and lowest pH has to be identified

Concept Information:

The equilibrium expression for the ionization of weak base B will be,

Kb=[HB+][OH-][B]

pOH definition:

The pOH of a solution is defined as the negative base-10 logarithm of the hydroxide ion [OH-] concentration. pOH scale is analogous to pH scale.

pOH=-log[OH-]

Relationship between pH and pOH

pH+pOH=14,at25oC

To Identify: The aqueous solution which is having highest and lowest pH values.

Explanation

0.1 M NH3 aqueous solution

pH of 0.1 M NH3 solution

The pH value for a 1.0 M solution of  NH3 is calculated as follows,

The concentration of hydroxide ion for NH3 base can be found out from its equilibrium reaction with water.

Construct an equilibrium table and express the equilibrium concentration of each species in terms of x

  NH3(aq)+H2O(l)NH4+(aq)+OH(aq)
Initial (M)

0.1

x

0.1-x

0.00 0.00
Change (M) +x +x
Equilibrium (M) x x

Thebaseionizationconstantfor ammonia is 1.8×105       Kb =[NH4+][OH-][NH3]  1.8×105 =x20.1x x20.1     x =1.34×103M

Therefore, the concentration of hydroxide ion is 1.34×103M

Calculation of pH:

The pOH can be calculated as follows,

pOH =-log[OH-] =log(1.34×103) =2.87pH+pOH =14pH =14.00-2.87 =11.13

Therefore, the pH of the given 0.1 M NH3 solution is 11.13

0.1 M NH4Br aqueous solution

pH of 0.1 M NH4Br solution

NH4Br has NH4+ and Br- ions. The NH4+ ion is the conjugate acid of ammonia (NH3)

The NH4+ ions undergoes hydrolysis as follows,

NH4+ + H2 NH3 + H3O+

The Kb for ammonia is 1.8×105

The Ka value for NH4+ can be calculated as follows,

Ka =KwKb =1.0×10-141.8×105 =5.56×1010

Therefore, the Ka value for NH4+ is 5.56×1010

Hydrolysis of NH4+ ion and calculation of concentration of hydronium ion:

The NH4+ ions undergo hydrolysis.

The equilibrium reaction will be as follows.

  NH4+    +   H2O           NH3   +    H3O+
Initial (M)

0.1

x

0.1-x

0.00 0.00
Change (M) +x +x
Equilibrium (M) x x

Solve for x

Write the equilibrium concentrations in terms of the equilibrium expression.

Knowing the value of the equilibrium constant Ka , solve for x .

     Ka =[NH3][H3O+][NH4+]5

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