   Chapter 16, Problem 16.76QP General Chemistry - Standalone boo...

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

Solutions

Chapter
Section General Chemistry - Standalone boo...

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

A buffer is prepared by adding 115 mL of 0.30 M NH3 to 135 mL of 0.15 M NH4NO3. What is the pH of the final solution?

Interpretation Introduction

Interpretation:

The pH of the buffer prepared by adding 115 mL of 0.30 M NH3 to 135 mL of 0.15 M

NH4NO3 has to be calculated

Concept Introduction:

The equilibrium expression for the weak base is given below.

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

Where,

Kb is base ionization constant,

[OH] is concentration of hydroxide ion

[BH+] is concentration of conjugate acid cation

[B] is concentration of the base

pOH definition:

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

pOH=-log[OH]

Relationship between pH and pOH:

pH + pOH = 14

Explanation

To Calculate:  The pH of the buffer prepared by adding 115 mL of 0.30 M NH3 to 135 mL of 0.15 M NH4NO3

Given data:

The amount of NH3 = 115 mL

The amount of NH4NO3 = 135 mL

The concentration of NH3 = 0.30 M

The concentration of NH4NO3 = 0.15 M

The total volume of the buffer is calculated as,

Total volume = 115 mL + 135 mL =250 mL =0.250 LThe moles of NH3 and NH4+ in the buffer aremol NH3  = 0.30 M x 0.115 L  = 0.0345 mol mol NH4+  = 0.15 M x 0.135 L  = 0.02025 mol

The concentrations of NH3 and NH4+ in the buffer are[NH3] =0.0345 mol0.250 L  = 0.138 M[NH4+] =0.02025 mol0.250 L  = 0.081 M

Now, substitute these concentrations in the equilibrium table by considering it as initial concentrations.

 NH3         +       H2O  ⇌     NH4+   +     OH− Initial (M) 0

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