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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

What is the pH of a buffer solution that is 0.10 M NH3 and 0.10 M NH4+? What is the pH if 12 mL of 0.20 M hydrochloric acid is added to 125 mL of buffer?

Interpretation Introduction

Interpretation:

The pH of a buffer solution that is 0.10 M NH3 and 0.10 M NH4+ has to be calculated.

The pH of a solution resulting from addition of 12.0 mL of 0.20 M Hydrochloric acid to the 125 mL of the buffer solution has to be calculated.

Concept Information:

Base ionization constant, Kb :

The equilibrium expression for the weak base is given below.

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

Where,

Kb is base ionization constant,

[OH] is concentration of hydroxide ion

[BH+] is concentration of conjugate acid

[B] is concentration of the base

Relationship between pH and pOH:

pH + pOH = 14

Explanation

To Calculate: The pH of a buffer solution that is 0.10 M NH3 and 0.10 M NH4+

Given data:

The concentration of NH3 = 0.10 M

The concentration of NH4+ = 0.10 M

pH Calculation:

Construct an equilibrium table with assumption of x as unknown concentration.

Consider the concentration of NH3 from 0.10 M NH3 solution and concentration of NH4+ from 0.10 M NH4+ solution

The hydrolysis reaction of NH3 is as follows,

  NH3     +     H2O       NH4+   +   OH
Initial (M)

0.10

x

0.10-x

0.10 0.00
Change (M) +x +x
Equilibrium (M) 0.10+x x

The base ionization constant (Kb) for NH3 is 1.8×105

Substitute the equilibrium concentrations into the equilibrium-constant expression.

      Kb =[NH4+][OH-][NH3] =(0.10+x)(x)(0.10x)   1.8×105 (0.10)(x)(0.10)      x =1.8×105 M

Here x gives the concentration of hydroxide ion, [OH] .

The pOH is calculated from the obtained [OH] as follows,

pOH = -log[OH] =log(1.8×105) =4.744

The pH is calculated from pOH as follows,

pH + pOH =14 pH =14 - 4.744 =9.26

Therefore, the pH of the buffer solution of 0.10 M NH3 and 0.10 M NH4+ (before addition of HCl ) is 9.26

To Calculate: The pH of a solution resulting from addition of 12.0 mL of 0.20 M Hydrochloric acid to the 125 mL of the buffer solution

Given data:

The concentration of Hydrochloric acid is 0.20 M

The volume of Hydrochloric acid is 12.0 mL

The volume of the buffer is 125 mL

Explanation:

12. mL of 0.20 M HCl is added to the 125 mL buffer, noting that the H3O+ ion reacts with NH3 to form NH4+

The total volume of the buffer is calculated as,

Total volume = 125 mL + 12 mL =0.125 L + 0.012 L =0.137 L

The moles of HCl added is calculated as:

mol of HCl =(0

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Chapter 16 Solutions

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Sect-16.4 P-16.9ESect-16.4 P-16.3CCSect-16.5 P-16.4CCSect-16.5 P-16.10ESect-16.5 P-16.11ESect-16.6 P-16.12ESect-16.6 P-16.13ESect-16.6 P-16.5CCSect-16.6 P-16.6CCSect-16.7 P-16.14ESect-16.7 P-16.15ESect-16.7 P-16.16ECh-16 P-16.1QPCh-16 P-16.2QPCh-16 P-16.3QPCh-16 P-16.4QPCh-16 P-16.5QPCh-16 P-16.6QPCh-16 P-16.7QPCh-16 P-16.8QPCh-16 P-16.9QPCh-16 P-16.10QPCh-16 P-16.11QPCh-16 P-16.12QPCh-16 P-16.13QPCh-16 P-16.14QPCh-16 P-16.15QPCh-16 P-16.16QPCh-16 P-16.17QPCh-16 P-16.18QPCh-16 P-16.19QPCh-16 P-16.20QPCh-16 P-16.21QPCh-16 P-16.22QPCh-16 P-16.23QPCh-16 P-16.24QPCh-16 P-16.25QPCh-16 P-16.26QPCh-16 P-16.27QPCh-16 P-16.28QPCh-16 P-16.29QPCh-16 P-16.30QPCh-16 P-16.31QPCh-16 P-16.32QPCh-16 P-16.33QPCh-16 P-16.34QPCh-16 P-16.35QPCh-16 P-16.36QPCh-16 P-16.37QPCh-16 P-16.38QPCh-16 P-16.39QPCh-16 P-16.40QPCh-16 P-16.41QPCh-16 P-16.42QPCh-16 P-16.43QPCh-16 P-16.44QPCh-16 P-16.45QPCh-16 P-16.46QPCh-16 P-16.47QPCh-16 P-16.48QPCh-16 P-16.49QPCh-16 P-16.50QPCh-16 P-16.51QPCh-16 P-16.52QPCh-16 P-16.53QPCh-16 P-16.54QPCh-16 P-16.55QPCh-16 P-16.56QPCh-16 P-16.57QPCh-16 P-16.58QPCh-16 P-16.59QPCh-16 P-16.60QPCh-16 P-16.61QPCh-16 P-16.62QPCh-16 P-16.63QPCh-16 P-16.64QPCh-16 P-16.65QPCh-16 P-16.66QPCh-16 P-16.67QPCh-16 P-16.68QPCh-16 P-16.69QPCh-16 P-16.70QPCh-16 P-16.71QPCh-16 P-16.72QPCh-16 P-16.73QPCh-16 P-16.74QPCh-16 P-16.75QPCh-16 P-16.76QPCh-16 P-16.77QPCh-16 P-16.78QPCh-16 P-16.79QPCh-16 P-16.80QPCh-16 P-16.81QPCh-16 P-16.82QPCh-16 P-16.83QPCh-16 P-16.84QPCh-16 P-16.85QPCh-16 P-16.86QPCh-16 P-16.87QPCh-16 P-16.88QPCh-16 P-16.89QPCh-16 P-16.90QPCh-16 P-16.91QPCh-16 P-16.92QPCh-16 P-16.93QPCh-16 P-16.94QPCh-16 P-16.95QPCh-16 P-16.96QPCh-16 P-16.97QPCh-16 P-16.98QPCh-16 P-16.99QPCh-16 P-16.100QPCh-16 P-16.101QPCh-16 P-16.102QPCh-16 P-16.103QPCh-16 P-16.104QPCh-16 P-16.105QPCh-16 P-16.106QPCh-16 P-16.107QPCh-16 P-16.108QPCh-16 P-16.109QPCh-16 P-16.110QPCh-16 P-16.111QPCh-16 P-16.112QPCh-16 P-16.113QPCh-16 P-16.114QPCh-16 P-16.115QPCh-16 P-16.116QPCh-16 P-16.117QPCh-16 P-16.118QPCh-16 P-16.119QPCh-16 P-16.120QPCh-16 P-16.121QPCh-16 P-16.122QPCh-16 P-16.123QPCh-16 P-16.124QPCh-16 P-16.125QPCh-16 P-16.126QPCh-16 P-16.127QPCh-16 P-16.128QPCh-16 P-16.129QPCh-16 P-16.130QPCh-16 P-16.131QPCh-16 P-16.132QPCh-16 P-16.133QPCh-16 P-16.134QPCh-16 P-16.135QPCh-16 P-16.136QPCh-16 P-16.137QPCh-16 P-16.138QPCh-16 P-16.139QPCh-16 P-16.140QPCh-16 P-16.141QPCh-16 P-16.142QPCh-16 P-16.143QPCh-16 P-16.144QPCh-16 P-16.145QPCh-16 P-16.146QPCh-16 P-16.147QPCh-16 P-16.148QPCh-16 P-16.149QPCh-16 P-16.150QPCh-16 P-16.151QPCh-16 P-16.152QPCh-16 P-16.153QPCh-16 P-16.154QPCh-16 P-16.155QPCh-16 P-16.156QPCh-16 P-16.157QP

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