Chemistry: The Molecular Science
Chemistry: The Molecular Science
5th Edition
ISBN: 9781285199047
Author: John W. Moore, Conrad L. Stanitski
Publisher: Cengage Learning
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Chapter 15, Problem 45QRT

(a)

Interpretation Introduction

Interpretation:

pH of 50mLof0.150MNaOH solution has to be calculated.

Concept Introduction:

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+]

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-]

(a)

Expert Solution
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Explanation of Solution

Given,

Volume of acid in liters:

  =50mL×1L1000mL=0.05000L

Original mole H3O+added:

  =0.05000L×0.150molH3O+1L=0.00750molH3O+

The given solution contains only0.150 MHCl, which ionizes to make 0.150 MH3O+.

  pH=log[H3O+]=-log(0.150)=0.824

(b)

Interpretation Introduction

Interpretation:

pH of the solution has to be calculated, after addition of 25.0mLof 0.150MHCl.

Concept Introduction:

Refer part (a).

(b)

Expert Solution
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Explanation of Solution

After the addition of the titrant,

To determine the equilibrium concentration of H3O+

  H3O+=originalmolesacid-totalmolesbaseaddedvolumeofacid(L)+volumeofbase(L)volumeofbaseinliters=25.00mL×1L1000mL=0.02500LtotalmolOH-added=0.02500mL×0.150molOH-1L=0.00375LmolOH- [H3O+]=0.00750mol-0.00375mol0.05000L+0.02500L=0.0500M pH=-log[H3O+] =-log(0.050) =1.30

(c)

Interpretation Introduction

Interpretation:

pH of the solution has to be calculated, after addition of49.9 mLof 0.150MHCl.

Concept Introduction:

Refer part (a).

(c)

Expert Solution
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Explanation of Solution

Used the equation in (b), again

Volume of base (L)

  49.9mL×1L1000mL=0.0490LtotalmolOH-added=0.0499L×0.150molOH-1L=0.007485molOH- =0.00749molOH-

   [H3O+]=0.00750mol-0.007485mol0.05000L+0.0499L =0.000015mol0.0999L =0.00015M0.0002M pH=-log[H3O+] =-log(0.00015) 3.8.

(d)

Interpretation Introduction

Interpretation:

pH of the solution has to be calculated, after addition of50.0mLof 0.150MHCl.

Concept Introduction:

Refer part (a).

(d)

Expert Solution
Check Mark

Explanation of Solution

The volume and total moles of base can be calculated as

  volumeofbase(L)=50mL×1L1000mL=0.05000LtotalmolOH-added=0.05000L×0.150molOH-1L=0.00750molH3O+

Total moles of H3O+added is equal to the original moles OH-,so the solution is neutral, and the pH = 7.000.

(e)

Interpretation Introduction

Interpretation:

pH of the solution has to be calculated, after addition of 50.1mLof 0.150MHCl.

Concept Introduction:

Refer part (a).

(e)

Expert Solution
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Explanation of Solution

At equivalent point, the moles of base begin to exceed the moles of acid, so we adapt the expression we used in (b) for a solution with excess base to determine the equilibrium concentration of OH.

   [OH-]=totalmolesbaseadded-originalmolesacidvolumeofacid(L)+volumeofbase(L) volumeofbase(L)=50.1mL×1L1000mL=0.0501LtotalmolOH-added=0.0501L×0.150molOHG1L=0.00750molOHG [OH-]=0.007515mol-0.00750mol0.05000L+0.0501L=0.000015mol0.1001L =0.00015M pH=-log[H3O+] =-log(0.000015) =3.8pH=14.00-pOH=14.00-3.8=10.2

The solution has a pH of 10.2.

(f)

Interpretation Introduction

Interpretation:

pH of the solution has to be calculated, after addition of75.0mLof 0.150MHCl.

Concept Introduction:

Refer part (a).

(f)

Expert Solution
Check Mark

Explanation of Solution

The pH of equivalent point of titrant can be calculated as

volumeofbase(L)=75mL×1L1000mL=0.0750LtotalmolOH-added=0.0750L×0.150molOH-1L=0.0113molH3O+

   [H3O+]=0.0113mol-0.00750mol0.05000L+0.0750L =0.00038mol0.1250L =0.030M pH=-log[H3O+] =-log(0.030) =1.52pH=14.00-pOH=14.00-1.52=12.48

Hence, the results in a pH verses volume of titrant to make a titration curve, including equivalence point:

Chemistry: The Molecular Science, Chapter 15, Problem 45QRT

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Chapter 15 Solutions

Chemistry: The Molecular Science

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