Chemistry: The Molecular Nature of Matter and Change - Standalone book
Chemistry: The Molecular Nature of Matter and Change - Standalone book
7th Edition
ISBN: 9780073511177
Author: Martin Silberberg Dr., Patricia Amateis Professor
Publisher: McGraw-Hill Education
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Chapter 18, Problem 18.152P
Interpretation Introduction

Interpretation:

A chemist makes four successive ten-fold dilutions of 1.0×10-5 M HCl.  The pH of the original solution and of each diluted solution has to be calculated (through 1.0×10-9 M HCl).

Concept Introduction:

Molarity:

The number of moles of the substance per litre of liquid is said to be the molarity of the solution.

Molarity can be used to calculate the concentration of solutions.  The expression used is

  M1V1 = M2V2where, M1 = molarity of initial solution M2 = molarity of diluted solution V1 = volume of initial solution V2 = volume of diluted solution

Expert Solution & Answer
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Explanation of Solution

As HCl is a strong acid, it dissociates completely in water.  The concentration of H3O+ is same as the concentration of HCl.

  [HCl] = [H3O+]

The pH of the original solution can be calculated as

  pH = -log (1.0×10-5)pH=5.00

Dilution-1:

The original solution of HCl is diluted to 10 mL.  The concentration can be calculated as follows:

  M1V1 = M2V2(1.0×10-5 M)(1.0 mL) = x (10 mL)[H3O+]HCl = 1.0×10-6 M H3O+

The pH of the solution can be calculated as

  pH = -log (1.0×10-6)pH=6.00

Dilution-2:

The diluted solution of HCl is diluted to 10 mL.  The concentration can be calculated as follows:

  M1V1 = M2V2(1.0×10-6 M)(1.0 mL) = x (10 mL)[H3O+]HCl = 1.0×10-7 M H3O+

The concentration of strong acid is close to the concentration of H3O+ in water ionization.  So, the concentration of H3O+ in the solution does not be the same as the initial concentration.

Hence, the concentration of H3O+ has to be calculated based on the ionization of water equilibrium.

  H2O(l)+ H2O(l)H3O+(aq)+ OH-(aq) Kw=1.0×10-14at25°C

The dilution-2 gives the concentration of H3O+ as 1.0×10-7M.  A reaction table is set up as follows to calculate the concentration of H3O+.

  Concentration(M) 2H2O(l) H3O+(aq)+ OH-(aq)Initial 1.0×10-7 0Change +x +x_Equilibrium 1.0×10-7+x x

  Kw    = [H3O+][OH-]1.0×10-14 = (1.0×10-7+ x)(x)

A quadratic equation is given as

  x2 + 1.0×10-7 x - 1.0×10-14 = 0a=1 b=1.0×10-7 c=-1.0×10-14x = -1.0×10-7±(1.0×10-7)2- 4(1)(-1.0×10-14)2(1)x = 6.18034×10-8

Using the x value, the concentration of H3O+ is calculated as

  [H3O+] = (1.0×10-7+ x) M = (1.0×10-7+ 6.18034×10-8)M = 1.618034×10-7 M H3O+

The pH of the solution can be calculated as

  pH = -log[H3O+]pH = -log (1.618034×10-7)pH=6.79101 = 6.79

Dilution-3:

The diluted solution of HCl is diluted to 10 mL.  The concentration can be calculated as follows:

  M1V1 = M2V2(1.0×10-7 M)(1.0 mL) = x (10 mL)[H3O+]HCl = 1.0×10-8 M H3O+

The dilution-3 gives the concentration of H3O+ as 1.0×10-8M.  A reaction table is set up as follows to calculate the concentration of H3O+.

  Concentration(M) 2H2O(l) H3O+(aq)+ OH-(aq)Initial 1.0×10-8 0Change +x +x_Equilibrium 1.0×10-8+x x

  Kw    = [H3O+][OH-]1.0×10-14 = (1.0×10-8+ x)(x)

A quadratic equation is given as

  x2 + 1.0×10-8 x - 1.0×10-14 = 0a=1 b=1.0×10-8 c=-1.0×10-14x = -1.0×10-8±(1.0×10-8)2- 4(1)(-1.0×10-14)2(1)x = 9.51249×10-8

Using the x value, the concentration of H3O+ is calculated as

  [H3O+] = (1.0×10-8+ x) M = (1.0×10-8+ 9.51249×10-8)M = 1.051249×10-7 M H3O+

The pH of the solution can be calculated as

  pH = -log[H3O+]pH = -log (1.051249×10-7)pH=6.97829 = 6.98

Dilution-4:

The diluted solution of HCl is diluted to 10 mL.  The concentration can be calculated as follows:

  M1V1 = M2V2(1.0×10-8 M)(1.0 mL) = x (10 mL)[H3O+]HCl = 1.0×10-9 M H3O+

The dilution-4 gives the concentration of H3O+ as 1.0×10-9M.  A reaction table is set up as follows to calculate the concentration of H3O+.

  Concentration(M) 2H2O(l) H3O+(aq)+ OH-(aq)Initial 1.0×10-9 0Change +x +x_Equilibrium 1.0×10-9+x x

  Kw    = [H3O+][OH-]1.0×10-14 = (1.0×10-9+ x)(x)

A quadratic equation is given as

  x2 + 1.0×10-9 x - 1.0×10-14 = 0a=1 b=1.0×10-9 c=-1.0×10-14x = -1.0×10-9±(1.0×10-9)2- 4(1)(-1.0×10-14)2(1)x = 9.95012×10-8

Using the x value, the concentration of H3O+ is calculated as

  [H3O+] = (1.0×10-9+ x) M = (1.0×10-9+ 9.95012×10-8)M = 1.005012×10-7 M H3O+

The pH of the solution can be calculated as

  pH = -log[H3O+]pH = -log (1.005012×10-7)pH=6.9978 = 7.00

The pH of the original and diluted solution are

pH of the original solution: 5.00

pH of the dilution-1: 6.00

pH of the dilution-2: 6.79

pH of the dilution-3: 6.98

pH of the dilution-4: 7.00

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Chapter 18 Solutions

Chemistry: The Molecular Nature of Matter and Change - Standalone book

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