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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

The pH of Mixtures of Acid, Base, and Salt Solutions

  1. a When 0.10 mol of the ionic solid NaX, where X is an unknown anion, is dissolved in enough water to make 1.0 L of solution, the pH of the solution is 9.12. When 0.10 mol of the ionic solid ACl, where A is an unknown cation, is dissolved in enough water to make 1.0 L of solution, the pH of the solution is 7.00. What would be the pH of 1.0 L of solution that contained 0.10 mol of AX? Be sure to document how you arrived at your answer.
  2. b In the AX solution prepared above, is there any OH present? If so, compare the [OH] in the solution to the [H3O+].
  3. c From the information presented in part a, calculate Kb for the X(aq) anion and Ka for the conjugate acid of X(aq).
  4. d To 1.0 L of solution that contains 0.10 mol of AX, you add 0.025 mol of HCl. How will the pH of this solution compare to that of the solution that contained only NaX? Use chemical reactions as part of your explanation; you do not need to solve for a numerical answer.
  5. e Another 1.0 L sample of solution is prepared by mixing 0.10 mol of AX and 0.10 mol of HCl. The pH of the resulting solution is found to be 3.12. Explain why the pH of this solution is 3.12.
  6. f Finally, consider a different 1.0-L sample of solution that contains 0.10 mol of AX and 0.1 mol of NaOH. The pH of this solution is found to be 13.00. Explain why the pH of this solution is 13.00.
  7. g Some students mistakenly think that a solution that contains 0.10 mol of AX and 0.10 mol of HCl should have a pH of 1.00. Can you come up with a reason why students have this misconception? Write an approach that you would use to help these students understand what they are doing wrong.

(a)

Interpretation Introduction

Interpretation:

The steps for the calculation of pH of 1.0 L solution of 0.1 mol of unknown AX has to be explained

Concept Introduction:

Salt hydrolysis:

Salt hydrolysis is a reaction in which the ion of salt reacts with water and produce either hydronium ion or hydroxide ion.

Based on pH of the solution, salt solutions can be classified as

  • Acidic-(pH will be less than seven)
  • Basic -(pH will be more than seven)
  • Neutral -(pH will be equal to seven)

To Explain: The steps for the calculation of pH of 1.0 L solution of 0.1 mol of unknown AX

Explanation

Given data:

A 1.0 L of solution is obtained by dissolving 0.1 mol of the ionic solid NaX , in water. The pH of this solution is 9.12 and X is an unknown anion.

A 1.0 L of solution is obtained by dissolving 0.1 mol of the ionic solid ACl , in water. The pH of this solution is 7.00 and A is an unknown cation.

A 1.0 L of solution is obtained by dissolving 0.1 mol of the ionic solid AX , in water

Calculation of pH of AX solution:

The ionic compound NaX dissociates into Na+ and X ion. X is the conjugate base of a weak acid  whereas Na+ is a neutral ion

(b)

Interpretation Introduction

Interpretation:

Does the AX solution prepared in part (a) has OH ions has to be explained and if present, [OH] has to be compared with the [H3O+]

Concept Introduction:

Salt hydrolysis:

Salt hydrolysis is a reaction in which the ion of salt reacts with water and produce either hydronium ion or hydroxide ion.

Based on pH of the solution, salt solutions can be classified as

  • Acidic-(pH will be less than seven)
  • Basic -(pH will be more than seven)
  • Neutral -(pH will be equal to seven)

(c)

Interpretation Introduction

Interpretation:

Using part (a) information, the Kb for the X anion and Ka for the conjugate acid of X ion has to be calculated.

Concept Introduction:

Relationship between Ka and Kb

Ka×Kb=Kw   orKa=KwKb

Where Kw is the auto-ionization of water. (1.0×1014)

To Calculate: Using part (a) information, the Kb for the X anion and Ka for the conjugate acid of X ion

(d)

Interpretation Introduction

Interpretation:

The pH of the 1.0 L of 0.1 mol AX solution containing 0.025 mol of HCl has to be compared with the solution containing only NaX

To Compare: The pH of the 1.0 L of 0.1 mol AX solution containing 0.025 mol of HCl with the solution containing only NaX

(e)

Interpretation Introduction

Interpretation:

The pH of the 1.0 L solution prepared by addition of 0.1 mol AX and 0.1 mol HCl is 3.12 which has to be explained

To Explain: The pH of the 1.0 L solution prepared by addition of 0.1 mol AX and 0.1 mol HCl is 3.12

(e)

Interpretation Introduction

Interpretation:

The pH of the 1.0 L solution prepared by addition of 0.1 mol AX and 0.1 mol NaOH is 13.00 which has to be explained

Concept Introduction:

The relationship between pH and pOH is gives as,

pH + pOH =14

To Explain: The pH of the 1.0 L solution prepared by addition of 0.1 mol AX and 0.1 mol NaOH is 13.00

(g)

Interpretation Introduction

Interpretation:

The wrong assumption of pH as 1.00 for a solution containing 0.1 mol of AX and 0.1 mol of HCl has to be explained.

To explain: The wrong assumption of pH as 1.00 for a solution containing 0.1 mol of AX and 0.1 mol of HCl

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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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