Living by Chemistry
Living by Chemistry
2nd Edition
ISBN: 9781464142314
Author: Angelica M. Stacy
Publisher: W. H. Freeman
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Chapter U4.89, Problem 6E

(a)

Interpretation Introduction

Interpretation:

The concentration of an acid and the pH of the solution should be predicted.

Concept Introduction:

The full form of pH is the potential of pH. In other words, it explains the strength of the hydrogen ion in a solution. It is given by the formula,

  pH=log[H+]

(a)

Expert Solution
Check Mark

Answer to Problem 6E

The concentration of an acid and pH of the solution is 1.7 M and -0.23 respectively.

Explanation of Solution

Given information:

  • 0.95 M NaOH
  • 46 mL NaOH
  • 25 mL of acid

The concentration of an acid is calculated as follows:

  C1V1(NaOH)=C2V2(acid)

  0.95M(46mL)=C2(25mL)

        C2=0.95M(46mL)25mL

        C2=1.7M

The pH of the solution is calculated as follows:

  pH=log[H+]

    =log[1.7]

    =0.23

Therefore, the concentration of an acid and pH of the solution is 1.7 M and -0.23 respectively.

(b)

Interpretation Introduction

Interpretation:

The concentration of an acid and the pH of the solution should be predicted.

Concept Introduction:

The full form of pH is the potential of pH. In other words, it explains the strength of the hydrogen ion in a solution. It is given by the formula,

  pH=log[H+]

(b)

Expert Solution
Check Mark

Answer to Problem 6E

The concentration of an acid and pH of the solution is 1.6 M and -0.20 respectively.

Explanation of Solution

Given information:

  • 0.95 M NaOH
  • 17 mL NaOH
  • 10 mL of acid

The concentration of an acid is calculated as follows:

  C1V1(NaOH)=C2V2(acid)

   0.95M(17mL)=C2(10mL)

        C2=0.95M(17mL)10mL

        C2=1.6M

The pH of the solution is calculated as follows:

   pH=log[H+]

    =log[1.6]

    =0.20

Therefore, the concentration of an acid and pH of the solution is 1.6 M and -0.20 respectively.

(c)

Interpretation Introduction

Interpretation:

The concentration of an acid and the pH of the solution should be predicted.

Concept Introduction:

The full form of pH is the potential of pH. In other words, it explains the strength of the hydrogen ion in a solution. It is given by the formula,

  pH=log[H+]

(c)

Expert Solution
Check Mark

Answer to Problem 6E

The concentration of an acid and pH of the solution is 0.46 M and 0.34 respectively.

Explanation of Solution

Given information:

  • 0.95 M NaOH
  • 12 mL NaOH
  • 25 mL of acid

The concentration of an acid is calculated as follows:

  C1V1(NaOH)=C2V2(acid)

  0.95M(12mL)=C2(25mL)

        C2=0.95M(12mL)25mL

        C2=0.46M

The pH of the solution is calculated as follows:

  pH=log[H+]

    =log[0.46]

    =0.34

Therefore, the concentration of an acid and pH of the solution is 0.46 M and 0.34 respectively.

Chapter U4 Solutions

Living by Chemistry

Ch. U4.69 - Prob. 3ECh. U4.69 - Prob. 4ECh. U4.69 - Prob. 5ECh. U4.70 - Prob. 1TAICh. U4.70 - Prob. 1ECh. U4.70 - Prob. 2ECh. U4.70 - Prob. 3ECh. U4.70 - Prob. 4ECh. U4.70 - Prob. 6ECh. U4.71 - Prob. 1TAICh. U4.71 - Prob. 1ECh. U4.71 - Prob. 2ECh. U4.71 - Prob. 3ECh. U4.71 - Prob. 4ECh. U4.71 - Prob. 5ECh. U4.71 - Prob. 6ECh. U4.71 - Prob. 7ECh. U4.72 - Prob. 1TAICh. U4.72 - Prob. 1ECh. U4.72 - Prob. 2ECh. U4.72 - Prob. 3ECh. U4.72 - Prob. 4ECh. U4.73 - Prob. 1TAICh. U4.73 - Prob. 1ECh. U4.73 - Prob. 2ECh. U4.73 - Prob. 3ECh. U4.73 - Prob. 5ECh. U4.73 - Prob. 6ECh. U4.73 - Prob. 7ECh. U4.74 - Prob. 1TAICh. U4.74 - Prob. 1ECh. U4.74 - Prob. 2ECh. U4.74 - Prob. 4ECh. U4.74 - Prob. 5ECh. U4.75 - Prob. 1TAICh. U4.75 - Prob. 1ECh. U4.75 - Prob. 2ECh. U4.75 - Prob. 4ECh. U4.75 - Prob. 5ECh. U4.75 - Prob. 6ECh. U4.75 - Prob. 7ECh. U4.75 - Prob. 8ECh. U4.75 - Prob. 9ECh. U4.75 - Prob. 10ECh. U4.75 - Prob. 11ECh. U4.76 - Prob. 1TAICh. U4.76 - Prob. 1ECh. U4.76 - Prob. 2ECh. U4.76 - Prob. 3ECh. U4.76 - Prob. 4ECh. U4.76 - Prob. 5ECh. U4.76 - Prob. 6ECh. U4.76 - Prob. 7ECh. U4.76 - Prob. 8ECh. U4.77 - Prob. 1TAICh. U4.77 - Prob. 1ECh. U4.77 - Prob. 2ECh. U4.77 - Prob. 3ECh. U4.77 - Prob. 4ECh. U4.77 - Prob. 5ECh. U4.77 - Prob. 6ECh. U4.77 - Prob. 7ECh. U4.77 - Prob. 8ECh. U4.78 - Prob. 1TAICh. U4.78 - Prob. 1ECh. U4.78 - Prob. 2ECh. U4.78 - Prob. 3ECh. U4.78 - Prob. 4ECh. U4.78 - Prob. 5ECh. U4.78 - Prob. 6ECh. U4.78 - Prob. 7ECh. U4.78 - Prob. 8ECh. U4.79 - Prob. 1TAICh. U4.79 - Prob. 1ECh. U4.79 - Prob. 2ECh. U4.79 - Prob. 3ECh. U4.79 - Prob. 4ECh. U4.80 - Prob. 1TAICh. U4.80 - Prob. 1ECh. U4.80 - Prob. 2ECh. U4.80 - Prob. 3ECh. U4.80 - Prob. 4ECh. U4.80 - Prob. 5ECh. U4.80 - Prob. 6ECh. U4.80 - Prob. 7ECh. U4.80 - Prob. 8ECh. U4.80 - Prob. 9ECh. U4.80 - Prob. 10ECh. U4.81 - Prob. 1TAICh. U4.81 - Prob. 1ECh. U4.81 - Prob. 2ECh. U4.81 - Prob. 3ECh. U4.81 - Prob. 4ECh. U4.81 - Prob. 5ECh. U4.81 - Prob. 6ECh. U4.81 - Prob. 7ECh. U4.81 - Prob. 8ECh. U4.81 - Prob. 9ECh. U4.82 - Prob. 1TAICh. U4.82 - Prob. 1ECh. U4.82 - Prob. 2ECh. U4.82 - Prob. 3ECh. U4.82 - Prob. 4ECh. U4.82 - Prob. 5ECh. U4.82 - Prob. 6ECh. U4.82 - Prob. 7ECh. U4.82 - Prob. 8ECh. U4.83 - Prob. 1TAICh. U4.83 - Prob. 1ECh. U4.83 - Prob. 2ECh. U4.83 - Prob. 3ECh. U4.83 - Prob. 4ECh. U4.83 - Prob. 5ECh. U4.83 - Prob. 6ECh. U4.83 - Prob. 7ECh. U4.84 - Prob. 1TAICh. U4.84 - Prob. 1ECh. U4.84 - Prob. 2ECh. U4.84 - Prob. 3ECh. U4.84 - Prob. 5ECh. U4.84 - Prob. 6ECh. U4.84 - Prob. 7ECh. U4.85 - Prob. 1TAICh. U4.85 - Prob. 1ECh. U4.85 - Prob. 2ECh. U4.85 - Prob. 3ECh. U4.85 - Prob. 4ECh. U4.85 - Prob. 5ECh. U4.85 - Prob. 6ECh. U4.85 - Prob. 7ECh. U4.85 - Prob. 8ECh. U4.86 - Prob. 1TAICh. U4.86 - Prob. 1ECh. U4.86 - Prob. 2ECh. U4.86 - Prob. 3ECh. U4.86 - Prob. 4ECh. U4.86 - Prob. 6ECh. U4.86 - Prob. 7ECh. U4.86 - Prob. 8ECh. U4.87 - Prob. 1TAICh. U4.87 - Prob. 1ECh. U4.87 - Prob. 2ECh. U4.87 - Prob. 3ECh. U4.87 - Prob. 4ECh. U4.87 - Prob. 5ECh. U4.87 - Prob. 6ECh. U4.87 - Prob. 7ECh. U4.87 - Prob. 8ECh. U4.88 - Prob. 1TAICh. U4.88 - Prob. 1ECh. U4.88 - Prob. 2ECh. U4.88 - Prob. 4ECh. U4.88 - Prob. 5ECh. U4.88 - Prob. 6ECh. U4.88 - Prob. 7ECh. U4.88 - Prob. 8ECh. U4.89 - Prob. 1TAICh. U4.89 - Prob. 1ECh. U4.89 - Prob. 2ECh. U4.89 - Prob. 3ECh. U4.89 - Prob. 4ECh. U4.89 - Prob. 5ECh. U4.89 - Prob. 6ECh. U4.90 - Prob. 1ECh. U4.90 - Prob. 2ECh. U4.90 - Prob. 3ECh. U4.90 - Prob. 4ECh. U4.90 - Prob. 5ECh. U4.90 - Prob. 6ECh. U4.90 - Prob. 7ECh. U4.91 - Prob. 1ECh. U4.91 - Prob. 2ECh. U4.91 - Prob. 3ECh. U4.91 - Prob. 5ECh. U4.91 - Prob. 6ECh. U4.92 - Prob. 1TAICh. U4.92 - Prob. 1ECh. U4.92 - Prob. 2ECh. U4.92 - Prob. 3ECh. U4.92 - Prob. 4ECh. U4.93 - Prob. 1TAICh. U4.93 - Prob. 1ECh. U4.93 - Prob. 2ECh. U4.93 - Prob. 4ECh. U4.93 - Prob. 5ECh. U4.93 - Prob. 6ECh. U4 - Prob. C13.3RECh. U4 - Prob. C13.4RECh. U4 - Prob. C14.1RECh. U4 - Prob. C14.2RECh. U4 - Prob. C14.3RECh. U4 - Prob. C14.5RECh. U4 - Prob. C14.6RECh. U4 - Prob. C15.1RECh. U4 - Prob. C15.2RECh. U4 - Prob. C15.3RECh. U4 - Prob. C15.4RECh. U4 - Prob. C15.5RECh. U4 - Prob. C15.6RECh. U4 - Prob. C15.7RECh. U4 - Prob. C15.8RECh. U4 - Prob. C16.1RECh. U4 - Prob. C16.2RECh. U4 - Prob. C16.3RECh. U4 - Prob. C16.4RECh. U4 - Prob. C17.1RECh. U4 - Prob. C17.2RECh. U4 - Prob. C17.3RECh. U4 - Prob. 1RECh. U4 - Prob. 4RECh. U4 - Prob. 5RECh. U4 - Prob. 6RECh. U4 - Prob. 7RECh. U4 - Prob. 8RECh. U4 - Prob. 9RECh. U4 - Prob. 10RECh. U4 - Prob. 11RECh. U4 - Prob. 12RECh. U4 - Prob. 1STPCh. U4 - Prob. 2STPCh. U4 - Prob. 3STPCh. U4 - Prob. 4STPCh. U4 - Prob. 5STPCh. U4 - Prob. 6STPCh. U4 - Prob. 7STPCh. U4 - Prob. 8STPCh. U4 - Prob. 9STPCh. U4 - Prob. 10STPCh. U4 - Prob. 11STPCh. U4 - Prob. 12STPCh. U4 - Prob. 13STPCh. U4 - Prob. 14STPCh. U4 - Prob. 15STPCh. U4 - Prob. 16STPCh. U4 - Prob. 17STPCh. U4 - Prob. 18STPCh. U4 - Prob. 19STPCh. U4 - Prob. 20STP
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