ORGANIC CHEMISTRY SAPLING ACCESS + ETEX
ORGANIC CHEMISTRY SAPLING ACCESS + ETEX
6th Edition
ISBN: 9781319306977
Author: LOUDON
Publisher: INTER MAC
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Chapter 20, Problem 20.35AP
Interpretation Introduction

Interpretation:

The validation corresponding to the fact that 0.1M solution of acetic acid (pKa=4.76) turns litmus red, and that a 0.1M solution of phenol (pKa=9.95) does not is to be stated.

Concept introduction:

The pH of a solution is represented as the negative logarithm of the concentration of proton.

pH=log[H+]

This scale is defined from 1 to 14. It helps to determine whether the given solution is acidic, basic or neutral. The solutions with pH<7 are acidic in nature, with pH=7 are neutral in nature and with pH>7 are basic in nature.

Expert Solution & Answer
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Answer to Problem 20.35AP

The value of pH for 0.1M solution of acetic acid is below 3 that is 2.89 and the value of pH for 0.1M solution of phenol is above 3 that is 5.48. So, 0.1M solution of acetic acid (pKa=4.76) turns litmus red, and that a 0.1M solution of phenol (pKa=9.95) does not.

Explanation of Solution

The given pKa value 0.1M solution of acetic acid is 4.76.

So, the value of Ka for 0.1M solution of acetic acid is 104.76=0.0000174.

The ionization table for acetic acid for initial and final concentration is given below.

 CH3CO2HCH3CO2+H+Initialconcentration0.100Final concentration0.1xxx

The expression to calculate the value of Ka is given below.

Ka=[H+][CH3CO2][CH3CO2H]

Substitute the values of respective concentrations and Ka in the above expression.

0.0000174=[x][x][0.1x]0=x2+0.0000174x0.00000174x=0.0013

The value of pH solution is represented as follows.

pH=log[H+] … (1)

Substitute the value of H+ as 0.0013 for 0.1M solution of acetic acid in equation (1).

pH=log[0.0013]=2.89

Therefore, the value of pH for 0.1M solution of acetic acid is 2.89 that is below the pH value 3 and turns the litmus paper red.

The given pKa value 0.1M solution of phenol is 9.95.

So, the value of Ka for 0.1M solution of phenol is 109.95=1.1×1010.

The ionization table for phenol for initial and final concentration is given below.

 PhOHH++PhOInitialconcentration0.100Final concentration0.1xxx

The expression to calculate the value of Ka is given below.

Ka=[H+][PhO][H+]

Substitute the values of respective concentrations and Ka in the above expression.

1.1×1010=[x][x][0.1x]0=x2+1.1×1010x1.1×1011x=3.3×106

Substitute the value of H+ as 3.3×106 for 0.1M solution of phenol in equation (1).

pH=log[3.3×106]=5.48

Therefore, the value of pH for 0.1M solution of phenol is 5.48 that is above the pH value 3 and does not turns the litmus paper red.

Conclusion

The 0.1M solution of acetic acid (pKa=4.76) turns litmus red, and that a 0.1M solution of phenol (pKa=9.95) does not because acetic acid has its pH value below 3 whereas, phenol has its pH value above 3.

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