General Chemistry
General Chemistry
7th Edition
ISBN: 9780073402758
Author: Chang, Raymond/ Goldsby
Publisher: McGraw-Hill College
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Chapter 16, Problem 16.48QP

(a)

Interpretation Introduction

Interpretation:

The percent ionization of the given 0.20M monoprotic acetylsalicylic acid solution has to be calculated.

Concept Information:

Acid ionization constant Ka:

Acids ionize in water.  Strong acids ionize completely whereas weak acids ionize to some limited extent.

The degree to which a weak acid ionizes depends on the concentration of the acid and the equilibrium constant for the ionization.

The ionization of a weak acid HA can be given as follows,

  HA(aq)H+(aq)+A-(aq)

The equilibrium expression for the above reaction is given below.

  Ka=[H+][A-][HA]

Where,

  Ka is acid ionization constant,

  [H+]  is concentration of hydrogen ion

  [A-]  is concentration of acid anion

  [HA] is concentration of the acid

Percent ionization:

A quantitative measure of the degree of ionization is percent ionization.

For a weak, monoprotic acid HA percent ionization can be calculated as follows,

percentionization=[H3O+][HA]×100%

(a)

Expert Solution
Check Mark

Answer to Problem 16.48QP

The percent ionization of given 0.20M acetylsalicylic acid solution is 3.9%

Explanation of Solution

From the given concentrations of acetylsalicylic acid solution and its Ka value, the hydrogen ion concentration can be found out.

Then the percent ionization can be calculated from the obtained hydrogen ion concentration as follows,

Construct an equilibrium table and express the equilibrium concentration of each species in terms of x

C9H8O4(aq)H(aq)++C9H8O4(aq)Initial(M)0.2000Change(M)x+x+xEquilibrium(M)(0.20x)xx

The Ka of acetylsalicylic acid is 3.0×104

  Ka=[H+][C9H7O4][C9H7O4]3.0×104=x2(0.20-x)

Assuming that x is small compared to 0.20 , we neglect it in the denominator:

  3.0×104=x2(0.20)x=(0.20)×3.0×104=7.7×103M[H+]=7.7×103M

The percent ionization can be calculated as follows,

  percentionization=[H+][HA]×100%=[H+][C9H8O4]×100%=7.7×103M0.20M×100%=3.9%

Therefore, the percent ionization of given 0.20M acetylsalicylic acid solution is 3.9%

(b)

Interpretation Introduction

Interpretation:

The percent ionization of acetylsalicylic acid solution under the given condition has to be calculated.

Concept Information:

Acid ionization constant Ka:

Acids ionize in water.  Strong acids ionize completely whereas weak acids ionize to some limited extent.

The degree to which a weak acid ionizes depends on the concentration of the acid and the equilibrium constant for the ionization.

The ionization of a weak acid HA can be given as follows,

  HA(aq)H+(aq)+A-(aq)

The equilibrium expression for the above reaction is given below.

  Ka=[H+][A-][HA]

Where,

  Ka is acid ionization constant,

  [H+]  is concentration of hydrogen ion

  [A-]  is concentration of acid anion

  [HA] is concentration of the acid

Percent ionization:

A quantitative measure of the degree of ionization is percent ionization.

For a weak, monoprotic acid HA percent ionization can be calculated as follows,

  percentionization=[H3O+][HA]×100%

pH of a solution is the negative of the base -10 logarithm of the hydronium ion or hydrogen ion concentration.

  pH=-log[H+]

  ConcentrationofH+=10-pH

(b)

Expert Solution
Check Mark

Answer to Problem 16.48QP

The percent ionization of acetylsalicylic acid solution under given condition is 0.30%

Explanation of Solution

Given,

The pH of gastric juice in the stomach of a certain individual is 1.00.after a few aspirin tablet have been swallowed, the concentration of aspirin in the stomach is 0.20M .

The concentration of H+ at pH 1 can be determined as,

  ConcentrationofH+=10-pH=10-1=0.10M

According to the Lechatelier’s principle, the position of equilibrium is shifted in the direction of the unionized acid.

Constructing an equilibrium table using the known equilibrium concentration of each species as follows,

C9H8O4(aq)H(aq)++C9H8O4(aq)Initial(M)0.200.100Change(M)x+x+xEquilibrium(M)(0.20x)0.10+xx

The Ka of acetylsalicylic acid is 3.0×104

  Ka=[H+][C9H7O4][C9H7O4]3.0×104=x×(0.10+x)(0.20-x)

Assuming that (0.20x)0.20and(0.10+x)0.10

  3.0×104=x×0.100.20x=0.20×(3.0×104)0.10=6.0×104M[H+]=6.0×104M

The percent ionization can be calculated as follows,

  percentionization=[H+][HA]×100%=[H+][C9H7O4]×100%=6.0×1040.20M×100%=0.30%

The percent ionization of acetylsalicylic acid solution under given condition is 0.30%

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Chapter 16 Solutions

General Chemistry

Ch. 16.4 - Prob. 1RCCh. 16.5 - Prob. 1PECh. 16.5 - Prob. 2PECh. 16.5 - Prob. 1RCCh. 16.5 - Prob. 3PECh. 16.5 - Prob. 2RCCh. 16.6 - Prob. 1PECh. 16.6 - Prob. 1RCCh. 16.7 - Prob. 1RCCh. 16.8 - Prob. 1PECh. 16.8 - Rank the following acids from strongest to...Ch. 16.9 - Prob. 1PECh. 16.9 - Practice Exercise Predict whether the following...Ch. 16.9 - Prob. 1RCCh. 16.10 - Prob. 1RCCh. 16.11 - Prob. 1PECh. 16.11 - Prob. 1RCCh. 16 - Prob. 16.1QPCh. 16 - Prob. 16.2QPCh. 16 - Prob. 16.3QPCh. 16 - Prob. 16.4QPCh. 16 - Prob. 16.5QPCh. 16 - Prob. 16.6QPCh. 16 - Prob. 16.7QPCh. 16 - Prob. 16.8QPCh. 16 - Prob. 16.9QPCh. 16 - Prob. 16.10QPCh. 16 - Prob. 16.12QPCh. 16 - Prob. 16.13QPCh. 16 - Prob. 16.14QPCh. 16 - 16.15 Calculate the hydrogen ion concentration for...Ch. 16 - 16.16 Calculate the hydrogen ion concentration in...Ch. 16 - Prob. 16.17QPCh. 16 - Prob. 16.18QPCh. 16 - 16.19 Complete this table for a...Ch. 16 - Prob. 16.20QPCh. 16 - Prob. 16.21QPCh. 16 - Prob. 16.22QPCh. 16 - Prob. 16.23QPCh. 16 - Prob. 16.24QPCh. 16 - Prob. 16.25QPCh. 16 - Prob. 16.26QPCh. 16 - Prob. 16.27QPCh. 16 - Prob. 16.28QPCh. 16 - Prob. 16.29QPCh. 16 - Prob. 16.30QPCh. 16 - Prob. 16.31QPCh. 16 - Prob. 16.32QPCh. 16 - Prob. 16.33QPCh. 16 - Prob. 16.34QPCh. 16 - Prob. 16.35QPCh. 16 - Prob. 16.36QPCh. 16 - Prob. 16.37QPCh. 16 - Prob. 16.38QPCh. 16 - Prob. 16.39QPCh. 16 - 16.40 Which of the following solutions has the...Ch. 16 - Prob. 16.41QPCh. 16 - Prob. 16.42QPCh. 16 - Prob. 16.43QPCh. 16 - Prob. 16.44QPCh. 16 - Prob. 16.45QPCh. 16 - Prob. 16.46QPCh. 16 - 16.47 A 0.040 M solution of a monoprotic acid is...Ch. 16 - Prob. 16.48QPCh. 16 - Prob. 16.49QPCh. 16 - 16.50 Write all the species (except water) that...Ch. 16 - Prob. 16.51QPCh. 16 - Prob. 16.52QPCh. 16 - Prob. 16.53QPCh. 16 - Prob. 16.54QPCh. 16 - Prob. 16.55QPCh. 16 - Prob. 16.56QPCh. 16 - 16.57 What is the original molarity of a solution...Ch. 16 - Prob. 16.58QPCh. 16 - Prob. 16.59QPCh. 16 - Prob. 16.60QPCh. 16 - Prob. 16.61QPCh. 16 - Prob. 16.62QPCh. 16 - Prob. 16.63QPCh. 16 - Prob. 16.64QPCh. 16 - Prob. 16.65QPCh. 16 - Prob. 16.66QPCh. 16 - Prob. 16.67QPCh. 16 - Prob. 16.68QPCh. 16 - Prob. 16.69QPCh. 16 - Prob. 16.70QPCh. 16 - Prob. 16.71QPCh. 16 - Prob. 16.72QPCh. 16 - Prob. 16.73QPCh. 16 - Prob. 16.74QPCh. 16 - Prob. 16.75QPCh. 16 - Prob. 16.76QPCh. 16 - Prob. 16.77QPCh. 16 - Prob. 16.78QPCh. 16 - Prob. 16.79QPCh. 16 - Prob. 16.80QPCh. 16 - Prob. 16.81QPCh. 16 - Prob. 16.82QPCh. 16 - Prob. 16.83QPCh. 16 - Prob. 16.84QPCh. 16 - Prob. 16.85QPCh. 16 - Prob. 16.86QPCh. 16 - Prob. 16.87QPCh. 16 - Prob. 16.88QPCh. 16 - Prob. 16.89QPCh. 16 - Prob. 16.90QPCh. 16 - Prob. 16.91QPCh. 16 - Prob. 16.92QPCh. 16 - 16.93 Most of the hydrides of Group 1A and Group...Ch. 16 - Prob. 16.94QPCh. 16 - Prob. 16.95QPCh. 16 - Prob. 16.96QPCh. 16 - Prob. 16.97QPCh. 16 - Prob. 16.98QPCh. 16 - Prob. 16.99QPCh. 16 - 16.100 Hydrocyanic acid (HCN) is a weak acid and a...Ch. 16 - Prob. 16.101QPCh. 16 - Prob. 16.102QPCh. 16 - Prob. 16.103QPCh. 16 - Prob. 16.104QPCh. 16 - 16.105 You are given two beakers containing...Ch. 16 - Prob. 16.106QPCh. 16 - Prob. 16.107QPCh. 16 - Prob. 16.108QPCh. 16 - Prob. 16.109QPCh. 16 - Prob. 16.110QPCh. 16 - Prob. 16.111QPCh. 16 - Prob. 16.112QPCh. 16 - Prob. 16.113QPCh. 16 - Prob. 16.114QPCh. 16 - Prob. 16.115QPCh. 16 - Prob. 16.116QPCh. 16 - Prob. 16.117QPCh. 16 - Prob. 16.118QPCh. 16 - Prob. 16.119QPCh. 16 - Prob. 16.120QPCh. 16 - Prob. 16.121SPCh. 16 - Prob. 16.122SPCh. 16 - Prob. 16.123SPCh. 16 - Prob. 16.124SPCh. 16 - Prob. 16.125SPCh. 16 - Prob. 16.126SPCh. 16 - Prob. 16.127SPCh. 16 - Prob. 16.128SPCh. 16 - Prob. 16.129SPCh. 16 - 16.130 Use the data in Appendix 2 to calculate the...
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