EBK GENERAL CHEMISTRY: THE ESSENTIAL CO
EBK GENERAL CHEMISTRY: THE ESSENTIAL CO
7th Edition
ISBN: 8220106637203
Author: Chang
Publisher: YUZU
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Chapter 16, Problem 16.46QP

(a)

Interpretation Introduction

Interpretation:

The percent ionization of the given 0.60M hydrofluoric 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 16.46QP

The percent ionization of given 0.60M hydrofluoric acid solution is 3.5%

Explanation of Solution

From the given concentrations of hydrofluoric 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

HF(aq)H(aq)++F(aq)Initial(M)0.6000Change(M)x+x+xEquilibrium(M)(0.60x)xx

The Ka of hydrofluoric acid is 7.1×104

  Ka=[H+][F-][HF]7.1×10-4=x2(0.60-x)

Assuming that x is small compared to 0.60 , we neglect it in the denominator:

  7.1×10-4=x2(0.60)x=(0.60)×7.1×10-4=0.021M[H+]=0.021M

The percent ionization can be calculated as follows,

  percentionization=[H+][HA]×100%=[H+][HF]×100%=0.021M0.60M×100%=3.5%

Therefore, the percent ionization of given 0.60M hydrofluoric acid solution is 3.5%

(b)

Interpretation Introduction

Interpretation:

The percent ionization of the given 0.080M  hydrofluoric 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%

(b)

Expert Solution
Check Mark

Answer to Problem 16.46QP

The percent ionization of given 0.080M hydrofluoric acid solution is 3.5%

Explanation of Solution

From the given concentrations of hydrofluoric 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

HF(aq)H(aq)++F(aq)Initial(M)0.08000Change(M)x+x+xEquilibrium(M)(0.080x)xx

The Ka of hydrofluoric acid is 7.1×104

  Ka=[H+][F-][HF]7.1×10-4=x2(0.080-x)

Assuming that x is small compared to 0.60, we neglect it in the denominator:

  7.1×10-4=x2(0.080)x2=0.0000568x=0.0000568[H+]=0.0075M

The percent ionization can be calculated as follows,

  percentionization=[H+][HA]×100%=[H+][HF]×100%=0.0075M0.080M×100%=9.4%

Therefore, the percent ionization of given 0.080M hydrofluoric acid solution is 9.4%

(c)

Interpretation Introduction

Interpretation:

The percent ionization of the given 0.0046M  hydrofluoric 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%

(c)

Expert Solution
Check Mark

Answer to Problem 16.46QP

The percent ionization of given 0.0046M hydrofluoric acid solution is 33%

Explanation of Solution

From the given concentrations of hydrofluoric 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

HF(aq)H(aq)++F(aq)Initial(M)0.004600Change(M)x+x+xEquilibrium(M)(0.0046x)xx

The Ka of hydrofluoric acid is 7.1×104

  Ka=[H+][F-][HF]7.1×10-4=x2(0.0046M-x)

Solving the above quadratic equation,

  x2+(7.1×10-4)x-(3.3×10-6)=0x=1.5×10-3M[H+]=1.5×10-3M

The percent ionization can be calculated as follows,

  percentionization=[H+][HA]×100%=[H+][HF]×100%=1.5×10-3M0.0046M×100%=33%

Therefore, the percent ionization of given 0.0046M hydrofluoric acid solution is 33%

(d)

Interpretation Introduction

Interpretation:

The percent ionization of the given 0.00028M  hydrofluoric 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%

(d)

Expert Solution
Check Mark

Answer to Problem 16.46QP

The percent ionization of given 0.00028M hydrofluoric acid solution is 79%

Explanation of Solution

From the given concentrations of hydrofluoric 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

HF(aq)H(aq)++F(aq)Initial(M)0.0002800Change(M)x+x+xEquilibrium(M)(0.00028x)xx

The Ka of hydrofluoric acid is 7.1×104

  Ka=[H+][F-][HF]7.1×10-4=x2(0.00028M-x)

Solving the above quadratic equation,

  x2+(7.1×10-4)x-(2.0×10-7)=0x=2.2×10-4M[H+]=2.2×10-4M

The percent ionization can be calculated as follows,

  percentionization=[H+][HA]×100%=[H+][HF]×100%=2.2×10-40.00028M×100%=79%

The percent ionization of given 0.00028M hydrofluoric acid solution is 79%

As the solution becomes more dilute, the percent ionization increases.

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Chapter 16 Solutions

EBK GENERAL CHEMISTRY: THE ESSENTIAL CO

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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