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Chapter 9, Problem 9.84E
Interpretation Introduction

(a)

Interpretation:

The dissociation reactions and Ka expressions for the given weak acids are to be stated.

Concept introduction:

The strength of acids and bases can be determined on the basis of their dissociation reactions and their dissociation constants. Those acids or bases which dissociate almost completely are strong acids or strong bases and those which dissociate to smaller extents are weak or moderately weak acids and weak bases.

Expert Solution
Check Mark

Answer to Problem 9.84E

The dissociation reaction for the given weak acid, HSe is,

HSe(aq)H+(aq)+Se2(aq)

The dissociation constant for the given reaction is,

Ka=[H+][Se2][HSe]

Explanation of Solution

The dissociation reaction for the given weak acid, HSe is,

HSe(aq)H+(aq)+Se2(aq)

The dissociation constant for the given reaction is,

Ka=[H+][Se2][HSe]

The [H+] and [Se2] are the equilibrium concentrations of proton and conjugate base of the given acid respectively and [HSe] represents the equilibrium concentration of acid. For weak acids, the values of the concentrations of proton and conjugate base is smaller than that of the concentration of acid, such that the value of Ka, that is, the dissociation constant is smaller.

Conclusion

The dissociation reaction for the given weak acid, HSe is,

HSe(aq)H+(aq)+Se2(aq)

The dissociation constant for the given reaction is,

Ka=[H+][Se2][HSe]

Interpretation Introduction

(b)

Interpretation:

The dissociation reactions and Ka expressions for the given weak acids are to be stated.

Concept introduction:

The strength of acids and bases can be determined on the basis of their dissociation reactions and their dissociation constants. Those acids or bases which dissociate almost completely are strong acids or strong bases and those which dissociate to smaller extents are weak or moderately weak acids and weak bases.

Expert Solution
Check Mark

Answer to Problem 9.84E

The dissociation reaction for the given weak acid, H2BO3 (1st H only) is,

H2BO3(aq)H+(aq)+HBO32(aq)

The dissociation constant for the given reaction is,

Ka=[H+][HBO32][H2BO3]

Explanation of Solution

The dissociation reaction for the given weak acid, H2BO3 (1st H only) is,

H2BO3(aq)H+(aq)+HBO32(aq)

The dissociation constant for the given reaction is,

Ka=[H+][HBO32][H2BO3]

The [H+] and [HBO32] are the equilibrium concentrations of proton and conjugate base of the given acid respectively and [H2BO3] represents the equilibrium concentration of acid. For weak acids, the values of the concentrations of proton and conjugate base is smaller than that of the concentration of acid, such that the value of Ka, that is, the dissociation constant is smaller.

Conclusion

The dissociation reaction for the given weak acid, H2BO3 (1st H only) is,

H2BO3(aq)H+(aq)+HBO32(aq)

The dissociation constant for the given reaction is,

Ka=[H+][HBO32][H2BO3]

Interpretation Introduction

(c)

Interpretation:

The dissociation reactions and Ka expressions for the given weak acids are to be stated.

Concept introduction:

The strength of acids and bases can be determined on the basis of their dissociation reactions and their dissociation constants. Those acids or bases which dissociate almost completely are strong acids or strong bases and those which dissociate to smaller extents are weak or moderately weak acids and weak bases.

Expert Solution
Check Mark

Answer to Problem 9.84E

The dissociation reaction for the given weak acid, HBO32 is,

HBO32(aq)H+(aq)+BO33(aq)

The dissociation constant for the given reaction is,

Ka=[H+][BO33][HBO32]

Explanation of Solution

The dissociation reaction for the given weak acid, HBO32 is,

HBO32(aq)H+(aq)+BO33(aq)

The dissociation constant for the given reaction is,

Ka=[H+][BO33][HBO32]

The [H+] and [BO33] are the equilibrium concentrations of proton and conjugate base of the given acid respectively and [HBO32] represents the equilibrium concentration of acid. For weak acids, the values of the concentrations of proton and conjugate base is smaller than that of the concentration of acid, such that the value of Ka, that is, the dissociation constant is smaller.

Conclusion

The dissociation reaction for the given weak acid, HBO32 is,

HBO32(aq)H+(aq)+BO33(aq)

The dissociation constant for the given reaction is,

Ka=[H+][BO33][HBO32]

Interpretation Introduction

(d)

Interpretation:

The dissociation reactions and Ka expressions for the given weak acids are to be stated.

Concept introduction:

The strength of acids and bases can be determined on the basis of their dissociation reactions and their dissociation constants. Those acids or bases which dissociate almost completely are strong acids or strong bases and those which dissociate to smaller extents are weak or moderately weak acids and weak bases.

Expert Solution
Check Mark

Answer to Problem 9.84E

The dissociation reaction for the given weak acid, HAsO42 is,

HAsO42(aq)H+(aq)+AsO43(aq)

The dissociation constant for the given reaction is,

Ka=[H+][AsO43][HAsO42]

Explanation of Solution

The dissociation reaction for the given weak acid, HAsO42 is,

HAsO42(aq)H+(aq)+AsO43(aq)

The dissociation constant for the given reaction is,

Ka=[H+][AsO43][HAsO42]

The [H+] and [AsO43] are the equilibrium concentrations of proton and conjugate base of the given acid respectively and [HAsO42] represents the equilibrium concentration of acid. For weak acids, the values of the concentrations of proton and conjugate base is smaller than that of the concentration of acid, such that the value of Ka, that is, the dissociation constant is smaller.

Conclusion

The dissociation reaction for the given weak acid, HAsO42 is,

HAsO42(aq)H+(aq)+AsO43(aq)

The dissociation constant for the given reaction is,

Ka=[H+][AsO43][HAsO42]

Interpretation Introduction

(e)

Interpretation:

The dissociation reactions and Ka expressions for the given weak acids are to be stated.

Concept introduction:

The strength of acids and bases can be determined on the basis of their dissociation reactions and their dissociation constants. Those acids or bases which dissociate almost completely are strong acids or strong bases and those which dissociate to smaller extents are weak or moderately weak acids and weak bases.

Expert Solution
Check Mark

Answer to Problem 9.84E

The dissociation reaction for the given weak acid, HClO is,

HClO(aq)H+(aq)+ClO(aq)

The dissociation constant for the given reaction is,

Ka=[H+][ClO][HClO]

Explanation of Solution

The dissociation reaction for the given weak acid, HClO is,

HClO(aq)H+(aq)+ClO(aq)

The dissociation constant for the given reaction is,

Ka=[H+][ClO][HClO]

The [H+] and [ClO] are the equilibrium concentrations of proton and conjugate base of the given acid respectively and [HClO] represents the equilibrium concentration of acid. For weak acids, the values of the concentrations of proton and conjugate base is smaller than that of the concentration of acid, such that the value of Ka, that is, the dissociation constant is smaller.

Conclusion

The dissociation reaction for the given weak acid, HClO is,

HClO(aq)H+(aq)+ClO(aq)

The dissociation constant for the given reaction is,

Ka=[H+][ClO][HClO]

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Chapter 9 Solutions

Bundle: Chemistry for Today: General, Organic, and Biochemistry, Loose-Leaf Version, 9th + LMS Integrated OWLv2, 4 terms (24 months) Printed Access Card

Ch. 9 - Write a formula for the conjugate base formed when...Ch. 9 - Write a formula for the conjugate base formed when...Ch. 9 - Prob. 9.13ECh. 9 - Prob. 9.14ECh. 9 - The following reactions illustrate Brnsted...Ch. 9 - Prob. 9.16ECh. 9 - Write equations to illustrate the acid-base...Ch. 9 - Prob. 9.18ECh. 9 - Prob. 9.19ECh. 9 - Prob. 9.20ECh. 9 - Prob. 9.21ECh. 9 - Prob. 9.22ECh. 9 - The acid H3C6H5O7 forms the citrate ion, C6H5O73,...Ch. 9 - The acid H2C4H4O4 forms the succinate ion,...Ch. 9 - Prob. 9.25ECh. 9 - Prob. 9.26ECh. 9 - Calculate the molar concentration of OH in water...Ch. 9 - Calculate the molar concentration of OH in water...Ch. 9 - Calculate the molar concentration of H3O+ in water...Ch. 9 - Prob. 9.30ECh. 9 - Classify the solutions represented in Exercises...Ch. 9 - Classify the solutions represented in Exercises...Ch. 9 - Prob. 9.33ECh. 9 - Prob. 9.34ECh. 9 - Determine the pH of water solutions with the...Ch. 9 - Prob. 9.36ECh. 9 - Prob. 9.37ECh. 9 - Determine the pH of water solutions with the...Ch. 9 - Determine the [H+] value for solutions with the...Ch. 9 - Determine the [H+] value for solutions with the...Ch. 9 - Prob. 9.41ECh. 9 - Prob. 9.42ECh. 9 - The pH values listed in Table 9.1 are generally...Ch. 9 - Prob. 9.44ECh. 9 - Prob. 9.45ECh. 9 - Prob. 9.46ECh. 9 - Prob. 9.47ECh. 9 - Using the information in Table 9.4, describe how...Ch. 9 - Write balanced molecular equations to illustrate...Ch. 9 - Write balanced molecular equations to illustrate...Ch. 9 - Prob. 9.51ECh. 9 - Prob. 9.52ECh. 9 - Prob. 9.53ECh. 9 - Prob. 9.54ECh. 9 - Write balanced molecular, total ionic, and net...Ch. 9 - Prob. 9.56ECh. 9 - Prob. 9.57ECh. 9 - Prob. 9.58ECh. 9 - Prob. 9.59ECh. 9 - Prob. 9.60ECh. 9 - Prob. 9.61ECh. 9 - Prob. 9.62ECh. 9 - Prob. 9.63ECh. 9 - Prob. 9.64ECh. 9 - Prob. 9.65ECh. 9 - Prob. 9.66ECh. 9 - Prob. 9.67ECh. 9 - Prob. 9.68ECh. 9 - Prob. 9.69ECh. 9 - Prob. 9.70ECh. 9 - Determine the number of moles of each of the...Ch. 9 - Prob. 9.72ECh. 9 - Prob. 9.73ECh. 9 - Determine the number of equivalents and...Ch. 9 - Determine the number of equivalents and...Ch. 9 - Prob. 9.76ECh. 9 - Prob. 9.77ECh. 9 - Prob. 9.78ECh. 9 - Prob. 9.79ECh. 9 - The Ka values have been determined for four acids...Ch. 9 - Prob. 9.81ECh. 9 - Prob. 9.82ECh. 9 - Prob. 9.83ECh. 9 - Prob. 9.84ECh. 9 - Prob. 9.85ECh. 9 - Prob. 9.86ECh. 9 - Arsenic acid (H3AsO4) is a moderately weak...Ch. 9 - Explain the purpose of doing a titration.Ch. 9 - Prob. 9.89ECh. 9 - Prob. 9.90ECh. 9 - Prob. 9.91ECh. 9 - Prob. 9.92ECh. 9 - Prob. 9.93ECh. 9 - Prob. 9.94ECh. 9 - Prob. 9.95ECh. 9 - Prob. 9.96ECh. 9 - A 25.00-mL sample of gastric juice is titrated...Ch. 9 - A 25.00-mL sample of H2C2O4 solution required...Ch. 9 - Prob. 9.99ECh. 9 - Prob. 9.100ECh. 9 - The following acid solutions were titrated to the...Ch. 9 - The following acid solutions were titrated to the...Ch. 9 - Prob. 9.103ECh. 9 - Prob. 9.104ECh. 9 - Prob. 9.105ECh. 9 - Prob. 9.106ECh. 9 - Prob. 9.107ECh. 9 - Predict the relative pH greater than 7, less than...Ch. 9 - Prob. 9.109ECh. 9 - Explain why the hydrolysis of salts makes it...Ch. 9 - How would the pH values of equal molar solutions...Ch. 9 - Write equations similar to Equations 9.48 and 9.49...Ch. 9 - Prob. 9.113ECh. 9 - Prob. 9.114ECh. 9 - Prob. 9.115ECh. 9 - a.Calculate the pH of a buffer that is 0.1M in...Ch. 9 - Which of the following acids and its conjugate...Ch. 9 - Prob. 9.118ECh. 9 - Prob. 9.119ECh. 9 - What ratio concentrations of NaH2PO4 and Na2HPO4...Ch. 9 - Prob. 9.121ECh. 9 - Prob. 9.122ECh. 9 - Prob. 9.123ECh. 9 - Prob. 9.124ECh. 9 - Prob. 9.125ECh. 9 - Prob. 9.126ECh. 9 - Prob. 9.127ECh. 9 - Prob. 9.128ECh. 9 - Prob. 9.129ECh. 9 - Bottles of ketchup are routinely left on the...Ch. 9 - Prob. 9.131ECh. 9 - Prob. 9.132ECh. 9 - Prob. 9.133ECh. 9 - Prob. 9.134ECh. 9 - Prob. 9.135ECh. 9 - Prob. 9.136ECh. 9 - Prob. 9.137ECh. 9 - A base is a substance that dissociates in water...Ch. 9 - Prob. 9.139ECh. 9 - Prob. 9.140ECh. 9 - What is the formula of the hydronium ion? a.H+...Ch. 9 - Which of the following substances has a pH closest...Ch. 9 - Dissolving H2SO4 in water creates an acid solution...Ch. 9 - Prob. 9.144ECh. 9 - A common detergent has a pH of 11.0, so the...Ch. 9 - Prob. 9.146ECh. 9 - The pH of a blood sample is 7.40 at room...Ch. 9 - Prob. 9.148ECh. 9 - Prob. 9.149ECh. 9 - Prob. 9.150ECh. 9 - Prob. 9.151ECh. 9 - Which of the following compounds would be...Ch. 9 - A substance that functions to prevent rapid,...Ch. 9 - Which one of the following equations represents...Ch. 9 - Which reaction below demonstrates a neutralization...Ch. 9 - In titration of 40.0mL of 0.20MNaOH with 0.4MHCl,...Ch. 9 - When titrating 50mL of 0.2MHCl, what quantity of...
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