FUND.OF GEN CHEM CHAP 1-13 W/ACCESS
FUND.OF GEN CHEM CHAP 1-13 W/ACCESS
16th Edition
ISBN: 9781323406038
Author: McMurry
Publisher: PEARSON C
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Chapter 10, Problem 10.105CP

(a)

Interpretation Introduction

Interpretation:

The given combination should be identified that whether it produce an effective buffer solution and assuming equal concentration of each acid and its conjugate base pH of that buffer solution should be identified.

Concept Introduction:

pH: pH is the logarithm of the reciprocal of the concentration of H3O+ in a solution. The pH of a solution is a measure of concentration of hydrogen ion in a solution. Lower pH due to more hydrogen ions and higher pH due to less concentration of hydrogen ions.

pH=pKa+log[conjugate base][acid] is Henderson-Hasselbalch equation.

Henderson-Hasselbalch equation

pKa=pH+log[HA][A-]

Henderson-Hasselbalch equation explains the relationship between pH of solution and pKa of acid. For a dissociation of acid (HA) in aqueous solution,

HA+H2OH3O++A

pKa=pH+log[HA][A]

During a dissociation of acid in aqueous solution,

  • If pH=pKa, the concentration of compound in its acidic and basic form is equal.
  • If pH<pKa, the compound exist in its acidic form.
  • If pH>pKa, the compound exist in its basic form.

Buffer solutions: Buffer solution is defined as a solution that oppose changes in pH while adding little amount of either an acid or a base. Buffer solution is a combination of a weak acid and its conjugate base or a weak base and its conjugate acid.

Ka: Acid dissociation constant Ka represents how strong the acid is in a solution.

pKa=log[Ka] It is negative logarithm of acid dissociation constant Ka.

(b)

Interpretation Introduction

Interpretation:

The given combination should be identified that whether it produce an effective buffer solution and assuming equal concentration of each acid and its conjugate base pH of that buffer solution should be identified.

Concept Introduction:

pH: pH is the logarithm of the reciprocal of the concentration of H3O+ in a solution. The pH of a solution is a measure of concentration of hydrogen ion in a solution. Lower pH due to more hydrogen ions and higher pH due to less concentration of hydrogen ions.

pH=pKa+log[conjugate base][acid] is Henderson-Hasselbalch equation.

Henderson-Hasselbalch equation

pKa=pH+log[HA][A-]

Henderson-Hasselbalch equation explains the relationship between pH of solution and pKa of acid. For a dissociation of acid (HA) in aqueous solution,

HA+H2OH3O++A

pKa=pH+log[HA][A]

During a dissociation of acid in aqueous solution,

  • If pH=pKa, the concentration of compound in its acidic and basic form is equal.
  • If pH<pKa, the compound exist in its acidic form.
  • If pH>pKa, the compound exist in its basic form.

Buffer solutions: Buffer solution is defined as a solution that oppose changes in pH while adding little amount of either an acid or a base. Buffer solution is a combination of a weak acid and its conjugate base or a weak base and its conjugate acid.

Ka: Acid dissociation constant Ka represents how strong the acid is in a solution.

pKa=log[Ka] It is negative logarithm of acid dissociation constant Ka.

(c)

Interpretation Introduction

Interpretation:

The given combination should be identified that whether it produce an effective buffer solution and assuming equal concentration of each acid and its conjugate base pH of that buffer solution should be identified.

Concept Introduction:

pH: pH is the logarithm of the reciprocal of the concentration of H3O+ in a solution. The pH of a solution is a measure of concentration of hydrogen ion in a solution. Lower pH due to more hydrogen ions and higher pH due to less concentration of hydrogen ions.

pH=pKa+log[conjugate base][acid] is Henderson-Hasselbalch equation.

Henderson-Hasselbalch equation

pKa=pH+log[HA][A-]

Henderson-Hasselbalch equation explains the relationship between pH of solution and pKa of acid. For a dissociation of acid (HA) in aqueous solution,

HA+H2OH3O++A

pKa=pH+log[HA][A]

During a dissociation of acid in aqueous solution,

  • If pH=pKa, the concentration of compound in its acidic and basic form is equal.
  • If pH<pKa, the compound exist in its acidic form.
  • If pH>pKa, the compound exist in its basic form.

Buffer solutions: Buffer solution is defined as a solution that oppose changes in pH while adding little amount of either an acid or a base. Buffer solution is a combination of a weak acid and its conjugate base or a weak base and its conjugate acid.

Ka: Acid dissociation constant Ka represents how strong the acid is in a solution.

pKa=log[Ka] It is negative logarithm of acid dissociation constant Ka.

(d)

Interpretation Introduction

Interpretation:

The given combination should be identified that whether it produce an effective buffer solution and assuming equal concentration of each acid and its conjugate base pH of that buffer solution should be identified.

Concept Introduction:

pH: pH is the logarithm of the reciprocal of the concentration of H3O+ in a solution. The pH of a solution is a measure of concentration of hydrogen ion in a solution. Lower pH due to more hydrogen ions and higher pH due to less concentration of hydrogen ions.

pH=pKa+log[conjugate base][acid] is Henderson-Hasselbalch equation.

Henderson-Hasselbalch equation

pKa=pH+log[HA][A-]

Henderson-Hasselbalch equation explains the relationship between pH of solution and pKa of acid. For a dissociation of acid (HA) in aqueous solution,

HA+H2OH3O++A

pKa=pH+log[HA][A]

During a dissociation of acid in aqueous solution,

  • If pH=pKa, the concentration of compound in its acidic and basic form is equal.
  • If pH<pKa, the compound exist in its acidic form.
  • If pH>pKa, the compound exist in its basic form.

Buffer solutions: Buffer solution is defined as a solution that oppose changes in pH while adding little amount of either an acid or a base. Buffer solution is a combination of a weak acid and its conjugate base or a weak base and its conjugate acid.

Ka: Acid dissociation constant Ka represents how strong the acid is in a solution.

pKa=log[Ka] It is negative logarithm of acid dissociation constant Ka.

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Chapter 10 Solutions

FUND.OF GEN CHEM CHAP 1-13 W/ACCESS

Ch. 10.2 - Prob. 10.8PCh. 10.2 - Prob. 10.9KCPCh. 10.3 - Prob. 10.10PCh. 10.4 - Prob. 10.11PCh. 10.5 - Prob. 10.12PCh. 10.5 - Prob. 10.13PCh. 10.5 - Prob. 10.14PCh. 10.6 - Identify the following solutions as acidic or...Ch. 10.6 - Calculate the pH of the following solutions and...Ch. 10.6 - What is the pH of a 0.0025 M solution of HCl?Ch. 10.6 - Prob. 10.4CIAPCh. 10.6 - Prob. 10.5CIAPCh. 10.7 - How many equivalents are in the following? (a) 5.0...Ch. 10.7 - Prob. 10.19PCh. 10.8 - Maalox, an over-the-counter antacid, contains...Ch. 10.8 - Prob. 10.21PCh. 10.8 - Prob. 10.22PCh. 10.8 - Show how ethylamine (C2H5NH2) reacts with...Ch. 10.9 - Predict whether the following salts produce an...Ch. 10.10 - What is the pH of 1.00 L of the 0.100 M...Ch. 10.10 - Prob. 10.26PCh. 10.10 - Prob. 10.27PCh. 10.10 - A buffer solution is prepared using CN-(from NaCN...Ch. 10.11 - A titration is carried out to determine the...Ch. 10.11 - Prob. 10.30PCh. 10.11 - Prob. 10.31PCh. 10.11 - Prob. 10.32PCh. 10.11 - Prob. 10.6CIAPCh. 10.11 - Prob. 10.7CIAPCh. 10 - Prob. 10.33UKCCh. 10 - Prob. 10.34UKCCh. 10 - The following pictures represent aqueous acid...Ch. 10 - Prob. 10.36UKCCh. 10 - Prob. 10.37UKCCh. 10 - Prob. 10.38APCh. 10 - What happens when a weak acid such as CH3CO2H is...Ch. 10 - What happens when a strong base such as KOH solved...Ch. 10 - Prob. 10.41APCh. 10 - Prob. 10.42APCh. 10 - Prob. 10.43APCh. 10 - Prob. 10.44APCh. 10 - Prob. 10.45APCh. 10 - Prob. 10.46APCh. 10 - Label the BrnstedLowry acids and bases in the...Ch. 10 - Write the formulas of the conjugate acids of the...Ch. 10 - Write the formulas of the conjugate bases of the...Ch. 10 - Prob. 10.50APCh. 10 - Prob. 10.51APCh. 10 - Prob. 10.52APCh. 10 - Prob. 10.53APCh. 10 - Prob. 10.54APCh. 10 - Write the expressions for the acid dissociation...Ch. 10 - Based on the Ka values in Table 10.3, rank the...Ch. 10 - Prob. 10.57APCh. 10 - A 0.10 M solution of the deadly poison hydrogen...Ch. 10 - Prob. 10.59APCh. 10 - Prob. 10.60APCh. 10 - What is the approximate pH of a 0.02 M solution of...Ch. 10 - Calculate the pOH of each solution in Problems...Ch. 10 - Prob. 10.63APCh. 10 - What are the OH concentration and pOH for each...Ch. 10 - What are the H3O+ and OH concentrations of...Ch. 10 - Prob. 10.66APCh. 10 - Prob. 10.67APCh. 10 - Write balanced equations for proton-transfer...Ch. 10 - Sodium bicarbonate (NaHCO3), also known as baking...Ch. 10 - Refer to Section 10.8 to write balanced equations...Ch. 10 - Prob. 10.71APCh. 10 - For each of the following salts, indicate if the...Ch. 10 - Which salt solutions in problem 10.72 could be...Ch. 10 - Prob. 10.74APCh. 10 - Prob. 10.75APCh. 10 - Prob. 10.76APCh. 10 - Which of the following buffer systems would you...Ch. 10 - What is the pH of a buffer system that contains...Ch. 10 - Consider 1.00 L of the buffer system described in...Ch. 10 - Prob. 10.80APCh. 10 - Prob. 10.81APCh. 10 - Prob. 10.82APCh. 10 - How does normality compare to molarity for...Ch. 10 - Prob. 10.84APCh. 10 - Prob. 10.85APCh. 10 - Prob. 10.86APCh. 10 - Prob. 10.87APCh. 10 - Prob. 10.88APCh. 10 - Prob. 10.89APCh. 10 - Prob. 10.90APCh. 10 - Prob. 10.91APCh. 10 - Titration of a 12.0 mL solution of HCl requires...Ch. 10 - Prob. 10.93APCh. 10 - Titration of a 10.0 mL solution of NH3 requires...Ch. 10 - If 35.0 mL of a 0.100 N acid solution is needed to...Ch. 10 - For the titrations discussed in Problems 10.92 and...Ch. 10 - Prob. 10.97APCh. 10 - Prob. 10.98CPCh. 10 - Prob. 10.99CPCh. 10 - Prob. 10.100CPCh. 10 - Prob. 10.101CPCh. 10 - Prob. 10.102CPCh. 10 - Prob. 10.103CPCh. 10 - Prob. 10.104CPCh. 10 - Prob. 10.105CPCh. 10 - Prob. 10.106CPCh. 10 - Prob. 10.107CPCh. 10 - Prob. 10.108CPCh. 10 - Obtain a package of Alka-Seltzer, an antacid, from...Ch. 10 - Prob. 10.110GPCh. 10 - Prob. 10.111GP
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