Smartwork5 Printed Access Card for Use with Chemistry: The Science in Context 5th Edition (SmartWork Access Printed Access Card)
Smartwork5 Printed Access Card for Use with Chemistry: The Science in Context 5th Edition (SmartWork Access Printed Access Card)
5th Edition
ISBN: 9780393615296
Author: Rein V. Kirss (Author), Natalie Foster (Author), Geoffrey Davies (Author) Thomas R. Gilbert (Author)
Publisher: W. W. Norton
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Chapter 16, Problem 16.43QP
Interpretation Introduction

Interpretation: The pH of the solution of acetic acid and sodium hydroxide at different volume of base is to be calculated.

Concept introduction: The pH of the solution is calculated by using the formula,

14pH=log[OH]

To determine: The pH of the solution of acetic acid and sodium hydroxide at different volume of base.

Expert Solution & Answer
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Answer to Problem 16.43QP

Solution

The pH of the solution at 10.0mL volume of base is 4.762_ .

The pH of the solution at 20.0mL volume of base is 8.75_ .

The pH of the solution at 30.0mL volume of base is 12.36_ .

Explanation of Solution

Explanation

Given

The volume of acetic acid is 25.0mL .

The concentration of acetic acid is 0.100M .

The concentration of sodium hydroxide is 0.125M .

The volume of sodium hydroxide added is 10.0mL,20.0mL and 30.0mL .

At 10.0mL volume of sodium hydroxide:

The concentration of sodium acetate formed after addition of sodium hydroxide is calculated by multiplying the ratio of sodium hydroxide present in the solution with the concentration of sodium hydroxide solution.

ConcentrationofCH3COONa=10.010.0+25.0×0.125M=0.036M

The concentration of acetic acid left in the solution after formation of sodium acetate is calculated by subtracting the concentration of sodium acetate from the volume of acetic acid in the solution.

ConcentrationofCH3COOH(insolution)=25.025.0+10.0×0.100M0.036M=0.035M

The pH of the given buffer solution is calculated by using the formula,

pH=pKa+log[CH3COONa][CH3COOH]

Where,

  • pH is the hydrogen content of solution.
  • pKa is the dissociation constant of acid.
  • [CH3COONa] is the concentration of sodium acetate solution.
  • [CH3COOH] is the concentration of acetic acid.

Substitute the values of [CH3COONa] , [CH3COOH] and pKa in the above equation.

pH=4.75+log0.0360.035pH=4.762_

At 20.0mL volume of sodium hydroxide:

At 20.0mL volume of sodium hydroxide the whole acid present in the solution gets neutralized and the solution will reach at the equivalence point. The equivalent concentration of sodium acetate salt in the solution is calculated as,

ConcentrationofCH3COONa=20.020.0+25.0×0.125M=0.056M

The concentration of hydroxyl ions in the given solution is calculated by using the dissociation constant of sodium acetate by using the following equation.

CH3COO(aq)+H2O(l)CH3COOH(aq)+OH(aq)

The concentration of acetic acid and hydroxyl ions formed in the given equilibrium reaction is considered to be x . The dissociation constant of the sodium acetate is calculated by using the formula,

Kb=[OH][CH3COOH][CH3COO]

Where,

  • Kb is the dissociation constant of the acid.
  • [OH] is the hydroxyl ion concentration.
  • [CH3COOH] is the concentration of acetic acid.
  • [CH3COO] is the concentration of acetate ion.

Substitute the values of Kb , [OH] , [CH3COOH] and [CH3COO] in the above equation.

5.55×1010=(x)(x)(0.056x)x=5.6×106

Therefore, the concentration of hydroxyl ions in the solution is 5.6×106 .

The pH of the solution is calculated by using the formula,

14pH=log[OH]

Where,

  • [OH] is the concentration of hydroxyl ions in the solution.

Substitute the values of [OH] in the above equation.

14pH=log(5.6×106)pH=8.75_

At 30.0mL volume of sodium hydroxide:

At 30.0mL volume of sodium hydroxide the volume of sodium hydroxide is present in excess in the solution. Therefore, the pH of the solution corresponds to the excess base present in the solution. The concentration of excess base present in the solution is calculated as,

ConcentrationofNaOH(excess)=10.030.0+25.0×0.125M=0.023M

The pH of the solution is calculated by using the formula,

14pH=log[OH]

Where,

  • [OH] is the concentration of hydroxyl ions in the solution.

Substitute the values of [OH] in the above equation.

14pH=log(0.023)pH=12.36_

Conclusion

The pH of the solution at 10.0mL volume of base is 4.762_ .

The pH of the solution at 20.0mL volume of base is 8.75_ .

The pH of the solution at 30.0mL volume of base is 12.36_

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Chapter 16 Solutions

Smartwork5 Printed Access Card for Use with Chemistry: The Science in Context 5th Edition (SmartWork Access Printed Access Card)

Ch. 16.4 - Prob. 11PECh. 16.4 - Prob. 12PECh. 16.5 - Prob. 13PECh. 16.6 - Prob. 14PECh. 16.8 - Prob. 15PECh. 16.8 - Prob. 16PECh. 16.8 - Prob. 17PECh. 16.8 - Prob. 18PECh. 16.8 - Prob. 19PECh. 16 - Prob. 16.1VPCh. 16 - Prob. 16.2VPCh. 16 - Prob. 16.3VPCh. 16 - Prob. 16.4VPCh. 16 - Prob. 16.5VPCh. 16 - Prob. 16.6VPCh. 16 - Prob. 16.7VPCh. 16 - Prob. 16.8VPCh. 16 - Prob. 16.9VPCh. 16 - Prob. 16.10VPCh. 16 - Prob. 16.11QPCh. 16 - Prob. 16.12QPCh. 16 - Prob. 16.13QPCh. 16 - Prob. 16.14QPCh. 16 - Prob. 16.15QPCh. 16 - Prob. 16.16QPCh. 16 - Prob. 16.17QPCh. 16 - Prob. 16.18QPCh. 16 - Prob. 16.19QPCh. 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 - Prob. 16.40QPCh. 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 - Prob. 16.47QPCh. 16 - Prob. 16.48QPCh. 16 - Prob. 16.49QPCh. 16 - Prob. 16.50QPCh. 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 - Prob. 16.57QPCh. 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 - Prob. 16.93QPCh. 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 - Prob. 16.100QPCh. 16 - Prob. 16.101QPCh. 16 - Prob. 16.102QPCh. 16 - Prob. 16.103QPCh. 16 - Prob. 16.104QPCh. 16 - Prob. 16.105QPCh. 16 - Prob. 16.106QPCh. 16 - Prob. 16.107QPCh. 16 - Prob. 16.108QPCh. 16 - Prob. 16.109QPCh. 16 - Prob. 16.110QPCh. 16 - Prob. 16.111APCh. 16 - Prob. 16.112APCh. 16 - Prob. 16.113APCh. 16 - Prob. 16.114APCh. 16 - Prob. 16.115APCh. 16 - Prob. 16.116APCh. 16 - Prob. 16.117APCh. 16 - Prob. 16.118AP
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