Chemistry: The Science in Context (Fifth Edition)
Chemistry: The Science in Context (Fifth Edition)
5th Edition
ISBN: 9780393614046
Author: Thomas R. Gilbert, Rein V. Kirss, Natalie Foster, Stacey Lowery Bretz, Geoffrey Davies
Publisher: W. W. Norton & Company
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Chapter 16, Problem 16.35QP

(a)

Interpretation Introduction

Interpretation: The pH of the given buffer solutions at the given conditions is to be calculated.

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

pH=pKa+log[ConjugateBase][Acid]

To determine: The pH of the given buffer solution consisting of HNO2 and NaNO2 .

(a)

Expert Solution
Check Mark

Answer to Problem 16.35QP

Solution

The pH of the given buffer solution consisting of HNO2 and NaNO2 is 3.49_ .

Explanation of Solution

Explanation

Given

The concentration of HNO2 solution is 0.120M .

The concentration of NaNO2 solution is 0.150M .

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

pH=pKa+log[NaNO2][HNO2]

Where,

  • pH is the hydrogen content of solution.
  • pKa is the dissociation constant of acid.
  • [NaNO2] is the concentration of sodium nitrate solution.
  • [HNO2] is the concentration of nitrous acid.

Substitute the values of [NaNO2] , [HNO2] and pKa in the above equation.

pH=3.39+log0.1500.120pH=3.49_

(b)

Interpretation Introduction

To determine: The pH of the solution after addition of 1.00mL of given hydrochloric acid to the given buffer solution.

(b)

Expert Solution
Check Mark

Answer to Problem 16.35QP

Solution

The pH of the buffer after addition of hydrochloric acid solution is 3.41_ .

Explanation of Solution

Explanation

Given

The volume of hydrochloric acid in the solution is 1.00mL(=0.001L)

The concentration of hydrochloric acid solution is 11.6M .

The concentration of HNO2 solution is 0.120M .

The concentration of NaNO2 solution is 0.150M .

The number of moles of hydrochloric acid present in the given solution is calculated by using the formula,

M=nVn=MV

Where,

  • M is the molarity of the solution.
  • n is the number of moles of solute.
  • V is the volume of solution in liters.

Substitute the values of M and V in above equation.

n=11.6M×0.001L=0.0116mol

The addition of acid to the given buffer solution increases the concentration of acid by the given number of moles and reduces the concentration of salt equal to the added moles of acid. Therefore, the concentration of HNO2 and NaNO2 after addition of hydrochloric acid is calculated as,

For concentration of NaNO2 ,

ConcentrationofNaNO2=(0.1500.0116)M=0.1384M

For concentration of HNO2 ,

ConcentrationofHNO2=(0.120+0.0116)M=0.1316M

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

pH=pKa+log[NaNO2][HNO2]

Where,

  • pH is the hydrogen content of solution.
  • pKa is the dissociation constant of acid.
  • [NaNO2] is the concentration of sodium nitrate solution.
  • [HNO2] is the concentration of nitrous acid.

Substitute the values of [NaNO2] , [HNO2] and pKa in the above equation.

pH=3.39+log0.13840.1316pH=3.41_

Conclusion

  1. a. The pH of the given buffer solution consisting of HNO2 and NaNO2 is 3.49_ .
  2. b. The pH of the buffer after addition of hydrochloric acid solution is 3.41_

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Chapter 16 Solutions

Chemistry: The Science in Context (Fifth Edition)

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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