CHEM:ATOM FOC 2E CL (TEXT)
CHEM:ATOM FOC 2E CL (TEXT)
2nd Edition
ISBN: 9780393284218
Author: Stacey Lowery Bretz, Natalie Foster, Thomas R. Gilbert, Rein V. Kirss
Publisher: WW Norton & Co
Question
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Chapter 16, Problem 16.45QA
Interpretation Introduction

To sketch:

A titration curve for the titration of 50.0 mL of 0.200 MHNO2 with 1.00 MNaOH. Find the pH of the sample after the addition of the titrant?

a) 2.50 mL

b) 5.00 mL

c) 7.50 mL

d) 10.00 mL

Expert Solution & Answer
Check Mark

Answer to Problem 16.45QA

Solution:

a) pH of solution after addition of 2.50 mL of 1.00 MNaOH is 2.92.

b) pH of solution after addition of 5.00 mL of 1.00 MNaOH is 3.40.

c) pH of solution after addition of 7.50 mL of 1.00 MNaOH is 3.88.

d) pH of solution after addition of 10.00 mL of 1.00 MNaOH is 8.31.

Explanation of Solution

1) Concept:

The given titration is the titration of a weak acid and a strong base. The volume of the sample of the acid and its molarity is given. We can find the initial moles of the acid present using these values. When HNO2 reacts with N aOH, it is an acid base reaction to form NaNO2. Calculating moles of NaOH at each addition and the moles of the acid remaining in the solution, we can find pH ofthe solution for each step. The acid base reaction between HNO2 and NaOH is written as:

HNO2aq+OH-aqNO2-aq+H2O(l)

The initial moles of HNO2 and NaOH, [HNO2] and [NO2-aq] can be determined in the final solution and pH can be found by using the Henderson–Hasselbalch equation:

pH=pKa+logBase[Acid]

Where, pKa=-logKa

The Ka value for HNO2 can be calculated using the relation between Ka and Kb

Ka×Kb=Kw(where,Kw=1.0×10-14)

The relation for pH calculation is

pH=-log[H3O+]

2) Formula:

i) pH=pKa+logBase[Acid]

ii) Molarity=molesofsolutevolumeinL

iii) Ka×Kb=Kw

iv) pH=-log[H3O+]

v) pKa=-logKa

3) Given:

i) Molarity of HNO2=0.200 M

ii) Volume of HNO2=50.0 mL=0.050L

iii) Molarity of NaOH= 1.00 M

iv) Ka=4.0×10-4 (Appendix 5 Table A5.1)

4) Calculation:

Calculating the initial moles of HNO2

MolofHNO2=0.200 molL×0.050 L=0.010 mol

a) Calculating the pH of solution when  2.50 mL 0.0025 Lof 1.00 MNaOH is added

Calculating the moles of NaOH

MolesofNaOH=1.00 molL×0.00250 L=0.00250 mol

Set up the RICE table to determine the reacted moles of HNO2 and how many moles of CH3COO- are formed

Reaction HNO2(aq) + OH-aq NO2-aq + H2O(l)
HNO2 (mol) OH-mol NO2-(mol)
Initial 0.010 0.00250 0
Change -0.00250 -0.00250 +0.00250
Final 0.00750 0 0.00250

Totalvolume = 0.050 L+0.00250 L=0.0525 L

Calculating the concentration of HNO2 and NO2-

HNO2=0.00750 mol0.0525 L=0.142857 M

NO2-=0.00250 mol0.0525 L=0.047619 M

pKa=-log(4.0×10-4)=3.40

Plug in the values in the Henderson–Hasselbalch equation,

pH=3.40+log0.047619[0.142857]

pH=3.40+log(0.33333)

pH=3.40-0.4771

pH=2.92

b) Calculating the pH of solution when  5.0 mL 0.005 Lof 1.00 MNaOH is added

Calculating the moles of NaOH

MolofNaOH=1.00 molL×0.0050 L=0.0050 mol

Set up the RICE table to determine the reacted moles of HNO2 and how many moles of NO2- are formed,

Reaction HNO2(aq) + OH-aq NO2-aq + H2O(l)
HNO2 (mol) OH-mol NO2-(mol)
Initial 0.010 0.0050 0
Change -0.0050 -0.0050 +0.0050
Final 0.0050 0 0.0050

Totalvolume = 0.050 L+0.0050 L=0.055 L

Calculating the concentration of HNO2 and NO2-

HNO2=NO2-=0.0050 mol0.055 L=0.09091 M

Plug in the values in the Henderson–Hasselbalch equation,

pH=3.40+log0.09091 [0.09091 ]

pH=3.40+log(1)

pH=3.40

c) Calculating the pH of solution when  7.50 mL 0.0075 Lof 1.00 MNaOH is added:

Calculating the moles of NaOH

MolesofNaOH=1.00 molL×0.00750 L=0.00750 mol

Set up the RICE table to determine the reacted moles of HNO2 and how many moles of NO2- are formed,

Reaction HNO2(aq) + OH-aq NO2-aq + H2O(l)
HNO2 (moles) OH-moles NO2-(moles)
Initial 0.010 0.00750 0
Change -0.00750 -0.00750 +0.00750
Final 0.00250 0 0.00750

Totalvolume = 0.050 L+0.00750 L=0.0575 L

Calculating the concentration of HNO2 and NO2-

HNO2=0.00250 mol0.0575 L=0.043478 M

NO2-=0.00750 mol0.0575 L= 0.1304349 M

pKa=-log(4.0×10-4)=3.40

Plug in the values in the Henderson–Hasselbalch equation,

pH=3.40+log0.1304349[0.043478]

pH=3.40+log(3.000)

pH=3.40+0.4771

pH=3.88

d) Calculating the pH of solution when  10 mL 0.010 Lof 1.00 MNaOH is added

Calculating the moles of NaOH

Mol NaOH=1.00 molL×0.0100 L=0.010 mol

Set up the RICE table to determine the reacted and remaining moles of HNO2 and moles of NO2- that are produced:

Reaction HNO2(aq) + OH-aq NO2-aq + H2O(l)
HNO2 (mol) OH-mol NO2-(mol)
Initial 0.010 0.010 0
Change -0.010 -0.010 +0.010
Final 0 0 0.010

Since the acid and added base have reacted completely, and an equivalent amount of conjugate base has formed, this represents the equivalence point.

Totalvolume = 0.050 L+0.010 L=0.060 L

Calculating the concentration of NO2- formed

NO2-=0.010 mol0.060 L=0.166667 M

Set up RICE table to calculate the [OH-]

Reaction NO2-aq + H2O(l) HNO2(aq) + OH-aq
Initial 0.166667 M 0 0
Change -x +x +x
Equilibrium (0.166667 x) x x

Kb=HNO2[OH-][NO2-]

Calculating Kb

Kb=KwKa

Kb=1.0×10-144.0×10-4

Kb=2.50×10-11

2.50×10-11=x2(0.166667 x)

2.50×10-11=x20.166667 

x=2.0412×10-6M=[OH-]

H3O+=1.0×10-142.0412×10-6

H3O+=4.8991×10-9 M

Calculating the pH

pH=-log[H3O+]

pH=-log(4.8991×10-9)

pH=8.31

A titration curve for the titration of 50.0 mL of 0.200 MHNO2 vs.mLof 1.00 MNaOH

CHEM:ATOM FOC 2E CL (TEXT), Chapter 16, Problem 16.45QA

Conclusion:

In the titration of a weak acid and a strong base, as you increase the volume of the strong base, the pH of the solution increases.

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Chapter 16 Solutions

CHEM:ATOM FOC 2E CL (TEXT)

Ch. 16 - Prob. 16.11QACh. 16 - Prob. 16.12QACh. 16 - Prob. 16.13QACh. 16 - Prob. 16.14QACh. 16 - Prob. 16.15QACh. 16 - Prob. 16.16QACh. 16 - Prob. 16.17QACh. 16 - Prob. 16.18QACh. 16 - Prob. 16.19QACh. 16 - Prob. 16.20QACh. 16 - Prob. 16.21QACh. 16 - Prob. 16.22QACh. 16 - Prob. 16.23QACh. 16 - Prob. 16.24QACh. 16 - Prob. 16.25QACh. 16 - Prob. 16.26QACh. 16 - Prob. 16.27QACh. 16 - Prob. 16.28QACh. 16 - Prob. 16.29QACh. 16 - Prob. 16.30QACh. 16 - Prob. 16.31QACh. 16 - Prob. 16.32QACh. 16 - Prob. 16.33QACh. 16 - Prob. 16.34QACh. 16 - Prob. 16.35QACh. 16 - Prob. 16.36QACh. 16 - Prob. 16.37QACh. 16 - Prob. 16.38QACh. 16 - Prob. 16.39QACh. 16 - Prob. 16.40QACh. 16 - Prob. 16.41QACh. 16 - Prob. 16.42QACh. 16 - Prob. 16.43QACh. 16 - Prob. 16.44QACh. 16 - Prob. 16.45QACh. 16 - Prob. 16.46QACh. 16 - Prob. 16.47QACh. 16 - Prob. 16.48QACh. 16 - Prob. 16.49QACh. 16 - Prob. 16.50QACh. 16 - Prob. 16.51QACh. 16 - Prob. 16.52QACh. 16 - Prob. 16.53QACh. 16 - Prob. 16.54QACh. 16 - Prob. 16.55QACh. 16 - Prob. 16.56QACh. 16 - Prob. 16.57QACh. 16 - Prob. 16.58QACh. 16 - Prob. 16.59QACh. 16 - Prob. 16.60QACh. 16 - Prob. 16.61QACh. 16 - Prob. 16.62QACh. 16 - Prob. 16.63QACh. 16 - Prob. 16.64QACh. 16 - Prob. 16.65QACh. 16 - Prob. 16.66QACh. 16 - Prob. 16.67QACh. 16 - Prob. 16.68QACh. 16 - Prob. 16.69QACh. 16 - Prob. 16.70QACh. 16 - Prob. 16.71QACh. 16 - Prob. 16.72QACh. 16 - Prob. 16.73QACh. 16 - Prob. 16.74QACh. 16 - Prob. 16.75QACh. 16 - Prob. 16.76QACh. 16 - Prob. 16.77QACh. 16 - Prob. 16.78QACh. 16 - Prob. 16.79QACh. 16 - Prob. 16.80QACh. 16 - Prob. 16.81QACh. 16 - Prob. 16.82QACh. 16 - Prob. 16.83QACh. 16 - Prob. 16.84QACh. 16 - Prob. 16.85QACh. 16 - Prob. 16.86QACh. 16 - Prob. 16.87QACh. 16 - Prob. 16.88QACh. 16 - Prob. 16.89QACh. 16 - Prob. 16.90QACh. 16 - Prob. 16.91QACh. 16 - Prob. 16.92QACh. 16 - Prob. 16.93QACh. 16 - Prob. 16.94QACh. 16 - Prob. 16.95QACh. 16 - Prob. 16.96QACh. 16 - Prob. 16.97QACh. 16 - Prob. 16.98QACh. 16 - Prob. 16.99QACh. 16 - Prob. 16.100QACh. 16 - Prob. 16.101QACh. 16 - Prob. 16.102QACh. 16 - Prob. 16.103QACh. 16 - Prob. 16.104QACh. 16 - Prob. 16.105QACh. 16 - Prob. 16.106QACh. 16 - Prob. 16.107QACh. 16 - Prob. 16.108QACh. 16 - Prob. 16.109QACh. 16 - Prob. 16.110QACh. 16 - Prob. 16.111QACh. 16 - Prob. 16.112QACh. 16 - Prob. 16.113QACh. 16 - Prob. 16.114QACh. 16 - Prob. 16.115QACh. 16 - Prob. 16.116QACh. 16 - Prob. 16.117QACh. 16 - Prob. 16.118QACh. 16 - Prob. 16.119QACh. 16 - Prob. 16.120QACh. 16 - Prob. 16.121QACh. 16 - Prob. 16.122QACh. 16 - Prob. 16.123QACh. 16 - Prob. 16.124QACh. 16 - Prob. 16.125QACh. 16 - Prob. 16.126QACh. 16 - Prob. 16.127QACh. 16 - Prob. 16.128QACh. 16 - Prob. 16.129QACh. 16 - Prob. 16.130QACh. 16 - Prob. 16.131QACh. 16 - Prob. 16.132QACh. 16 - Prob. 16.133QACh. 16 - Prob. 16.134QACh. 16 - Prob. 16.135QACh. 16 - Prob. 16.136QA
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