Chemistry Principles And Practice
Chemistry Principles And Practice
3rd Edition
ISBN: 9781305295803
Author: David Reger; Scott Ball; Daniel Goode
Publisher: Cengage Learning
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Chapter 16, Problem 16.23QE
Interpretation Introduction

Interpretation:

The pH for the titration of 1.00 mL of 0.240 M Ba(OH)2 with 0.200 M HNO3 after addition of 0, 0.50, 1.00, 2.40, and 3.00 mL HNO3 has to be calculated and also the titration curve has to be sketched.

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Explanation of Solution

Given,

    Titration of 1.00 mL of 0.240 M Ba(OH)2 with 0.200 M HNO3 after addition of 0, 0.50, 1.00, 2.40, and 3.00 mL HNO3

Addition of 0 mL HNO3:

Given system is titration of 1.00 mL of 0.240 M Ba(OH)2 with 0.200 M HNO3 after addition of 0 mL HNO3

The concentration of Ba(OH)2 is 2×0.24= 0.48 M,

    pOH = - log (OH) pOH = - log (0.48)pOH = 0.32pH=pH +pOH=14pH +0.32=14pH =13.68

Addition of 0.50 mL HNO3:

Given system is titration of 1.00 mL of 0.240 M Ba(OH)2 with 0.200 M HNO3 after addition of 0.50 mL HNO3.

The given titration is shown below,

    2HNO3 (aq) + Ba(OH)2 (aq)  Ba(NO3)2 (aq)   +  2H2O (l) 

Moles of HNO3 is calculated as follows,

  Moles of HNO3 = 5×104 Lof HNO3×0.200molHNO31LofHNO3Moles of HNO3 = 1×104moles

Moles of Ba(OH)2 is calculated as follows,

  Moles of Ba(OH)2 = 1×10-3 Lof Ba(OH)2×0.24molBa(OH)21LofBa(OH)2Moles of Ba(OH)2 = 2.4×10-4moles

Limiting reagent:

According to the balanced chemical equation, for 1 mole of HNO3 needed 0.5 mol of Ba(OH)2. So for 1×104moles of HNO3, needed (1×104moles / 2) = 5×105moles  of Ba(OH)2.

But there is excess amount of Ba(OH)2. Thus, HNO3 is the limiting reagent. The limiting reagent will be fully used after the completion of the reaction.

ICE table:

  2HNO3 (aq) + Ba(OH)2 (aq)  Ba(NO3)2 (aq)   +  2H2O (l) 

S (mol)1×104moles2.4×10-4moles0Excess
R (mol) -1×104moles-1×104moles1×104moles1×104moles
F(mol)01.4×104moles1×104molesExcess

The total volume of the given solution is 1 ml + 0.50 ml = 1.50 ml (or) 1.5×10-3L

One mole of Ba(OH)2 gives two moles of hydroxide (OH) ion, therefore,

1.4×104moles of Ba(OH)2 gives 2×1.4×10-4moles = 2.8×10-4moles of hydroxide (OH) ion.

  Concentration of base = MolesofexcessbaseTotal volumeConcentration of base = 2.8×10-4moles1.5×103LConcentration of base =0.19M

The concentration of Ba(OH)2 of 0.19 M,

    pOH = - log (OH) pOH = - log (0.19)pOH = 0.72pH=pH +pOH=14pH +0.72=14pH =13.28

Addition of 1.0 mL HNO3:

Given system is titration of 1.00 mL of 0.240 M Ba(OH)2 with 0.200 M HNO3 after addition of 1.0 mL HNO3.

The given titration is shown below,

    2HNO3 (aq) + Ba(OH)2 (aq)  Ba(NO3)2 (aq)   +  2H2O (l) 

Moles of HNO3 is calculated as follows,

  Moles of HNO3 = 1×103 Lof HNO3×0.200molHNO31LofHNO3Moles of HNO3 = 2×104moles

Moles of Ba(OH)2 is calculated as follows,

  Moles of Ba(OH)2 = 1×10-3 Lof Ba(OH)2×0.24molBa(OH)21LofBa(OH)2Moles of Ba(OH)2 = 2.4×10-4moles

Limiting reagent:

According to the balanced chemical equation, for 1 mole of HNO3 needed 0.5 mol of Ba(OH)2. So for 2×104moles of HNO3, needed (2×104moles / 2) = 1×104moles  of Ba(OH)2.

But there is excess amount of Ba(OH)2. Thus, HNO3 is the limiting reagent. The limiting reagent will be fully used after the completion of the reaction.

ICE table:

  2HNO3 (aq) + Ba(OH)2 (aq)  Ba(NO3)2 (aq)   +  2H2O (l) 

S (mol)2×104moles2.4×10-4moles0Excess
R (mol) -2×104moles-2×104moles2×104moles2×104moles
F(mol)00.4×104moles2×104molesExcess

The total volume of the given solution is 1 ml + 1 ml = 2 ml (or) 2×10-3L

One mole of Ba(OH)2 gives two moles of hydroxide (OH) ion, therefore,

0.4×104moles of Ba(OH)2 gives 2×0.4×10-4moles = 8×10-5moles of hydroxide (OH) ion.

  Concentration of base = MolesofexcessbaseTotal volumeConcentration of base = 8×10-5moles2×103LConcentration of base =0.04M

The concentration of Ba(OH)2 of 0.04 M,

    pOH = - log (OH) pOH = - log (0.04)pOH = 1.40pH=pH +pOH=14pH +1.40=14pH =12.60

Addition of 2.4 mL HNO3:

Given system is titration of 1.00 mL of 0.240 M Ba(OH)2 with 0.200 M HNO3 after addition of 2.4 mL HNO3.

The given titration is shown below,

    2HNO3 (aq) + Ba(OH)2 (aq)  Ba(NO3)2 (aq)   +  2H2O (l) 

Moles of HNO3 is calculated as follows,

  Moles of HNO3 = 2.4×103 Lof HNO3×0.200molHNO31LofHNO3Moles of HNO3 = 2.4×103moles

Moles of Ba(OH)2 is calculated as follows,

  Moles of Ba(OH)2 = 1×10-3 Lof Ba(OH)2×0.24molBa(OH)21LofBa(OH)2Moles of Ba(OH)2 = 2.4×10-4moles

ICE table:

  2HNO3 (aq) + Ba(OH)2 (aq)  Ba(NO3)2 (aq)   +  2H2O (l) 

S (mol)2.4×10-4moles2.4×10-4moles0Excess
R (mol) -2.4×10-4moles-2.4×10-4moles2.4×10-4moles2.4×10-4moles
F(mol)002.4×10-4molesExcess

The total volume of the given solution is 1 ml + 2.4 ml = 3.4 ml (or) 3.4×10-3L

The resulting solution is neutral. Therefore,

    Kw= 1.0×10-14

Hence,

    [H3O+] = [OH] = 1.0×10-7 M

The concentration of HCl 0.08 M,

    pH = - log (H3O+) pH = 7

Addition of 3.0 mL HNO3:

Given system is titration of 1.00 mL of 0.240 M Ba(OH)2 with 0.200 M HNO3 after addition of 3.0 mL HNO3.

The given titration is shown below,

    2HNO3 (aq) + Ba(OH)2 (aq)  Ba(NO3)2 (aq)   +  2H2O (l) 

Moles of HNO3 is calculated as follows,

  Moles of HNO3 = 3×103 Lof HNO3×0.200molHNO31LofHNO3Moles of HNO3 = 6×104moles

Moles of Ba(OH)2 is calculated as follows,

  Moles of Ba(OH)2 = 1×10-3 Lof Ba(OH)2×0.24molBa(OH)21LofBa(OH)2Moles of Ba(OH)2 = 2.4×10-4moles

Limiting reagent:

According to the balanced chemical equation, for 1 mole of HNO3 needed 0.5 mol of Ba(OH)2. So for 6×104moles of HNO3, needed (6×104moles / 2) = 3×104moles  of Ba(OH)2.

But there is less amount of Ba(OH)2. Thus, Ba(OH)2 is the limiting reagent. The limiting reagent will be fully used after the completion of the reaction.

ICE table:

  2HNO3 (aq) + Ba(OH)2 (aq)  Ba(NO3)2 (aq)   +  2H2O (l) 

S (mol)6×104moles2.4×10-4moles0Excess
R (mol) -2.4×10-4moles-2.4×10-4moles2.4×10-4moles2.4×10-4moles
F(mol)3.6×10-4moles02.4×10-4molesExcess

The total volume of the given solution is 1 ml + 3 ml = 4 ml (or) 4×10-3L

    Concentration of acid = MolesofexcessacidTotal volumeConcentration of acid = 3.6×104M4×10-3LConcentration of acid = 0.09M

The concentration of HNO3 of 0.09 M,

    pH = - log (H+) pH = - log (0.09)pH = 1.05

The titration curve is shown below,

Chemistry Principles And Practice, Chapter 16, Problem 16.23QE

Figure 1

The curve in Question 16.21 has the same general form, but the equivalence point occurs at half the volume the equivalence point occurs in Question 16.23.

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Chapter 16 Solutions

Chemistry Principles And Practice

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