Volume of 0 .108 M H 2 SO 4 solution needed to neutralize 15 .0 μ L solution of 0 .010 M Ca ( OH ) 2 has to be calculated. Concept Introduction: The dilution formula is given as follows: n 1 M 1 V 1 = n 2 M 2 V 2 Here, M 1 denotes the molarity of the solution before dilution. V 1 denotes the volume of the solution before dilution. M 2 denotes the molarity of solution after dilution. V 2 denotes the volume of the solution after dilution. n 1 and n 2 are n factors of acid and base respectively.

Question

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Answer 1

Question

Chapter 12, Problem 12.31P

(a)

Interpretation Introduction

Interpretation:

Volume of 0.108 M H2SO4 solution needed to neutralize 15.0 μL solution of 0.010 M Ca(OH)2 has to be calculated.

Concept Introduction:

The dilution formula is given as follows:

n1M1V1=n2M2V2

Here,

M1 denotes the molarity of the solution before dilution.

V1 denotes the volume of the solution before dilution.

M2 denotes the molarity of solution after dilution.

V2 denotes the volume of the solution after dilution.

n1 and n2 are n factors of acid and base respectively.

(a)

Expert Solution

Answer to Problem 12.31P

Volume of 0.108 M H2SO4 solution needed to neutralize 15.0 μL solution of 0.010 M Ca(OH)2 is 0.1851×10−6 L.

Explanation of Solution

The conversion factor to convert liters to is μL as follows:

1 μL=10−6 L

Hence convert 15 μL to L as follows:

Volume=(15 μL)(10−6 L1 μL)=15×10−6 L

The dilution formula is given as follows:

n1M1V1=n2M2V2 (1)

Substitute 1 for n1, 2 for n2, 0.108 M for M1, 0.010 M for M2, 15×10−6 L for V2 in equation (1).

(1)(0.108 M)(V1)=(2)(0.010 M)(15×10−6 L) (2)

Rearrange equation (2) to calculate the value of V1.

V1=(2)(0.010 M)(15×10−6 L)(1)(0.108 M)=0.1851×10−6 L

(b)

Interpretation Introduction

Interpretation:

Volume of 0.300 M H2SO4 solution needed to neutralize 25.0 mL solution of 0.200 M NaOH has to be calculated.

Concept Introduction:

Refer to part (a).

(b)

Expert Solution

Answer to Problem 12.31P

Volume of 0.300 M H2SO4 solution needed to neutralize 25.0 mL solution of 0.200 M NaOH is 8.333 mL.

Explanation of Solution

The dilution formula is given as follows:

n1M1V1=n2M2V2 (1)

Substitute 2 for n1, 1 for n2, 0.300 M for M1, 0.200 M for M2, 25.0 mL for V2 in equation (1).

(2)(0.300 M)(V1)=(1)(0.200 M)(25.0 mL) (3)

Rearrange equation (3) to calculate the value of V1.

V1=(0.200 M)(25.0 mL)(2)0.300 M=8.333 mL

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Answer 2

Question

Chapter 12, Problem 12.31P

(a)

Interpretation Introduction

Interpretation:

Volume of 0.108 M H2SO4 solution needed to neutralize 15.0 μL solution of 0.010 M Ca(OH)2 has to be calculated.

Concept Introduction:

The dilution formula is given as follows:

n1M1V1=n2M2V2

Here,

M1 denotes the molarity of the solution before dilution.

V1 denotes the volume of the solution before dilution.

M2 denotes the molarity of solution after dilution.

V2 denotes the volume of the solution after dilution.

n1 and n2 are n factors of acid and base respectively.

(a)

Expert Solution

Answer to Problem 12.31P

Volume of 0.108 M H2SO4 solution needed to neutralize 15.0 μL solution of 0.010 M Ca(OH)2 is 0.1851×10−6 L.

Explanation of Solution

The conversion factor to convert liters to is μL as follows:

1 μL=10−6 L

Hence convert 15 μL to L as follows:

Volume=(15 μL)(10−6 L1 μL)=15×10−6 L

The dilution formula is given as follows:

n1M1V1=n2M2V2 (1)

Substitute 1 for n1, 2 for n2, 0.108 M for M1, 0.010 M for M2, 15×10−6 L for V2 in equation (1).

(1)(0.108 M)(V1)=(2)(0.010 M)(15×10−6 L) (2)

Rearrange equation (2) to calculate the value of V1.

V1=(2)(0.010 M)(15×10−6 L)(1)(0.108 M)=0.1851×10−6 L

(b)

Interpretation Introduction

Interpretation:

Volume of 0.300 M H2SO4 solution needed to neutralize 25.0 mL solution of 0.200 M NaOH has to be calculated.

Concept Introduction:

Refer to part (a).

(b)

Expert Solution

Answer to Problem 12.31P

Volume of 0.300 M H2SO4 solution needed to neutralize 25.0 mL solution of 0.200 M NaOH is 8.333 mL.

Explanation of Solution

The dilution formula is given as follows:

n1M1V1=n2M2V2 (1)

Substitute 2 for n1, 1 for n2, 0.300 M for M1, 0.200 M for M2, 25.0 mL for V2 in equation (1).

(2)(0.300 M)(V1)=(1)(0.200 M)(25.0 mL) (3)

Rearrange equation (3) to calculate the value of V1.

V1=(0.200 M)(25.0 mL)(2)0.300 M=8.333 mL

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Answer 3

Question

Chapter 12, Problem 12.31P

(a)

Interpretation Introduction

Interpretation:

Volume of 0.108 M H2SO4 solution needed to neutralize 15.0 μL solution of 0.010 M Ca(OH)2 has to be calculated.

Concept Introduction:

The dilution formula is given as follows:

n1M1V1=n2M2V2

Here,

M1 denotes the molarity of the solution before dilution.

V1 denotes the volume of the solution before dilution.

M2 denotes the molarity of solution after dilution.

V2 denotes the volume of the solution after dilution.

n1 and n2 are n factors of acid and base respectively.

(a)

Expert Solution

Answer to Problem 12.31P

Volume of 0.108 M H2SO4 solution needed to neutralize 15.0 μL solution of 0.010 M Ca(OH)2 is 0.1851×10−6 L.

Explanation of Solution

The conversion factor to convert liters to is μL as follows:

1 μL=10−6 L

Hence convert 15 μL to L as follows:

Volume=(15 μL)(10−6 L1 μL)=15×10−6 L

The dilution formula is given as follows:

n1M1V1=n2M2V2 (1)

Substitute 1 for n1, 2 for n2, 0.108 M for M1, 0.010 M for M2, 15×10−6 L for V2 in equation (1).

(1)(0.108 M)(V1)=(2)(0.010 M)(15×10−6 L) (2)

Rearrange equation (2) to calculate the value of V1.

V1=(2)(0.010 M)(15×10−6 L)(1)(0.108 M)=0.1851×10−6 L

(b)

Interpretation Introduction

Interpretation:

Volume of 0.300 M H2SO4 solution needed to neutralize 25.0 mL solution of 0.200 M NaOH has to be calculated.

Concept Introduction:

Refer to part (a).

(b)

Expert Solution

Answer to Problem 12.31P

Volume of 0.300 M H2SO4 solution needed to neutralize 25.0 mL solution of 0.200 M NaOH is 8.333 mL.

Explanation of Solution

The dilution formula is given as follows:

n1M1V1=n2M2V2 (1)

Substitute 2 for n1, 1 for n2, 0.300 M for M1, 0.200 M for M2, 25.0 mL for V2 in equation (1).

(2)(0.300 M)(V1)=(1)(0.200 M)(25.0 mL) (3)

Rearrange equation (3) to calculate the value of V1.

V1=(0.200 M)(25.0 mL)(2)0.300 M=8.333 mL

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Answer 4

Question

Chapter 12, Problem 12.31P

(a)

Interpretation Introduction

Interpretation:

Volume of 0.108 M H2SO4 solution needed to neutralize 15.0 μL solution of 0.010 M Ca(OH)2 has to be calculated.

Concept Introduction:

The dilution formula is given as follows:

n1M1V1=n2M2V2

Here,

M1 denotes the molarity of the solution before dilution.

V1 denotes the volume of the solution before dilution.

M2 denotes the molarity of solution after dilution.

V2 denotes the volume of the solution after dilution.

n1 and n2 are n factors of acid and base respectively.

(a)

Expert Solution

Answer to Problem 12.31P

Volume of 0.108 M H2SO4 solution needed to neutralize 15.0 μL solution of 0.010 M Ca(OH)2 is 0.1851×10−6 L.

Explanation of Solution

The conversion factor to convert liters to is μL as follows:

1 μL=10−6 L

Hence convert 15 μL to L as follows:

Volume=(15 μL)(10−6 L1 μL)=15×10−6 L

The dilution formula is given as follows:

n1M1V1=n2M2V2 (1)

Substitute 1 for n1, 2 for n2, 0.108 M for M1, 0.010 M for M2, 15×10−6 L for V2 in equation (1).

(1)(0.108 M)(V1)=(2)(0.010 M)(15×10−6 L) (2)

Rearrange equation (2) to calculate the value of V1.

V1=(2)(0.010 M)(15×10−6 L)(1)(0.108 M)=0.1851×10−6 L

(b)

Interpretation Introduction

Interpretation:

Volume of 0.300 M H2SO4 solution needed to neutralize 25.0 mL solution of 0.200 M NaOH has to be calculated.

Concept Introduction:

Refer to part (a).

(b)

Expert Solution

Answer to Problem 12.31P

Volume of 0.300 M H2SO4 solution needed to neutralize 25.0 mL solution of 0.200 M NaOH is 8.333 mL.

Explanation of Solution

The dilution formula is given as follows:

n1M1V1=n2M2V2 (1)

Substitute 2 for n1, 1 for n2, 0.300 M for M1, 0.200 M for M2, 25.0 mL for V2 in equation (1).

(2)(0.300 M)(V1)=(1)(0.200 M)(25.0 mL) (3)

Rearrange equation (3) to calculate the value of V1.

V1=(0.200 M)(25.0 mL)(2)0.300 M=8.333 mL

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Answer 5

Chapter 12, Problem 12.31P

(a)

Interpretation Introduction

Interpretation:

Volume of 0.108 M H2SO4 solution needed to neutralize 15.0 μL solution of 0.010 M Ca(OH)2 has to be calculated.

Concept Introduction:

The dilution formula is given as follows:

n1M1V1=n2M2V2

Here,

M1 denotes the molarity of the solution before dilution.

V1 denotes the volume of the solution before dilution.

M2 denotes the molarity of solution after dilution.

V2 denotes the volume of the solution after dilution.

n1 and n2 are n factors of acid and base respectively.

(a)

Expert Solution

Answer to Problem 12.31P

Volume of 0.108 M H2SO4 solution needed to neutralize 15.0 μL solution of 0.010 M Ca(OH)2 is 0.1851×10−6 L.

Explanation of Solution

The conversion factor to convert liters to is μL as follows:

1 μL=10−6 L

Hence convert 15 μL to L as follows:

Volume=(15 μL)(10−6 L1 μL)=15×10−6 L

The dilution formula is given as follows:

n1M1V1=n2M2V2 (1)

Substitute 1 for n1, 2 for n2, 0.108 M for M1, 0.010 M for M2, 15×10−6 L for V2 in equation (1).

(1)(0.108 M)(V1)=(2)(0.010 M)(15×10−6 L) (2)

Rearrange equation (2) to calculate the value of V1.

V1=(2)(0.010 M)(15×10−6 L)(1)(0.108 M)=0.1851×10−6 L

(b)

Interpretation Introduction

Interpretation:

Volume of 0.300 M H2SO4 solution needed to neutralize 25.0 mL solution of 0.200 M NaOH has to be calculated.

Concept Introduction:

Refer to part (a).

(b)

Expert Solution

Answer to Problem 12.31P

Volume of 0.300 M H2SO4 solution needed to neutralize 25.0 mL solution of 0.200 M NaOH is 8.333 mL.

Explanation of Solution

The dilution formula is given as follows:

n1M1V1=n2M2V2 (1)

Substitute 2 for n1, 1 for n2, 0.300 M for M1, 0.200 M for M2, 25.0 mL for V2 in equation (1).

(2)(0.300 M)(V1)=(1)(0.200 M)(25.0 mL) (3)

Rearrange equation (3) to calculate the value of V1.

V1=(0.200 M)(25.0 mL)(2)0.300 M=8.333 mL

Answer 6

Chapter 12, Problem 12.31P

(a)

Interpretation Introduction

Interpretation:

Volume of 0.108 M H2SO4 solution needed to neutralize 15.0 μL solution of 0.010 M Ca(OH)2 has to be calculated.

Concept Introduction:

The dilution formula is given as follows:

n1M1V1=n2M2V2

Here,

M1 denotes the molarity of the solution before dilution.

V1 denotes the volume of the solution before dilution.

M2 denotes the molarity of solution after dilution.

V2 denotes the volume of the solution after dilution.

n1 and n2 are n factors of acid and base respectively.

(a)

Expert Solution

Answer to Problem 12.31P

Volume of 0.108 M H2SO4 solution needed to neutralize 15.0 μL solution of 0.010 M Ca(OH)2 is 0.1851×10−6 L.

Explanation of Solution

The conversion factor to convert liters to is μL as follows:

1 μL=10−6 L

Hence convert 15 μL to L as follows:

Volume=(15 μL)(10−6 L1 μL)=15×10−6 L

The dilution formula is given as follows:

n1M1V1=n2M2V2 (1)

Substitute 1 for n1, 2 for n2, 0.108 M for M1, 0.010 M for M2, 15×10−6 L for V2 in equation (1).

(1)(0.108 M)(V1)=(2)(0.010 M)(15×10−6 L) (2)

Rearrange equation (2) to calculate the value of V1.

V1=(2)(0.010 M)(15×10−6 L)(1)(0.108 M)=0.1851×10−6 L

Answer 7

Chapter 12, Problem 12.31P

(a)

Interpretation Introduction

Interpretation:

Volume of 0.108 M H2SO4 solution needed to neutralize 15.0 μL solution of 0.010 M Ca(OH)2 has to be calculated.

Concept Introduction:

The dilution formula is given as follows:

n1M1V1=n2M2V2

Here,

M1 denotes the molarity of the solution before dilution.

V1 denotes the volume of the solution before dilution.

M2 denotes the molarity of solution after dilution.

V2 denotes the volume of the solution after dilution.

n1 and n2 are n factors of acid and base respectively.

Answer 8

(a)

Expert Solution

Answer to Problem 12.31P

Volume of 0.108 M H2SO4 solution needed to neutralize 15.0 μL solution of 0.010 M Ca(OH)2 is 0.1851×10−6 L.

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

The conversion factor to convert liters to is μL as follows:

1 μL=10−6 L

Hence convert 15 μL to L as follows:

Volume=(15 μL)(10−6 L1 μL)=15×10−6 L

The dilution formula is given as follows:

n1M1V1=n2M2V2 (1)

Substitute 1 for n1, 2 for n2, 0.108 M for M1, 0.010 M for M2, 15×10−6 L for V2 in equation (1).

(1)(0.108 M)(V1)=(2)(0.010 M)(15×10−6 L) (2)

Rearrange equation (2) to calculate the value of V1.

V1=(2)(0.010 M)(15×10−6 L)(1)(0.108 M)=0.1851×10−6 L

Answer 13

(b)

Interpretation Introduction

Interpretation:

Volume of 0.300 M H2SO4 solution needed to neutralize 25.0 mL solution of 0.200 M NaOH has to be calculated.

Concept Introduction:

Refer to part (a).

(b)

Expert Solution

Answer to Problem 12.31P

Volume of 0.300 M H2SO4 solution needed to neutralize 25.0 mL solution of 0.200 M NaOH is 8.333 mL.

Explanation of Solution

The dilution formula is given as follows:

n1M1V1=n2M2V2 (1)

Substitute 2 for n1, 1 for n2, 0.300 M for M1, 0.200 M for M2, 25.0 mL for V2 in equation (1).

(2)(0.300 M)(V1)=(1)(0.200 M)(25.0 mL) (3)

Rearrange equation (3) to calculate the value of V1.

V1=(0.200 M)(25.0 mL)(2)0.300 M=8.333 mL

Answer 14

(b)

Interpretation Introduction

Interpretation:

Volume of 0.300 M H2SO4 solution needed to neutralize 25.0 mL solution of 0.200 M NaOH has to be calculated.

Concept Introduction:

Refer to part (a).

Answer 15

(b)

Expert Solution

Answer to Problem 12.31P

Volume of 0.300 M H2SO4 solution needed to neutralize 25.0 mL solution of 0.200 M NaOH is 8.333 mL.

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

The dilution formula is given as follows:

n1M1V1=n2M2V2 (1)

Substitute 2 for n1, 1 for n2, 0.300 M for M1, 0.200 M for M2, 25.0 mL for V2 in equation (1).

(2)(0.300 M)(V1)=(1)(0.200 M)(25.0 mL) (3)

Rearrange equation (3) to calculate the value of V1.

V1=(0.200 M)(25.0 mL)(2)0.300 M=8.333 mL

Answer 20

Ch. 12 - Prob. 12.1P Ch. 12 - Prob. 12.2P Ch. 12 - Prob. 12.3P Ch. 12 - Prob. 12.4P Ch. 12 - Prob. 12.5P Ch. 12 - Prob. 12.6P Ch. 12 - Prob. 12.7P Ch. 12 - Prob. 12.8P Ch. 12 - Prob. 12.9P Ch. 12 - Prob. 12.10P

Concept explainers

Videos

Answer to Problem 12.31P

Explanation of Solution

Interpretation:

Volume of 0.300 M H2SO4 solution needed to neutralize 25.0 mL solution of 0.200 M NaOH has to be calculated.

Concept Introduction:

Refer to part (a).

Answer to Problem 12.31P

Explanation of Solution

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Chapter 12 Solutions

Concept explainers

Videos

Answer to Problem 12.31P

Explanation of Solution

Interpretation: Volume of 0.300 M H2SO4 solution needed to neutralize 25.0 mL solution of 0.200 M NaOH has to be calculated. Concept Introduction: Refer to part (a).

Answer to Problem 12.31P

Explanation of Solution

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Chapter 12 Solutions

Interpretation:

Volume of 0.300 M H2SO4 solution needed to neutralize 25.0 mL solution of 0.200 M NaOH has to be calculated.

Concept Introduction:

Refer to part (a).