EBK FOUNDATIONS OF COLLEGE CHEMISTRY
EBK FOUNDATIONS OF COLLEGE CHEMISTRY
15th Edition
ISBN: 9781118930144
Author: Willard
Publisher: JOHN WILEY+SONS INC.
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Chapter 15, Problem 24PE

(a)

Interpretation Introduction

Interpretation:

Molar concentration of each ion found after 45.5 mL of 0.10 M NaCl is mixed with 60.5 mL of 0.35 M NaCl has to be calculated.

Concept Introduction:

Molarity is quantitatively defined as moles of solute in one liter of solution. For example 0.070 M AlCl3 indicates that in 1 L the moles of AlCl3 is 0.070 mol. The formula to evaluate volume from molarity is given as follows:

  M=nV

Where,

V represents volume.

M represents molarity.

n represents number of moles.

(a)

Expert Solution
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Explanation of Solution

Since in 1000 mL the moles of NaCl present is 0.10 mol so, moles of NaCl in 45.5 mL is calculated as follows:

  Moles of NaCl=(45.5 mL)(0.10 mol1000 mL)=0.0045 mol

Similarly for other solution, in 1000 mL moles of NaCl present is 0.35 mol so, moles of NaCl in 60.5 mL is calculated as follows:

  Moles of NaCl=(60.5 mL)(0.35 mol1000 mL)=0.02117 mol

Total moles of solution when two NaCl solutions is  mixed are calculated as follows:

  Total moles=0.0045 mol+0.02117 mol=0.02567mol

Total volume of solution when two NaCl solutions are mixed is calculated as follows:

  Total volume =45.5 mL+60.5 mL=106 mL

The formula to evaluate volume from molarity is given as follows:

  M=nV        (1)

Substitute 0.02567mol for n and 106 mL for V in equation (1).

  M=(0.02567mol106 mL)(1000 mL1 L)=0.2421 M

NaCl can be broken into 1 mol Na+ and 1 mol Cl as follows:

  NaClNa++Cl

Since 1 M Na+ is furnished by 1 M NaCl so molarity of Na+ ion due to 0.2421 M NaCl  is calculated as follows:

  Molarity of  Na+=(0.2421 M NaCl)(1 M Na+1 M NaCl)=0.2421 M Na+

Since 1 M Cl is furnished by 1 M NaCl so molarity of Cl ions due to 0.2421 MNaCl is calculated as follows:

  Molarity of Cl=(0.2421 M NaCl)(1 M Cl1 M NaCl)=0.2421 M Cl

Hence molar concentration of both Na+ and Cl is 0.2421 M and 0.2421 M respectively.

(b)

Interpretation Introduction

Interpretation:

Molar concentration of each ion found after 95.5 mL of 1.25 M HCl is mixed with 125.5 mL of 2.50 M HCl has to be calculated.

Concept Introduction:

Refer to part (a).

(b)

Expert Solution
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Explanation of Solution

1.25 M HCl indicates that in 1 L the moles of HCl is 1.25 mol, Similarly 2.50 M HCl indicates that in 1 L the moles of HCl is 2.50 mol.

Since in 1000 mL the moles of HCl present is 1.25 mol so, moles of HCl in 95.5 mL is calculated as follows:

  Moles of HCl=(95.5 mL)(1.25 mol1000 mL)=0.119375 mol

Similarly for other solution, in 1000 mL moles of HCl present is 2.50 mol so, moles of HCl in 125.5 mL is calculated as follows:

  Moles of HCl=(125.5 mL)(2.50 mol1000 mL)=0.31375 mol

Total volume of solution upon when two HCl solutions are mixed is calculated as follows:

  Total volume =95.5 mL+125.5 mL=221 mL

Substitute 0.119375 mol for n and 221 mL for V in equation (1).

  M=(0.119375 mol221 mL)(1000 mL1 L)=0.54015 M

Substitute 0.31375 mol for n and 221 mL for V in equation (1).

  M=(0.31375 mol221 mL)(1000 mL1 L)=1.41968 M

Thus total molar concentration of HCl is sum of concentrations calculated as follows:

  Total concentration of HCl=0.54015 M+1.41968 M=1.9598 M

HCl can be broken into 1 mol H+ and 1 mol Cl as follows:

  HClH++Cl

Since 1 M H+ is furnished by 1 M HCl so molarity of H+ ion due to 1.9598 M HCl is calculated as follows:

  Molarity of H+=(1.9598 M HCl)(1 M H+1 M HCl)=1.9598 M H+

Since 1 M Cl is furnished by 1 M HCl so molarity of Cl ions due to 1.9598 M HCl is calculated as follows:

  Molarity of Cl=(1.9598 M HCl)(1 M Cl1 M HCl)=1.9598 M Cl

Hence molar concentration of both H+ and Cl is 1.9598 M.

(c)

Interpretation Introduction

Interpretation:

Molar concentration each ion found after 15.5 mL of 0.10 M Ba(NO3)2 is mixed with 10.5 mL of 0.20 M AgNO3 has to be calculated.

Concept Introduction:

Refer to part (a).

(c)

Expert Solution
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Explanation of Solution

0.10 M Ba(NO3)2 indicates that in 1 L the moles of Ba(NO3)2 is 0.10 mol. Similarly 0.20 M AgNO3 indicates that in EBK FOUNDATIONS OF COLLEGE CHEMISTRY, Chapter 15, Problem 24PE the moles of AgNO3 is 0.20 mol.

Since in 1000 mL moles of Ba(NO3)2 present is 0.10 mol so, moles of Ba(NO3)2 in 15.5 mL is calculated as follows:

  Moles of Ba(NO3)2=(15.5 mL)(0.10 mol1000 mL)=0.00155 mol

Since in 1000 mL moles of AgNO3 present is 0.20 mol so, moles of Ba(NO3)2 in 10.5 mL is calculated as follows:

  Moles of AgNO3=(10.5 mL)(0.20 mol1000 mL)=0.0021 mol

Ba(NO3)2 can be broken into 1 mol Ba2+ and 2 mol NO3 as follows:

  Ba(NO3)2Ba2++2NO3

Since 1 M Ba2+ is furnished by 1 M Ba(NO3)2 so molarity of Ba2+ ion due to 0.00155 M Ba(NO3)2 is calculated as follows:

  Molarity of Ba2+=(0.00155 M Ba(NO3)2)(1 M Ba2+1 M Ba(NO3)2)=0.00155 M Ba2+

Since 2 M NO3 is furnished by 1 M Ba(NO3)2 so molarity of NO3 ions due to 0.00155 M Ba(NO3)2 is calculated as follows:

  Molarity of NO3=(0.00155 M Ba(NO3)2)(2 M NO31 M Ba(NO3)2)=0.0031 M NO3

AgNO3 can be broken into 1 mol Ag2+ and 1 mol NO3 as follows:

  AgNO3Ag++NO3

Since 1 M Ag+ is furnished by 1 M AgNO3 so molarity of Ag+ ion due to 0.0021 M AgNO3 is calculated as follows:

  Molarity of Ag+=(0.0021 M AgNO3)(1 M Ag+1 M AgNO3)=0.0021 M Ag+

Since 1 M Ag+ is furnished by 1 M AgNO3 so molarity of NO3 ion due to 0.0021 M AgNO3 is calculated as follows:

  Molarity of NO3=(0.0021 M AgNO3)(1 M NO31 M AgNO3)=0.0021 M NO3

Thus total molar concentration of NO3 is calculated as follows:

  Total molar concentration =0.0031 M+0.0021 M=0.0052 M

Hence molar concentration of Ba2+, Ag+ and NO3 are 0.00155 M, 0.0021 M and 0.0052 M.

(d)

Interpretation Introduction

Interpretation:

Molar concentration of each ion found after 25.5 mL of 0.25 M NaCl is mixed with 15.5 mL of 0.15 M Ca(C2H3O2)2 has to be calculated.

Concept Introduction:

Refer to part (a).

(d)

Expert Solution
Check Mark

Explanation of Solution

0.333 M NaCl indicates that in 1000 mL moles of NaCl is 0.25 mol. Similarly 0.15 M Ca(C2H3O2)2 indicates that in 1000 mL the moles of Ca(C2H3O2)2 is 0.15 mol.

Since in 1000 mL moles of NaCl present is 0.25 mol so, moles of NaCl in 25.5 mL is calculated as follows:

  Moles of NaCl=(25.5 mL)(0.25 mol1000 mL)=0.006375 mol

Since in 1000 mL moles of Ca(C2H3O2)2 present is 0.20 mol so, moles of Ca(C2H3O2)2 in 15.5 mL is calculated as follows:

  Moles of Ca(C2H3O2)2=(15.5 mL)(0.15 mol1000 mL)=0.002325 mol

NaCl can be broken into 1 mol Na+ and 1 mol Cl as follows:

  NaClNa++Cl

Since 1 M Na+ is furnished by 1 M NaCl so molarity of Na+ ion due to 0.006375 M NaCl  is calculated as follows:

  Molarity of  Na+=(0.006375 M NaCl)(1 M Na+1 M NaCl)=0.006375 M Na+

Since 1 M Cl is furnished by 1 M NaCl so molarity of Cl ions due to 0.006375 M NaCl is calculated as follows:

  Molarity of Cl=(0.006375 M NaCl)(1 M Cl1 M NaCl)=0.006375 M Cl

Ca(C2H3O2)2 can be broken into 1 mol Ca2+ and 2 mol CH3COO as follows:

  Ca(C2H3O2)2Ca2++2CH3COO

Since 1 M Ca2+ is furnished by 1 M Ca(C2H3O2)2 so molarity of Ca2+ ion due to 0.002325 M Ca(C2H3O2)2 is calculated as follows:

  Molarity of Ca2+=(0.002325 M Ca(C2H3O2)2)(1 M Ca2+1 M Ca(C2H3O2)2)=0.002325 M Ca2+

Since 2 M CH3COO is furnished by 1 M Ca(C2H3O2)2 so molarity of CH3COO ions due to 0.002325 M Ca(C2H3O2)2 is calculated as follows:

  Molarity of CH3COO=(0.002325 M Ca(C2H3O2)2)(2 M CH3COO1 M Ca(C2H3O2)2)=0.00465 M CH3COO

Hence molar concentration of Na+, Cl is 0.006375 M while molar concentration of Ca2+ and CH3COO ions are 0.002325 M and 0.00465 M respectively.

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Chapter 15 Solutions

EBK FOUNDATIONS OF COLLEGE CHEMISTRY

Ch. 15 - Prob. 3RQCh. 15 - Prob. 4RQCh. 15 - Prob. 5RQCh. 15 - Prob. 6RQCh. 15 - Prob. 7RQCh. 15 - Prob. 8RQCh. 15 - Prob. 9RQCh. 15 - Prob. 10RQCh. 15 - Prob. 11RQCh. 15 - Prob. 12RQCh. 15 - Prob. 13RQCh. 15 - Prob. 14RQCh. 15 - Prob. 15RQCh. 15 - Prob. 16RQCh. 15 - Prob. 17RQCh. 15 - Prob. 18RQCh. 15 - Prob. 19RQCh. 15 - Prob. 20RQCh. 15 - Prob. 21RQCh. 15 - Prob. 22RQCh. 15 - Prob. 23RQCh. 15 - Prob. 24RQCh. 15 - Prob. 25RQCh. 15 - Prob. 26RQCh. 15 - Prob. 27RQCh. 15 - Prob. 28RQCh. 15 - Prob. 1PECh. 15 - Prob. 2PECh. 15 - Prob. 3PECh. 15 - Prob. 4PECh. 15 - Prob. 5PECh. 15 - Prob. 6PECh. 15 - Prob. 7PECh. 15 - Prob. 8PECh. 15 - Prob. 9PECh. 15 - Prob. 10PECh. 15 - Prob. 11PECh. 15 - Prob. 12PECh. 15 - Prob. 13PECh. 15 - Prob. 14PECh. 15 - Prob. 15PECh. 15 - Prob. 16PECh. 15 - Prob. 17PECh. 15 - Prob. 18PECh. 15 - Prob. 19PECh. 15 - Prob. 20PECh. 15 - Prob. 21PECh. 15 - Prob. 22PECh. 15 - Prob. 23PECh. 15 - Prob. 24PECh. 15 - Prob. 25PECh. 15 - Prob. 26PECh. 15 - Prob. 27PECh. 15 - Prob. 28PECh. 15 - Prob. 29PECh. 15 - Prob. 30PECh. 15 - Prob. 31PECh. 15 - Prob. 32PECh. 15 - Prob. 33PECh. 15 - Prob. 34PECh. 15 - Prob. 35PECh. 15 - Prob. 36PECh. 15 - Prob. 37PECh. 15 - Prob. 38PECh. 15 - Prob. 39PECh. 15 - Prob. 40PECh. 15 - Prob. 41PECh. 15 - Prob. 42PECh. 15 - Prob. 43PECh. 15 - Prob. 44PECh. 15 - Prob. 45AECh. 15 - Prob. 46AECh. 15 - Prob. 47AECh. 15 - Prob. 48AECh. 15 - Prob. 49AECh. 15 - Prob. 50AECh. 15 - Prob. 51AECh. 15 - Prob. 52AECh. 15 - Prob. 53AECh. 15 - Prob. 54AECh. 15 - Prob. 55AECh. 15 - Prob. 56AECh. 15 - Prob. 57AECh. 15 - Prob. 58AECh. 15 - Prob. 59AECh. 15 - Prob. 60AECh. 15 - Prob. 61AECh. 15 - Prob. 62AECh. 15 - Prob. 63AECh. 15 - Prob. 64AECh. 15 - Prob. 65AECh. 15 - Prob. 66AECh. 15 - Prob. 67AECh. 15 - Prob. 68AECh. 15 - Prob. 69AECh. 15 - Prob. 70AECh. 15 - Prob. 71AECh. 15 - Prob. 72AECh. 15 - Prob. 73CECh. 15 - Prob. 74CE
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