Chemistry: Atoms First
Chemistry: Atoms First
2nd Edition
ISBN: 9780073511184
Author: Julia Burdge, Jason Overby Professor
Publisher: McGraw-Hill Education
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Chapter 17, Problem 17.106QP

(a)

Interpretation Introduction

Interpretation:

The pH of final solutions has to be calculated.

Concept introduction:

  • pH is the logarithm of the reciprocal of the concentration  of H3O+  in a solution.
  •   pH is used to determine the acidity of an aqueous solution.
  • pH=-log[H3O+]
  • pH=pKa+log[conjugate base][acid] is Henderson-Hasselbalch equation

To find: the pH of 0.10MCacodylic acid

(a)

Expert Solution
Check Mark

Answer to Problem 17.106QP

The pH of final solution is 3.60

Explanation of Solution

CacH(aq)     Cac-(aq) +H+(aq)Initialconcentration (M):   0.10         0               0Changeinconcentration (M):  -x     +x+xFinal concentration(M):  0.10-x        x       xKavalue for cacodylic acid is 6.4 ×10-7Ka=[H+ ][Cac- ][CacH]6.4 ×10-7=x2(0.10-x)x2(0.10)xisverysmallandneglectit,x = [H+] = 2.5 ×10-4MpH=-log[H+]=-log(2.5 ×10-4)pH=3.60

The concentration of hydrogen ion can be determined using acid dissociation constant.  From the concentration of hydrogen ion, the pH is calculated by taking negative log of hydrogen ion.

(b)

Interpretation Introduction

Interpretation:

The pH of final solutions has to be calculated.

Concept introduction:

  • pH is the logarithm of the reciprocal of the concentration  of H3O+  in a solution.
  •   pH is used to determine the acidity of an aqueous solution.
  • pH=-log[H3O+]
  • pH=pKa+log[conjugate base][acid] is Henderson-Hasselbalch equation

To find the pH of 0.15M(CH3)2AsO2Na

(b)

Expert Solution
Check Mark

Answer to Problem 17.106QP

The pH of final solution is 9.69

Explanation of Solution

Cac-(aq) +H2O(l)    CacH(aq) +OH-(aq)Initialconcentration (M):   0.15         0               0Changeinconcentration (M):  -x     +x+xEquilibriumconcentration(M):  0.15-x        x       xKbvalue for Cac- is 1.6 ×10-8Kb=KwKa=1.0 ×10-146.4×10-7=1.6 ×10-8Kb=[CacH ][OH- ][Cac-]1.6 ×10-8=x2(0.15-x)x2(0.15)xisverysmallandneglectit,x = [OH-] = 4.9 ×10-5MpOH=-log[OH-]=-log(4.9 ×10-5)pOH=4.31pH = 14.00 - 4.31 = 9.69

The concentration of hydroxide ion can be determined using base dissociation constant.  From the concentration of hydrogen ion, the pH is calculated by taking negative log of hydroxide ion and subtracting by fourteen.

NumberofmolesofNH3 in 10 mL=10.0 mL×0.300 mol1000mL=3.0×10-3molNumberofmolesofHCl in 10 mL=10.0 mL×0.100 mol1000mL=1.0×10-3mol

(c)

Interpretation Introduction

Interpretation:

The pH of final solutions has to be calculated.

Concept introduction:

  • pH is the logarithm of the reciprocal of the concentration  of H3O+  in a solution.
  •   pH is used to determine the acidity of an aqueous solution.
  • pH=-log[H3O+]
  • pH=pKa+log[conjugate base][acid] is Henderson-Hasselbalch equation

To find the number of moles of CacH and CacNa

(c)

Expert Solution
Check Mark

Answer to Problem 17.106QP

The pH of final solution is 6.07

Explanation of Solution

Numberofmolesof0.10MCacH in 50 mL=50.0 mLCacH×0.10 molCacH1000mL=5.0×10-3mol CacHNumberofmolesof0.10MCacNa in 25 mL=25.0 mLCacNa×0.150 molCacNa1000mL=3.8×10-3molCacNa

The number of moles of CacH and CacNa can be calculated using respective volume and given concentration.

To find the pH of mixed solution

We got buffer solution.  TheHenderson-Hasselbalch equationisusedtofindpHpH=pKa+log[conjugate base][acid]pH=pKa+log[Cac-][CacH]pH=-log(6.4×10-7)+log[3.8×10-3][5.0×10-3]=6.07

The Henderson-Hasselbalch equation can be used to calculate the pH because the solution acts as buffer system.  From the equation and using acid dissociation constant, the pH is calculated.

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Chapter 17 Solutions

Chemistry: Atoms First

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