Chemistry: The Molecular Science
Chemistry: The Molecular Science
5th Edition
ISBN: 9781285199047
Author: John W. Moore, Conrad L. Stanitski
Publisher: Cengage Learning
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Chapter 14, Problem ISP

Lactic acid, CH3CH(OH)COOH, is a weak monoprotic acid with a melting point of 53 °C. It exists as two enantiomers (Sec. 7-2f) that have slightly different Ka values. The D form has a Ka of 1.5 × 10−4 and the L form has a Ka of 1.6 × 10−4. The D form is synthesized by some bacteria. The L form is produced in muscle cells during anaerobic metabolism in which glucose molecules are broken down into lactic acid and molecules of adenosine triphosphate (ATP) are formed. When lactic acid builds up too rapidly in muscle tissue, severe pain results.

  1. (a) Which form of lactic acid (D or L) is the stronger acid? Explain your answer.
  2. (b) Determine the pKa that would be measured for a 50:50 mixture of the two forms of lactic acid in aqueous solution, pKa = −log Ka
  3. (c) A solution of D-lactic acid is prepared. Use HL as a general formula for lactic acid, and write the equation for the ionization of lactic acid in water.
  4. (d) If 0.100-M solutions of these two acids (D and L) were prepared, calculate what the pH of each solution would be.
  5. (e) Before any lactic acid dissolves in the water, what reaction determines the pH?
  6. (f) Calculate the pH of a solution made by dissolving 4.46 g D-lactic acid in 500. mL of water.
  7. (g) Calculate the volume (mL) of 1.15-M NaOH(aq) required to completely neutralize 4.46 g of pure lactic acid.
  8. (h) Calculate the pH of the solution when exactly enough NaOH was added to neutralize all of the lactic acid for (i) the D form; (ii) the L form; and (iii) a 50:50 mixture of the two forms.

(a)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation:

The stronger acid has to be chosen between D- form of lactic acid and L- form of lactic acid.

Explanation of Solution

The acid having larger Ka value is known as strong acid.

The Ka values of L- and D- form of lactic acid are 1.6×104and1.5×104 respectively.  The Ka value of L- form of lactic acid is slightly larger than that of D- form of lactic acid.  Thus, L- form of lactic acid is stronger acid.

(b)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation:

The value of pKa that would be measured for a 50:50 mixture of the two forms of lactic acid in aqueous solution has to be determined.

Answer to Problem ISP

The value of pKa of the mixture is 3.81.

Explanation of Solution

As the solution contains a mixture of two acids, the observed value of the Ka is the combination of the two Ka's.

When the reactions are added, Ka's are multiplied as they are logarithmic.

  HLD(aq)+H2O(l)LD(aq)+H3O+(aq)Ka,DHLL(aq)+H2O(l)LL(aq)+H3O+(aq)Ka,L_HLD(aq)+HLL(aq)+2H2O(l)LD(aq)+LL(aq)+2H3O+(aq)K=Ka,D.Ka,L

It can be further written as given below.

  12HLD(aq)+12HLL(aq)+H2O(l)12LD(aq)+12LL(aq)+H3O+(aq)Ka=(Ka,D.Ka,L)1/2pKa=logKa=log[(Ka,D.Ka,L)1/2]

On simplification,

  pKa=12{log[(Ka,D.Ka,L)]}=12[(logKa,D)(logKa,L)]pKa=12(pKa,D+pKa,L)

Now,

  pKa,L=logKa,L=log(1.6×104)=3.80pKa,D=logKa,D=log(1.5×104)=3.82pKa=12(pKa,D+pKa,L)=12(3.80+3.82)=3.81.

Therefore, the value of pKa of the mixture is 3.81.

(c)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation:

The equation for the ionization of lactic acid in water has to be written.

Explanation of Solution

The equation for the ionization of lactic acid in water is given below.  The ionization causes the transfer of the carboxyl hydrogen ion from lactic acid to water.

  HL(aq)+H2O(l)L(aq)+H3O+(aq)

(d)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation:

The pH of 0.100M solutions of D and L- lactic acids has to be determined.

Answer to Problem ISP

The pH of 0.100M solutions of D and L- lactic acids are 2.41and2.40 respectively.

Explanation of Solution

The equilibrium reaction is given below.

  HL(aq)+H2O(l)L(aq)+H3O+(aq)

The acid ionization constant can be written as given below.

  Kb=[L][H3O+][HL]

A table can be set up as shown below.

  HLLH3O+Initialconc.(M)0.1001×107Changeinconc.(M)x+x+xEquilibriumconc.(M)0.10xxx

The concentration of H3O+ comes from its autoionization that is 1.0×107.  It is assumed to be small compared to the change.

At equilibrium, Ka=(x)(x)(0.10x)

Assuming x is very small, it can be written as 0.10x0.10.

Then,

For L- form of the acid:

  x2=(Ka)(0.10)x=(1.6×104)(0.10)x=4.0×10-3M=[H3O+]

The pH of the solution can be calculated as shown below.

  pH=log[H3O+]=log(4.0×103)=2.40.

For D- form of the acid:

  x2=(Ka)(0.10)x=(1.5×104)(0.10)x=3.9×10-3M=[H3O+]

The pH of the solution can be calculated as shown below.

  pH=log[H3O+]=log(3.9×103)=2.41.

Therefore, the pH of 0.100M solutions of D and L- lactic acids are 2.41and2.40 respectively.

(e)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation:

The reaction which determines the pH before any lactic acid dissolves in the water has to be written.

Explanation of Solution

The pH is determined by the autoionization of water.  The reaction is given below.

  2H2O(l)OH(aq)+H3O+(aq)

(f)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation:

The pH of a solution made by dissolving 4.46g D- lactic acid in 500mL of water has to be calculated.

Answer to Problem ISP

The pH of solutions of D-lactic acid is 2.41.

Explanation of Solution

Calculation of initial concentration of D- lactic acid:

The number of moles of D-HL can be calculated as given below.

  Moles=MassMolarmass=4.46g90.0777g/mol=0.0495mol.

Then, the concentration of D-HL is given below.

  Molarity=MolesVolumeofsolutioninL=0.0495mol500×103L=0.0990M.Where,1L=1000mL.

Therefore, the initial concentration of D- lactic acid is 0.0990M.

The equilibrium reaction is given below.

  HL(aq)+H2O(l)L(aq)+H3O+(aq)

The acid ionization constant can be written as given below.

  Kb=[L][H3O+][HL]

A table can be set up as shown below.

  HLLH3O+Initialconc.(M)0.099001×107Changeinconc.(M)x+x+xEquilibriumconc.(M)0.0990xxx

The concentration of H3O+ comes from its autoionization that is 1.0×107.  It is assumed to be small compared to the change.

At equilibrium, Ka=(x)(x)(0.0990x)=1.5×104

Assuming x is very small, it can be written as 0.0990x0.0990.

Then,

  x2=(1.6×104)(0.10)x=3.9×10-3M=[H3O+]

The pH of the solution can be calculated as shown below.

  pH=log[H3O+]=log(3.9×103)=2.41.

Therefore, the pH of solutions of D-lactic acid is 2.41.

(g)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation:

The volume of 1.15MNaOH(aq) required to completely neutralize 4.46g of pure lactic acid has to be calculated.

Answer to Problem ISP

The volume of 1.15MNaOH(aq) required to completely neutralize 4.46g of pure lactic acid is 43.0mL.

Explanation of Solution

The neutralization reaction is given below.

  HL(aq)+OH(aq)L(aq)+H2O(l)

The stoichiometry of acid and the base is 1:1.

The number of moles of D-HL can be calculated as given below.

  Moles=MassMolarmass=4.46g90.0777g/mol=0.0495mol.

Hence, the number of moles of NaOH required is 0.0495mol.

Then, the volume of NaOH required can be calculated as given below.

  Molarity=MolesVolumeofsolutioninLVolume=MolesMolarity=0.0495mol1.15mol/L=0.0430L=0.0430×103mL=43.0mL.Where,1L=1000mL.

Therefore, the volume of 1.15MNaOH(aq) required to completely neutralize 4.46g of pure lactic acid is 43.0mL.

(h)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation:

The pH of the solution when exactly enough NaOH was added to neutralize all of the lactic acid for (i) the D form, (ii) the L form and (iii) a 50:50 mixture of the two forms has to be calculated.

Explanation of Solution

The equilibrium reaction is given below.

  L(aq)+H2O(l)HL(aq)+OH(aq)

The equilibrium constant for the above reaction can be written as given below.

  Kb=[HL][OH][L]

There is relationship exist as given below.

  Kw=KaKbKb=KwKa

Now,

For D form:

  Kb,D=1.0×10141.5×104=6.7×1011

For L form:

  Kb,L=1.0×10141.6×104=6.3×1011

For 50:50 mixture of the two forms:

  pKb,L=logKb,L=log(6.7×1011)=10.17pKb,D=logKb,D=log(6.3×1011)=10.20pKb=12(pKb,L+pKb,D)=12(10.17+10.20)=10.19Kb10pKb=1010.19=6.5×1011.

Calculation of concentration of lactate ion:

The number of moles of HL can be calculated as given below.

  Moles=MassMolarmass=4.46g90.0777g/mol=0.0495mol.

Then,

  Molarity=MolesVolumeofsolutioninL=0.0495mol43.0×103L=1.15M.

Therefore, the concentration of lactate ion is 1.15M.

A table can be set up as shown below.

  L(aq)HL(aq)OH(aq)Initialconc.(M)1.1501×107Changeinconc.(M)x+x+xEquilibriumconc.(M)1.15xxx

The concentration of OH comes from its autoionization that is 1.0×107.  It is assumed to be small compared to the change.

At equilibrium, Kb=(x)(x)(1.15x)

Assuming x is very small, it can be written as 1.15x1.15.

Then,

(i) For D form:

  x2=(6.7×1011)(1.15)x=8.8×10-6M=[OH]

The pH of the solution can be calculated as shown below.

  pOH=log[OH]=log(8.8×106)=5.06pH=14pOH=145.06=8.94.

Therefore, the pH of solutions of D-lactic acid is 8.94.

(ii) For L form:

  x2=(6.3×1011)(1.15)x=8.5×10-6M=[OH]

The pH of the solution can be calculated as shown below.

  pOH=log[OH]=log(8.5×106)=5.07pH=14pOH=145.07=8.93.

Therefore, the pH of solutions of L-lactic acid is 8.93.

(iii) For 50:50 mixture of the two forms:

  x2=(6.5×1011)(1.15)x=8.6×10-6M=[OH]

The pH of the solution can be calculated as shown below.

  pOH=log[OH]=log(8.6×106)=5.06pH=14pOH=145.06=8.94.

Therefore, the pH of 50:50 mixture of the two forms is 8.94.

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Chapter 14 Solutions

Chemistry: The Molecular Science

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