Chapter 6, Problem 6D.2E

(a)

Interpretation Introduction

Interpretation:

The $\text{pH}$, pOH and percentage deprotonation of 0.11 M aqueous lactic acid has to be calculated.

pH definition:

The concentration of hydrogen ion is measured using $\text{pH}$ scale.  The acidity of aqueous solution is expressed by $\text{pH}$ scale.

The $\text{pH}$ of a solution is defined as the negative base -10 logarithm of the hydrogen or hydronium ion concentration.

  $\text{pH}\text{=}{\text{-log[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}$

On rearranging, the concentration of hydrogen ion ${\text{[H}}^{+}\text{]}$ can be calculated using pH as follows,

  ${\text{[H}}^{+}\text{]}\text{=}{\text{10}}^{\text{-pH}}$

Answer to Problem 6D.2E

The $\text{pH}$, pOH and percentage deprotonation of 0.11 M aqueous lactic acid is 2.02, 11.98 and 8.73% respectively.

Explanation of Solution

Lactic acid is a weak acid when it is dissolved in water it ionized as positive and negative ions and it is given below.

  ${\text{CH}}_{\text{3}}{\text{COOH}}_{\text{(s)}}{\text{+H}}_{\text{2}}{\text{O}}_{\text{(l)}}\rightleftharpoons {\text{CH}}_{\text{3}}{\text{COO}}^{\text{-}}{}_{\text{(aq)}}{\text{+H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\text{(aq)}}$

The equilibrium expression for the above reaction is given below.

  ${\text{K}}_{\text{a}}\text{=}\frac{{\text{[CH}}_{\text{3}}{\text{CH(OH)COO}}^{\text{-}}\text{]}{\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}}{{\text{[CH}}_{\text{3}}\text{CH(OH)COOH]}}$

	${\text{CH}}_{\text{3}}\text{CH(OH)COOH}$	${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$	${\text{CH}}_{\text{3}}{\text{CH(OH)COO}}^{\text{-}}$
Initial concentration	0.11	0	0
Change in concentration	-x	+x	+x
Equilibrium concentration	0.11-x	x	x

The equilibrium concentration values are obtained in the above table and is substituted in above equation and is given below.

  ${\text{K}}_{\text{a}}\text{=}\frac{\text{x}\times \text{x}}{\text{0}\text{.11-x}}$

Lactic acid ${\text{K}}_{\text{a}}$ value is $8.4\times {10}^{-4}$ which is given in table 6C.1 and it is substituted in above equation and then the value of x is calculated.

  $\text{8}\text{.4}\times {\text{10}}^{-4}\text{=}\frac{{\text{x}}^2}{\text{0}\text{.11-x}}$

The above equation, assume that the x present in 0.11-x is very small than 0.11 then it can be negligible and as follows,

  $\begin{array}{l}\text{8}{\text{.4×10}}^{\text{-4}}\text{=}\frac{{\text{x}}^{\text{2}}}{\text{0}\text{.11}}\\ \\ {\text{x}}^{\text{2}}\text{=}\text{0}\text{.11×(8}{\text{.4×10}}^{\text{-4}}\text{)}\\ \\ \text{x}\text{=}\sqrt{0.11\times (8.4\times {10}^{-4})}\\ \\ \text{=}9.\text{6}\times {\text{10}}^{-3}\text{(}\because \text{x}\text{=}{\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{])}\end{array}$

Now, the $\text{pH}$ of the solution is calculated using given equation.

  $\begin{array}{l}\text{pH}\text{=}{\text{-log[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{] }\\ \\ \text{=}\text{-log(9}\text{.6}\times {10}^{-3}\text{)}\\ \\ \text{=}2.02\end{array}$

Therefore, the calculated $\text{pH}$ value of 0.11 M aqueous lactic acid is 2.02.

The general equilibrium expression to find out the pOH of the solution is given below,

  $\begin{array}{l}\text{pH+pOH = 14}\\ \\ \text{ pOH = 14-pH}\\ \\ \text{ = 14-2}\text{.02 }\\ \\ \text{ = 11}\text{.98}\\ \end{array}$

Therefore, the calculated $\text{pOH}$ value of 0.11 M lactic acid is 11.98.

The percentage deprotonation is calculated using the concentration of hydronium ion divided by the initial concentration of lactic acid and the respective equation is given below.

  $\text{Percentage}\text{deprotonated}\text{=}\frac{\text{concentration}\text{of}{\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}}{\text{initial}\text{concentration}\text{of}{\text{[CH}}_{\text{3}}\text{CH(OH)COOH]}}\text{×100\%}$

Concentration of ${\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}$ is $9.6\times {10}^{-3}$ M

Initial concentration of ${\text{[CH}}_{\text{3}}\text{CH(OH)COOH]}$ is $\text{0}\text{.11}\text{M}$

Substitute the obtained values in above equation

  $\begin{array}{l}\text{Percentage}\text{deprotonated =}\frac{9.6\times {10}^{-3}}{0.11}\text{×100\%}\\ \\ \text{= 8}\text{.73\%}\end{array}$

Therefore, the percentage deprotonation of 0.11 M aqueous lactic acid is 8.73%

(b)

Interpretation Introduction

Interpretation:

The $\text{pH}$, pOH and percentage deprotonation of $\text{3}{\text{.7×10}}^{\text{-3}}\text{M}$ aqueous lactic acid has to be calculated.

Concept introduction:

Refer to part (a).

Answer to Problem 6D.2E

The $\text{pH}$, pOH and percentage deprotonation of $\text{3}{\text{.7×10}}^{\text{-3}}\text{M}$ aqueous lactic acid is 3.75, 10.25 and 47.57% respectively.

Explanation of Solution

Lactic acid is a weak acid when it is dissolved in water it ionized as positive and negative ions and it is given below.

  ${\text{CH}}_{\text{3}}{\text{COOH}}_{\text{(s)}}{\text{+H}}_{\text{2}}{\text{O}}_{\text{(l)}}\leftrightarrow {\text{CH}}_{\text{3}}{\text{COO}}^{\text{-}}{}_{\text{(aq)}}{\text{+H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\text{(aq)}}$

The equilibrium expression for the above reaction is given below.

  ${\text{K}}_{\text{a}}\text{=}\frac{{\text{[CH}}_{\text{3}}{\text{CH(OH)COO}}^{\text{-}}\text{]}{\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}}{{\text{[CH}}_{\text{3}}\text{CH(OH)COOH]}}$

	${\text{CH}}_{\text{3}}\text{CH(OH)COOH}$	${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$	${\text{CH}}_{\text{3}}{\text{CH(OH)COO}}^{\text{-}}$
Initial concentration	$\text{3}{\text{.7×10}}^{\text{-3}}$	0	0
Change in concentration	-x	+x	+x
Equilibrium concentration	$\text{(3}{\text{.7×10}}^{\text{-3}})$-x	x	x

The equilibrium concentration values are obtained in the above table and is substituted in above equation and is given below.

  ${\text{K}}_{\text{a}}\text{=}\frac{\text{x}\times \text{x}}{\text{(3}{\text{.7×10}}^{\text{-3}}\text{)-x}}$

Lactic acid ${\text{K}}_{\text{a}}$ value is $8.4\times {10}^{-4}$ which is given in table 6C.1 and it is substituted in above equation and then the value of x is calculated.

  $\text{8}\text{.4}\times {\text{10}}^{-4}\text{=}\frac{{\text{x}}^2}{\text{(3}{\text{.7×10}}^{\text{-3}}\text{)-x}}$

The above equation, assume that the x present in $\text{(3}{\text{.7×10}}^{\text{-3}}\text{)-x}$ is very small than $\text{3}{\text{.7×10}}^{\text{-3}}$ then it can be negligible and as follows,

  $\begin{array}{l}\text{8}{\text{.4×10}}^{\text{-4}}\text{=}\frac{{\text{x}}^{\text{2}}}{\text{3}{\text{.7×10}}^{\text{-3}}}\\ \\ {\text{x}}^{\text{2}}\text{=}(\text{3}{\text{.7×10}}^{\text{-3}})\text{×(8}{\text{.4×10}}^{\text{-4}}\text{)}\\ \\ \text{x}\text{=}\sqrt{\text{(3}{\text{.7×10}}^{\text{-3}})\times (8.4\times {10}^{-4})}\\ \\ \text{=}1.76\times {\text{10}}^{-3}\text{(}\because \text{x}\text{=}{\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{])}\end{array}$

Now, the $\text{pH}$ of the solution is calculated using given equation.

  $\begin{array}{l}\text{pH}\text{=}{\text{-log[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{] }\\ \\ \text{=}\text{-log(1}\text{.76}\times {10}^{-4}\text{)}\\ \\ \text{=}3.75\end{array}$

Therefore, the calculated $\text{pH}$ value of $\text{3}{\text{.7×10}}^{\text{-3}}\text{M}$ aqueous lactic acid is 3.75.

The general equilibrium expression to find out the pOH of the solution is given below,

  $\begin{array}{l}\text{pH+pOH = 14}\\ \\ \text{ pOH = 14-pH}\\ \\ \text{ = 14-3}\text{.75 }\\ \\ \text{ = 10}\text{.25}\\ \end{array}$

Therefore, the calculated $\text{pOH}$ value of $\text{3}{\text{.7×10}}^{\text{-3}}\text{M}$ lactic acid is 10.25.

The percentage deprotonation is calculated using the concentration of hydronium ion divided by the initial concentration of lactic acid and the respective equation is given below.

  $\text{Percentage}\text{deprotonated}\text{=}\frac{\text{concentration}\text{of}{\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}}{\text{initial}\text{concentration}\text{of}{\text{[CH}}_{\text{3}}\text{CH(OH)COOH]}}\text{×100\%}$

Concentration of ${\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}$ is $1.76\times {10}^{-3}$ M

Initial concentration of ${\text{[CH}}_{\text{3}}\text{CH(OH)COOH]}$ is $\text{3}{\text{.7×10}}^{\text{-3}}\text{M}$

Substitute the obtained values in above equation

  $\begin{array}{l}\text{Percentage}\text{deprotonated =}\frac{1.76\times {10}^{-3}}{\text{3}{\text{.7×10}}^{\text{-3}}}\text{×100\%}\\ \\ \text{= 47}\text{.57\%}\end{array}$

Therefore, the percentage deprotonation of $\text{3}{\text{.7×10}}^{\text{-3}}\text{M}$ aqueous lactic acid is 47.57%

(c)

Interpretation Introduction

Interpretation:

The $\text{pH}$, pOH and percentage deprotonation of $\text{8}{\text{.2×10}}^{\text{-5}}\text{M}$ aqueous lactic acid has to be calculated.

Concept introduction:

Refer to part (a).

Answer to Problem 6D.2E

The $\text{pH}$, pOH and percentage deprotonation of $\text{8}{\text{.2×10}}^{\text{-5}}\text{M}$ aqueous lactic acid is 5.12, 8.88 and 92.80% respectively.

Explanation of Solution

Lactic acid is a weak acid when it is dissolved in water it ionized as positive and negative ions and it is given below.

  ${\text{CH}}_{\text{3}}{\text{COOH}}_{\text{(s)}}{\text{+H}}_{\text{2}}{\text{O}}_{\text{(l)}}\leftrightarrow {\text{CH}}_{\text{3}}{\text{COO}}^{\text{-}}{}_{\text{(aq)}}{\text{+H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\text{(aq)}}$

The equilibrium expression for the above reaction is given below.

  ${\text{K}}_{\text{a}}\text{=}\frac{{\text{[CH}}_{\text{3}}{\text{CH(OH)COO}}^{\text{-}}\text{]}{\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}}{{\text{[CH}}_{\text{3}}\text{CH(OH)COOH]}}$

	${\text{CH}}_{\text{3}}\text{CH(OH)COOH}$	${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$	${\text{CH}}_{\text{3}}{\text{CH(OH)COO}}^{\text{-}}$
Initial concentration	$\text{8}{\text{.2×10}}^{\text{-5}}$	0	0
Change in concentration	-x	+x	+x
Equilibrium concentration	$\text{8}{\text{.2×10}}^{\text{-5}}$-x	x	x

The equilibrium concentration values are obtained in the above table and is substituted in above equation and is given below.

  ${\text{K}}_{\text{a}}\text{=}\frac{\text{x}\times \text{x}}{\text{(8}{\text{.2×10}}^{\text{-5}}\text{)-x}}$

Lactic acid ${\text{K}}_{\text{a}}$ value is $8.4\times {10}^{-4}$ which is given in table 6C.1 and it is substituted in above equation and then the value of x is calculated.

  $\text{8}\text{.4}\times {\text{10}}^{-4}\text{=}\frac{{\text{x}}^2}{\text{(8}{\text{.2×10}}^{\text{-5}}\text{)-x}}$

The above equation, assume that the x present in $\text{(8}{\text{.2×10}}^{\text{-5}}\text{)-x}$ is very small than $\text{8}{\text{.2×10}}^{\text{-5}}$ then it can be negligible and as follows,

  $\begin{array}{l}\text{8}{\text{.4×10}}^{\text{-4}}\text{=}\frac{{\text{x}}^{\text{2}}}{\text{8}{\text{.2×10}}^{\text{-5}}}\\ \\ {\text{x}}^{\text{2}}\text{=}(\text{8}{\text{.2×10}}^{\text{-5}})\text{×(8}{\text{.4×10}}^{\text{-4}}\text{)}\\ \\ \text{x}\text{=}\sqrt{\text{(8}{\text{.2×10}}^{\text{-5}}\text{M})\times (8.4\times {10}^{-4})}\\ \\ \text{=}7.61\times {\text{10}}^{-6}\text{(}\because \text{x}\text{=}{\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{])}\end{array}$

Now, the $\text{pH}$ of the solution is calculated using given equation.

  $\begin{array}{l}\text{pH}\text{=}{\text{-log[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{] }\\ \\ \text{=}\text{-log(7}{\text{.61×10}}^{\text{-6}}\text{)}\\ \\ \text{=}\text{5}\text{.12}\end{array}$

Therefore, the calculated $\text{pH}$ value of $\text{8}{\text{.2×10}}^{\text{-5}}\text{M}$ aqueous lactic acid is 5.12.

The general equilibrium expression to find out the pOH of the solution is given below,

  $\begin{array}{l}\text{pH+pOH = 14}\\ \\ \text{ pOH = 14-pH}\\ \\ \text{ = 14-5}\text{.12 }\\ \\ \text{ = 8}\text{.88}\\ \end{array}$

Therefore, the calculated $\text{pOH}$ value of $\text{8}{\text{.2×10}}^{\text{-5}}\text{M}$ lactic acid is 8.88.

The percentage deprotonation is calculated using the concentration of hydronium ion divided by the initial concentration of lactic acid and the respective equation is given below.

  $\text{Percentage}\text{deprotonated}\text{=}\frac{\text{concentration}\text{of}{\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}}{\text{initial}\text{concentration}\text{of}{\text{[CH}}_{\text{3}}\text{CH(OH)COOH]}}\text{×100\%}$

Concentration of ${\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}$ is $\text{7}{\text{.61×10}}^{\text{-6}}$ M

Initial concentration of ${\text{[CH}}_{\text{3}}\text{CH(OH)COOH]}$ is $\text{8}{\text{.2×10}}^{\text{-5}}\text{M}$

Substitute the obtained values in above equation

  $\begin{array}{l}\text{Percentage}\text{deprotonated =}\frac{7.61\times {10}^{-5}}{\text{8}{\text{.2×10}}^{\text{-5}}}\text{×100\%}\\ \\ \text{= 92}\text{.80\%}\end{array}$

Therefore, the percentage deprotonation of $\text{8}{\text{.2×10}}^{\text{-5}}\text{M}$ aqueous lactic acid is 92.80%