Chapter 6, Problem 6D.1E

(a)

Interpretation Introduction

Interpretation:

The $\text{pH}$, pOH and percentage deprotonation of 0.20 M aqueous ${\text{CH}}_{\text{3}}\text{COOH}$ 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.1E

The $\text{pH}$, pOH and percentage deprotonation of 0.20 M aqueous ${\text{CH}}_{\text{3}}\text{COOH}$ is 2.72, 11.28 and 0.95% respectively.

Explanation of Solution

Acetic 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.

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

	${\text{CH}}_{\text{3}}\text{COOH}$	${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$	${\text{CH}}_{\text{3}}{\text{COO}}^{\text{-}}$
Initial concentration	0.20	0	0
Change in concentration	-x	+x	+x
Equilibrium concentration	0.20-x	x	x

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

  ${\text{K}}_{\text{a}}\text{=}\frac{\text{x}\times \text{x}}{\text{0}\text{.20-x}}$

Acetic acid ${\text{K}}_{\text{a}}$ value is $1.8\times {10}^{-5}$ which is given in table 6C.1 and it is substituted in above equation and then the value of x is calculated.

  $\text{1}\text{.8}\times {\text{10}}^{-5}\text{=}\frac{{\text{x}}^2}{\text{0}\text{.20-x}}$

The above equation, assume that the x present in 0.20-x is very small than 0.20 then it can be negligible and as follows,

  $\begin{array}{l}\text{1}{\text{.8×10}}^{\text{-5}}\text{=}\frac{{\text{x}}^{\text{2}}}{\text{0}\text{.20}}\\ \\ {\text{x}}^{\text{2}}\text{=}\text{0}\text{.20×(1}{\text{.8×10}}^{\text{-5}}\text{)}\\ \\ \text{x}\text{=}\sqrt{0.20\times (1.8\times {10}^{-5})}\\ \\ \text{=}1.9\times {\text{10}}^{-3}\text{(}\because \text{x}\text{=}{\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{])}\end{array}$

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

  $\begin{array}{l}\text{pH}\text{=}{\text{-log[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{] }\\ \\ \text{=}\text{-log(1}\text{.9}\times {10}^{-3}\text{)}\\ \\ \text{=}2.72\end{array}$

Therefore, the calculated $\text{pH}$ value of 0.20 M acetic acid is 2.72.

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

  $\begin{array}{l}\text{pH+pOH = 14}\\ \\ \text{ pOH = 14-pH}\\ \\ \text{ = 14-2}\text{.72 }\\ \\ \text{ = 11}\text{.28}\\ \end{array}$

Therefore, the calculated $\text{pOH}$ value of 0.20 M acetic acid is 11.28.

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.

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

Concentration of ${\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}$ is $1.9\times {10}^{-3}$ M

Initial concentration of ${\text{[CH}}_{\text{3}}\text{COOH]}$ is $\text{0}\text{.20}\text{M}$

Substitute the obtained values in above equation

  $\begin{array}{l}\text{Percentage}\text{deprotonated =}\frac{1.9\times {10}^{-3}}{0.20}\text{×100\%}\\ \\ \text{= 0}\text{.95\%}\end{array}$

Therefore, the percentage deprotonation of 0.20 M aqueous acetic acid is 0.95%

(b)

Interpretation Introduction

Interpretation:

The $\text{pH}$, pOH and percentage deprotonation of 0.20 M aqueous ${\text{CCl}}_{\text{3}}\text{COOH}$ has to be calculated.

Concept introduction:

Refer to part (a).

Answer to Problem 6D.1E

The $\text{pH}$, pOH and percentage deprotonation of 0.20 M aqueous ${\text{CCl}}_{\text{3}}\text{COOH}$ is 0.61, 13.39 and 122.5% respectively.

Explanation of Solution

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

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

The equilibrium expression for the above reaction is given below.

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

	${\text{CCl}}_{\text{3}}\text{COOH}$	${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$	${\text{CCl}}_{\text{3}}{\text{COO}}^{\text{-}}$
Initial concentration	0.20	0	0
Change in concentration	-x	+x	+x
Equilibrium concentration	0.20-x	x	x

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

  ${\text{K}}_{\text{a}}\text{=}\frac{\text{x}\times \text{x}}{\text{0}\text{.20-x}}$

Trichloroacetic acid ${\text{K}}_{\text{a}}$ value is $3.0\times {10}^{-1}$ which is given in table 6C.1 and it is substituted in above equation and then the value of x is calculated.

  $3.0\times {\text{10}}^{-1}\text{=}\frac{{\text{x}}^2}{\text{0}\text{.20-x}}$

The above equation, assume that the x present in 0.20-x is very small than 0.20 then it can be negligible and as follows,

  $\begin{array}{l}\text{3}{\text{.0×10}}^{\text{-1}}\text{=}\frac{{\text{x}}^{\text{2}}}{\text{0}\text{.20}}\\ \\ {\text{x}}^{\text{2}}\text{=}\text{0}\text{.20×(3}{\text{.0×10}}^{\text{-1}}\text{)}\\ \\ \text{x}\text{=}\sqrt{0.20\times (3.0\times {10}^{-1})}\\ \\ \text{=}0.\text{245}\text{(}\because \text{x}\text{=}{\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{])}\end{array}$

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

  $\begin{array}{l}\text{pH}\text{=}{\text{-log[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{] }\\ \\ \text{=}\text{-log(0}\text{.245)}\\ \\ \text{=}0.61\end{array}$

Therefore, the calculated $\text{pH}$ value of 0.20 M trichloroacetic acid is 0.611.

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

  $\begin{array}{l}\text{pH+pOH = 14}\\ \\ \text{ pOH = 14-pH}\\ \\ \text{ = 14-0}\text{.61 }\\ \\ \text{ = 13}\text{.39}\\ \end{array}$

Therefore, the calculated $\text{pOH}$ value of 0.20 M trichloroacetic acid is 13.39.

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.

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

Concentration of ${\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}$ is $\text{0}\text{.245}\text{M}$ M

Initial concentration of ${\text{[CCl}}_{\text{3}}\text{COOH]}$ is $\text{0}\text{.20}\text{M}$

Substitute the obtained values in above equation

  $\begin{array}{l}\text{Percentage}\text{deprotonated =}\frac{0.245}{0.20}\text{×100\%}\\ \\ \text{= 122}\text{.5\%}\end{array}$

Therefore, the percentage deprotonation of 0.20 M aqueous trichloroacetic acid is 122.5%

(c)

Interpretation Introduction

Interpretation:

The $\text{pH}$, pOH and percentage deprotonation of 0.20 M aqueous $\text{HCOOH}$ has to be calculated.

Concept introduction:

Refer to part (a).

Answer to Problem 6D.1E

The $\text{pH}$, pOH and percentage deprotonation of 0.20 M aqueous $\text{HCOOH}$ is 2.22, 11.78 and 3.0% respectively.

Explanation of Solution

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

  ${\text{HCOOH}}_{\text{(s)}}{\text{+H}}_{\text{2}}{\text{O}}_{\text{(l)}}\rightleftharpoons {\text{HCOO}}^{\text{-}}{}_{\text{(aq)}}{\text{+H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\text{(aq)}}$

The equilibrium expression for the above reaction is given below.

  ${\text{K}}_{\text{a}}\text{=}\frac{{\text{[HCOO}}^{\text{-}}\text{]}{\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}}{\text{[HCOOH]}}$

	$\text{HCOOH}$	${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$	${\text{HCOO}}^{\text{-}}$
Initial concentration	0.20	0	0
Change in concentration	-x	+x	+x
Equilibrium concentration	0.20-x	x	x

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

  ${\text{K}}_{\text{a}}\text{=}\frac{\text{x}\times \text{x}}{\text{0}\text{.20-x}}$

Formic acid ${\text{K}}_{\text{a}}$ value is $1.8\times {10}^{-4}$ which is given in table 6C.1 and it is substituted in above equation and then the value of x is calculated.

  $1.8\times {\text{10}}^{-4}\text{=}\frac{{\text{x}}^2}{\text{0}\text{.20-x}}$

The above equation, assume that the x present in 0.20-x is very small than 0.20 then it can be negligible and as follows,

  $\begin{array}{l}\text{1}{\text{.8×10}}^{\text{-4}}\text{=}\frac{{\text{x}}^{\text{2}}}{\text{0}\text{.20}}\\ \\ {\text{x}}^{\text{2}}\text{=}\text{0}\text{.20×(1}{\text{.8×10}}^{\text{-4}}\text{)}\\ \\ \text{x}\text{=}\sqrt{0.20\times (1.8\times {10}^{-4})}\\ \\ \text{=}0.006\text{(}\because \text{x}\text{=}{\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{])}\end{array}$

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

  $\begin{array}{l}\text{pH}\text{=}{\text{-log[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{] }\\ \\ \text{=}\text{-log(0}\text{.006)}\\ \\ \text{=}2.22\end{array}$

Therefore, the calculated $\text{pH}$ value of 0.20M formic acid is 2.22.

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

  $\begin{array}{l}\text{pH+pOH = 14}\\ \\ \text{ pOH = 14-pH}\\ \\ \text{ = 14-2}\text{.22 }\\ \\ \text{ = 11}\text{.78}\\ \end{array}$

Therefore, the calculated $\text{pOH}$ value of 0.20 M formic acid is 11.78.

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.

  $\text{Percentage}\text{deprotonated}\text{=}\frac{\text{concentration}\text{of}{\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}}{\text{initial}\text{concentration}\text{of}\text{[HCOOH]}}\text{×100\%}$

Concentration of ${\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}$ is $\text{0}\text{.006 M}$ M

Initial concentration of $\text{[HCOOH]}$ is $\text{0}\text{.20}\text{M}$

Substitute the obtained values in above equation

  $\begin{array}{l}\text{Percentage}\text{deprotonated =}\frac{0.006}{0.20}\text{×100\%}\\ \\ \text{= 3}\text{.0\%}\end{array}$

Therefore, the percentage deprotonation of 0.20 M aqueous formic acid is 3.0%

(d)

Interpretation Introduction

Interpretation:

The $\text{pH}$ differences on the basis of molecular has to be explained.

Concept introduction:

Electronegativity:

The electronegativity is a chemical property; it is measured by an atom attract the bonding pair of electrons towards itself.

Answer to Problem 6D.1E

The more electronegative groups present in the molecular structure makes the solution more acidic. Based on the molecular structure the increasing order of pH is given here, (b)<(c)<(a).

Explanation of Solution

pH of the solution is depending on the atoms or group attached to the carboxyl group.  Here, three different groups attached with carboxyl group such as ${\text{-CH}}_{\text{3}}\text{,}{\text{-CCl}}_{\text{3}}\text{and}\text{-H}$.  The chlorine is more electronegative than hydrogen and ${\text{-CH}}_{\text{3}}$ groups.  The ${\text{-CCl}}_{\text{3}}$ group present in trichloroacetic acid is more electron withdrawing group, so it withdraws electrons from carbon and oxygen present in the carboxyl group. Finally the $\text{O-H}$ bond gets weakened and it easily release ${\text{H}}^{\text{+}}$ ion. If the ${\text{H}}^{\text{+}}$ ion get removed easily then the acid strength increased. On another side the carboxylate ion stability is increased, therefore the acid strength also increased. The acid strength is also depending on the inductive effect of the atoms or group. Here, the chlorine has more inductive effect than hydrogen and ${\text{-CH}}_{\text{3}}$ group. Therefore, the acid strength also follows same order. Hence, the trichloroacetic acid is stronger than formic acid and acetic acid.

Based on the molecular structure increasing order of pH is given below.

(b)< (c) < (a)