Chapter 6, Problem 6D.5E

(a)

Interpretation Introduction

Interpretation:

The $\text{pH}$, pOH and percentage deprotonation of 0.057 M aqueous ammonia 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.5E

The $\text{pH}$, pOH and percentage deprotonation of 0.057 M aqueous ammonia is 11.0, 3.0 and 1.77% respectively.

Explanation of Solution

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

  ${\text{NH}}_3{}_{\text{(aq)}}{\text{+H}}_{\text{2}}{\text{O}}_{\text{(l)}}\rightleftharpoons {\text{NH}}_4{{}^{\text{+}}}_{\text{(aq)}}{\text{+OH}}^{-}{}_{\text{(aq)}}$

The equilibrium expression for the above reaction is given below.

  ${\text{K}}_{\text{b}}\text{=}\frac{{\text{[NH}}_4{}^{\text{+}}\text{]}{\text{[OH}}^{-}\text{]}}{{\text{[NH}}_{\text{3}}\text{]}}$

	${\text{NH}}_{\text{3}}$	${\text{NH}}_4{}^{\text{+}}$	${\text{OH}}^{\text{-}}$
Initial concentration	0.057	0	0
Change in concentration	-x	+x	+x
Equilibrium concentration	0.057-x	x	x

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

  ${\text{K}}_{\text{b}}\text{=}\frac{\text{x×x}}{\text{0}\text{.057-x}}$

Ammonia ${\text{K}}_{\text{b}}$ value is $\text{1}{\text{.8×10}}^{\text{-5}}$ which is given in table 6C.2 and it is substituted in above equation and then the value of x is calculated.  Assume that the x present in 0.057-x is very small than 0.057 then it can be negligible and as follows,

  $\begin{array}{l}\text{1}{\text{.8×10}}^{\text{-5}}\text{=}\frac{{\text{x}}^{\text{2}}}{\text{0}\text{.057}}\\ \\ {\text{x}}^{\text{2}}\text{=}\text{0}\text{.057×(1}{\text{.8×10}}^{\text{-5}}\text{)}\\ \\ \text{x}\text{=}\sqrt{\text{0}\text{.057×(1}{\text{.8×10}}^{\text{-5}}\text{)}}\\ \\ \text{=}\text{1}{\text{.01×10}}^{\text{-3}}\text{(}\because \text{x}\text{=}{\text{[OH}}^{\text{-}}\text{])}\end{array}$

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

  $\begin{array}{l}\text{pOH}\text{=}{\text{-log[OH}}^{-}\text{] }\\ \\ \text{=}\text{-log(1}\text{.01}\times {10}^{-3}\text{)}\\ \\ \text{=}3.0\end{array}$

Therefore, the calculated $\text{pOH}$ value of 0.057 M aqueous ammonia is 3.0.

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

  $\begin{array}{l}\text{pH+pOH = 14}\\ \\ \text{ pH = 14-pOH}\\ \\ \text{ = 14-3}\text{.0 }\\ \\ \text{ = 11}\text{.0}\\ \end{array}$

Therefore, the calculated $\text{pH}$ value of 0.057 M aqueous ammonia is 11.0.

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.

  $\text{Percentage}\text{deprotonated}\text{=}\frac{\text{concentration}\text{of}{\text{[OH}}^{-}\text{]}}{\text{initial}\text{concentration}\text{of}{\text{[NH}}_3\text{]}}\text{×100\%}$

Concentration of ${\text{[OH}}^{-}\text{]}$ is $1.01\times {10}^{-3}$ M

Initial concentration of ${\text{[NH}}_{\text{3}}\text{]}$ is $\text{0}\text{.057}\text{M}$

Substitute the obtained values in above equation

  $\begin{array}{l}\text{Percentage}\text{deprotonated =}\frac{1.01\times {10}^{-3}}{0.057}\text{×100\%}\\ \\ \text{= 1}\text{.77\%}\end{array}$

Therefore, the percentage deprotonation of 0.057 M aqueous ammonia is 1.77%

(b)

Interpretation Introduction

Interpretation:

The $\text{pH}$, pOH and percentage deprotonation of $\text{0}\text{.162}\text{M}$ aqueous hydroxylamine has to be calculated.

Concept introduction:

Refer to part (a).

Answer to Problem 6D.5E

The $\text{pH}$, pOH and percentage deprotonation of $\text{0}\text{.162}\text{M}$ aqueous hydroxylamine is 9.63, 4.37 and 0.026% respectively.

Explanation of Solution

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

  ${\text{NH}}_{\text{2}}{\text{OH}}_{\text{(aq)}}{\text{+H}}_{\text{2}}{\text{O}}_{\text{(l)}}\rightleftharpoons {\text{NH}}_{\text{3}}{\text{OH}}^{\text{+}}{}_{\text{(aq)}}{\text{+OH}}^{\text{-}}{}_{\text{(aq)}}$

The equilibrium expression for the above reaction is given below.

  ${\text{K}}_{\text{b}}\text{=}\frac{{\text{[NH}}_{\text{3}}{\text{OH}}^{\text{+}}\text{]}{\text{[OH}}^{\text{-}}\text{]}}{{\text{[NH}}_{\text{2}}\text{OH]}}$

	${\text{NH}}_{\text{2}}\text{OH}$	${\text{NH}}_{\text{3}}{\text{OH}}^{\text{+}}$	${\text{OH}}^{\text{-}}$
Initial concentration	0.162	0	0
Change in concentration	-x	+x	+x
Equilibrium concentration	0.162-x	x	x

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

  ${\text{K}}_{\text{b}}\text{=}\frac{\text{x×x}}{\text{0}\text{.162-x}}$

Hydroxylamine ${\text{K}}_{\text{b}}$ value is $\text{1}{\text{.1×10}}^{\text{-8}}$ which is given in table 6C.2 and it is substituted in above equation and then the value of x is calculated.  Assume that the x present in 0.162-x is very small than 0.162 then it can be negligible and as follows,

  $\begin{array}{l}\text{1}{\text{.1×10}}^{\text{-8}}\text{=}\frac{{\text{x}}^{\text{2}}}{\text{0}\text{.162}}\\ \\ {\text{x}}^{\text{2}}\text{=}\text{0}\text{.162×(1}{\text{.1×10}}^{\text{-8}}\text{)}\\ \\ \text{x}\text{=}\sqrt{\text{0}\text{.162×(1}{\text{.1×10}}^{\text{-8}}\text{)}}\\ \\ \text{=}4.22{\text{×10}}^{\text{-5}}\text{(}\because \text{x}\text{=}{\text{[OH}}^{\text{-}}\text{])}\end{array}$

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

  $\begin{array}{l}\text{pOH}\text{=}{\text{-log[OH}}^{-}\text{] }\\ \\ \text{=}\text{-log(4}\text{.22}\times {10}^{-5}\text{)}\\ \\ \text{=}4.37\end{array}$

Therefore, the calculated $\text{pOH}$ value of 0.162 M aqueous hydroxylamine is 4.37.

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

  $\begin{array}{l}\text{pH+pOH = 14}\\ \\ \text{ pH = 14-pOH}\\ \\ \text{ = 14-4}\text{.37 }\\ \\ \text{ = 9}\text{.63}\\ \end{array}$

Therefore, the calculated $\text{pH}$ value of 0.162 M aqueous hydroxylamine is 9.63.

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.

  $\text{Percentage}\text{deprotonated}\text{=}\frac{\text{concentration}\text{of}{\text{[OH}}^{-}\text{]}}{\text{initial}\text{concentration}\text{of}{\text{[NH}}_2\text{OH]}}\text{×100\%}$

Concentration of ${\text{[OH}}^{-}\text{]}$ is $4.22\times {10}^{-5}$ M

Initial concentration of ${\text{[NH}}_2\text{OH]}$ is $\text{0}\text{.162}\text{M}$

Substitute the obtained values in above equation

  $\begin{array}{l}\text{Percentage}\text{deprotonated =}\frac{4.22\times {10}^{-5}}{0.162}\text{×100\%}\\ \\ \text{= 0}\text{.026\%}\end{array}$

Therefore, the percentage deprotonation of 0.162 M aqueous hydroxylamine is 0.026%

(c)

Interpretation Introduction

Interpretation:

The $\text{pH}$, pOH and percentage deprotonation of $\text{0}\text{.35}\text{M}$ aqueous trimethylamine has to be calculated.

Concept introduction:

Refer to part (a).

Answer to Problem 6D.5E

The $\text{pH}$, pOH and percentage deprotonation of $\text{0}\text{.35}\text{M}$ aqueous trimethylamine is 11.68, 2.32 and 1.36% respectively.

Explanation of Solution

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

  ${{\text{(CH}}_{\text{3}}\text{)}}_{\text{3}}{\text{N}}_{\text{(aq)}}{\text{+H}}_{\text{2}}{\text{O}}_{\text{(l)}}\rightleftharpoons {{\text{(CH}}_{\text{3}}\text{)}}_{\text{3}}{\text{NH}}^{\text{+}}{}_{\text{(aq)}}{\text{+OH}}^{\text{-}}{}_{\text{(aq)}}$

The equilibrium expression for the above reaction is given below.

  ${\text{K}}_{\text{b}}\text{=}\frac{{\text{[(CH}}_3{\text{)}}_3{\text{NH}}^{\text{+}}\text{]}{\text{[OH}}^{\text{-}}\text{]}}{{\text{[(CH}}_3{\text{)}}_3\text{N]}}$

	${{\text{(CH}}_3\text{)}}_3\text{N}$	${{\text{(CH}}_3\text{)}}_3{\text{NH}}^{\text{+}}$	${\text{OH}}^{\text{-}}$
Initial concentration	0.35	0	0
Change in concentration	-x	+x	+x
Equilibrium concentration	0.35-x	x	x

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

  ${\text{K}}_{\text{b}}\text{=}\frac{\text{x×x}}{\text{0}\text{.35-x}}$

Trimethylamine ${\text{K}}_{\text{b}}$ value is $\text{6}{\text{.5×10}}^{\text{-5}}$ which is given in table 6C.2 and it is substituted in above equation and then the value of x is calculated.  Assume that the x present in 0.35-x is very small than 0.35 then it can be negligible and as follows,

  $\begin{array}{l}\text{6}{\text{.5×10}}^{\text{-5}}\text{=}\frac{{\text{x}}^{\text{2}}}{\text{0}\text{.35}}\\ \\ {\text{x}}^{\text{2}}\text{=}\text{0}\text{.35×(6}{\text{.5×10}}^{\text{-5}}\text{)}\\ \\ \text{x}\text{=}\sqrt{\text{0}\text{.35×(6}{\text{.5×10}}^{\text{-5}}\text{)}}\\ \\ \text{=}4.77{\text{×10}}^{\text{-3}}\text{(}\because \text{x}\text{=}{\text{[OH}}^{\text{-}}\text{])}\end{array}$

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

  $\begin{array}{l}\text{pOH}\text{=}{\text{-log[OH}}^{-}\text{] }\\ \\ \text{=}\text{-log(4}\text{.77}\times {10}^{-3}\text{)}\\ \\ \text{=}2.32\end{array}$

Therefore, the calculated $\text{pOH}$ value of 0.35 M aqueous trimethylamine is 2.32.

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

  $\begin{array}{l}\text{pH+pOH = 14}\\ \\ \text{ pH = 14-pOH}\\ \\ \text{ = 14-2}\text{.32 }\\ \\ \text{ = 11}\text{.68}\\ \end{array}$

Therefore, the calculated $\text{pH}$ value of 0.35 M aqueous trimethylamine is 11.68.

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.

  $\text{Percentage}\text{deprotonated}\text{=}\frac{\text{concentration}\text{of}{\text{[OH}}^{-}\text{]}}{\text{initial}\text{concentration}\text{of}{\text{[(CH}}_3{\text{)}}_3\text{N]}}\text{×100\%}$

Concentration of ${\text{[OH}}^{-}\text{]}$ is $4.77\times {10}^{-3}$ M

Initial concentration of ${\text{[NH}}_2\text{OH]}$ is $\text{0}\text{.35}\text{M}$

Substitute the obtained values in above equation

  $\begin{array}{l}\text{Percentage}\text{deprotonated =}\frac{4.77\times {10}^{-3}}{0.35}\text{×100\%}\\ \\ \text{= 1}\text{.36\%}\end{array}$

Therefore, the percentage deprotonation of 0.35 M aqueous trimethylamine is 1.36%

(d)

Interpretation Introduction

Interpretation:

The $\text{pH}$, pOH and percentage deprotonation of $\text{0}\text{.0073}\text{M}$ aqueous codeine has to be calculated using the $\text{pKa}$ value of 8.21.

Concept introduction:

Refer to part (a).

Answer to Problem 6D.5E

The $\text{pH}$, pOH and percentage deprotonation of $\text{0}\text{.0073}\text{M}$ aqueous codeine is 5.17, 8.83 and 0.092% respectively.

Explanation of Solution

Codeine is dissolved in water it ionized as positive and negative ions and it is given below.

  ${\text{C}}_{18}{\text{H}}_{21}{\text{NO}}_3{}_{\text{(aq)}}{\text{+H}}_{\text{2}}{\text{O}}_{\text{(l)}}\rightleftharpoons {\text{C}}_{18}{\text{H}}_{20}{\text{NO}}_3{{}^{-}}_{\text{(aq)}}{\text{+H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\text{(aq)}}$

The equilibrium expression for the above reaction is given below.

  ${\text{K}}_{\text{a}}\text{=}\frac{{\text{[C}}_{\text{18}}{\text{H}}_{\text{20}}{\text{NO}}_{\text{3}}{}^{\text{-}}\text{]}{\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}}{{\text{[C}}_{\text{18}}{\text{H}}_{\text{21}}{\text{NO}}_{\text{3}}\text{]}}$

	${\text{C}}_{18}{\text{H}}_{21}{\text{NO}}_3$	${\text{C}}_{18}{\text{H}}_{20}{\text{NO}}_3{}^{-}$	${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}$
Initial concentration	0.0073	0	0
Change in concentration	-x	+x	+x
Equilibrium concentration	0.0073-x	x	x

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

  ${\text{K}}_{\text{a}}\text{=}\frac{\text{x×x}}{\text{0}\text{.0073-x}}$

The $\text{pKa}$ value of codeine is 8.21 using this we can calculate the ${\text{K}}_{\text{a}}$ value and the following equation is given below,

  $\begin{array}{l}{\text{pK}}_{\text{a}}\text{=}{\text{-log(k}}_{\text{a}}\text{)}\\ \\ {\text{K}}_{\text{a}}\text{=}{\text{10}}^{{\text{-pK}}_{\text{a}}}\\ \\ ={10}^{-8.21}\\ \\ =6.17\times {10}^{-9}\end{array}$

  Therefore, the ${\text{K}}_{\text{a}}$ value of codeine is $\text{6}{\text{.17×10}}^{\text{-9}}$.

  The concentration of hydronium ion is calculated using the given formula as,

  ${\text{K}}_{\text{a}}\text{=}\frac{\text{x×x}}{\text{0}\text{.0073-x}}$

The obtained codeine ${\text{K}}_{\text{a}}$ ($\text{6}{\text{.17×10}}^{\text{-9}}$) value is substituted in the above equation and then the value of x is calculated.  Assume that the x present in 0.0073-x is very small than 0.0073 then it can be negligible and as follows,

  $\begin{array}{l}\text{6}{\text{.17×10}}^{\text{-9}}\text{=}\frac{{\text{x}}^{\text{2}}}{\text{0}\text{.0073}}\\ \\ {\text{x}}^{\text{2}}\text{=}\text{0}\text{.0073×(6}{\text{.17×10}}^{\text{-9}}\text{)}\\ \\ \text{x}\text{=}\sqrt{\text{0}\text{.0073×(6}{\text{.17×10}}^{\text{-9}}\text{)}}\\ \\ \text{=}\text{6}{\text{.71×10}}^{\text{-6}}\text{(}\because \text{x}\text{=}{\text{[H}}_{\text{3}}{\text{O}}^{+}\text{])}\end{array}$

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

  $\begin{array}{l}\text{pH}\text{=}{\text{-log[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{] }\\ \\ \text{=}\text{-log(6}{\text{.71×10}}^{\text{-6}}\text{)}\\ \\ \text{=}\text{5}\text{.17}\end{array}$

Therefore, the calculated $\text{pH}$ value of 0.0073 M aqueous codeine is 5.17.

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

  $\begin{array}{l}\text{pH+pOH = 14}\\ \\ \text{ pOH = 14-pH}\\ \\ \text{ = 14-5}\text{.17 }\\ \\ \text{ = 8}\text{.83}\\ \end{array}$

Therefore, the calculated $\text{pOH}$ value of 0.0073 M aqueous codeine is 8.83.

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.

  $\text{Percentage}\text{deprotonated}\text{=}\frac{\text{concentration}\text{of}{\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}}{\text{initial}\text{concentration}\text{of}{\text{[C}}_{\text{18}}{\text{H}}_{\text{21}}{\text{NO}}_{\text{3}}\text{]}}\text{×100\%}$

Concentration of ${\text{[H}}_{\text{3}}{\text{O}}^{\text{+}}\text{]}$ is $\text{6}{\text{.71×10}}^{\text{-6}}\text{M}$

Initial concentration of ${\text{[C}}_{18}{\text{H}}_{21}{\text{NO}}_3\text{]}$ is $\text{0}\text{.0073}\text{M}$

Substitute the obtained values in above equation

  $\begin{array}{l}\text{Percentage}\text{deprotonated =}\frac{6.71\times {10}^{-6}}{0.0073}\text{×100\%}\\ \\ \text{= 0}\text{.092\%}\end{array}$

Therefore, the percentage deprotonation of 0.0073 M aqueous codeine is 0.092%