Chapter 6, Problem 6D.6E

(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.6E

The $\text{pH}$, pOH and percentage deprotonation of 0.082 M aqueous pyridine is 9.08, 34.92 and 0.015% respectively.

Explanation of Solution

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

  ${\text{C}}_5{\text{H}}_{\text{5}}{\text{N}}_{\text{(aq)}}{\text{+H}}_{\text{2}}{\text{O}}_{\text{(l)}}\rightleftharpoons {\text{C}}_5{\text{H}}_{\text{5}}{\text{NH}}^{\text{+}}{}_{\text{(aq)}}{\text{+OH}}^{\text{-}}{}_{\text{(aq)}}$

The equilibrium expression for the above reaction is given below.

  ${\text{K}}_{\text{b}}\text{=}\frac{{\text{[C}}_5{\text{H}}_{\text{5}}{\text{NH}}^{\text{+}}\text{]}{\text{[OH}}^{-}\text{]}}{{\text{[C}}_5{\text{H}}_{\text{5}}\text{N]}}$

	${\text{C}}_5{\text{H}}_{\text{5}}\text{N}$	${\text{C}}_5{\text{H}}_{\text{5}}{\text{NH}}^{\text{+}}$	${\text{OH}}^{\text{-}}$
Initial concentration	0.082	0	0
Change in concentration	-x	+x	+x
Equilibrium concentration	0.082-x	x	x

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

  ${\text{K}}_{\text{b}}\text{=}\frac{\text{x×x}}{\text{0}\text{.082-x}}$

Pyridine ${\text{K}}_{\text{b}}$ value is $\text{1}{\text{.8×10}}^{\text{-9}}$ 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 0.082-x is very small than 0.082 then it can be negligible and as follows,

  $\begin{array}{l}\text{1}{\text{.8×10}}^{\text{-9}}\text{=}\frac{{\text{x}}^{\text{2}}}{\text{0}\text{.082}}\\ \\ {\text{x}}^{\text{2}}\text{=}\text{0}\text{.082×(1}{\text{.8×10}}^{\text{-9}}\text{)}\\ \\ \text{x}\text{=}\sqrt{\text{0}\text{.082×(1}{\text{.8×10}}^{\text{-9}}\text{)}}\\ \\ \text{=}\text{1}{\text{.21×10}}^{\text{-5}}\text{(}\because \text{x}\text{=}{\text{[OH}}^{\text{-}}\text{])}\end{array}$

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

  $\begin{array}{l}\text{pOH}\text{=}{\text{-log[OH}}^{-}\text{] }\\ \\ \text{=}\text{-log(1}\text{.21}\times {10}^{-5}\text{)}\\ \\ \text{=}4.92\end{array}$

Therefore, the calculated $\text{pOH}$ value of 0.082 M aqueous pyridine is 4.92.

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

  $\begin{array}{l}\text{pH+pOH = 14}\\ \\ \text{ pH = 14-pOH}\\ \\ \text{ = 14-4}\text{.92 }\\ \\ \text{ = 9}\text{.08}\\ \end{array}$

Therefore, the calculated $\text{pH}$ value of 0.082 M aqueous pyridine is 9.08.

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.

  $\text{Percentage}\text{deprotonated}\text{=}\frac{\text{concentration}\text{of}{\text{[OH}}^{-}\text{]}}{\text{initial}\text{concentration}\text{of}{\text{[C}}_5{\text{H}}_5\text{N]}}\text{×100\%}$

Concentration of ${\text{[OH}}^{-}\text{]}$ is $1.21\times {10}^{-5}$ M

Initial concentration of ${\text{[C}}_5{\text{H}}_5\text{N]}$ is $\text{0}\text{.082}\text{M}$

Substitute the obtained values in above equation

  $\begin{array}{l}\text{Percentage}\text{deprotonated =}\frac{1.21\times {10}^{-5}}{0.082}\text{×100\%}\\ \\ \text{= 0}\text{.015\%}\end{array}$

Therefore, the percentage deprotonation of 0.082 M aqueous pyridine is 0.015%

(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.6E

The $\text{pH}$, pOH and percentage deprotonation of $\text{0}\text{.0103}\text{M}$ aqueous nicotine is 10.03, 3.97 and 1.03% 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.

  ${\text{C}}_{\text{10}}{\text{H}}_{\text{14}}{\text{N}}_{\text{2}}{}_{\text{(aq)}}{\text{+H}}_{\text{2}}{\text{O}}_{\text{(l)}}\rightleftharpoons {\text{C}}_{\text{10}}{\text{H}}_{\text{14}}{\text{N}}_{\text{2}}{\text{H}}^{\text{+}}{}_{\text{(aq)}}{\text{+OH}}^{\text{-}}{}_{\text{(aq)}}$

The equilibrium expression for the above reaction is given below.

  ${\text{K}}_{\text{b}}\text{=}\frac{{\text{[C}}_{10}{\text{H}}_{14}{\text{N}}_2{\text{H}}^{\text{+}}\text{]}{\text{[OH}}^{\text{-}}\text{]}}{{\text{[C}}_{10}{\text{H}}_{14}{\text{N}}_2\text{]}}$

	${\text{C}}_{10}{\text{H}}_{14}{\text{N}}_2$	${\text{C}}_{10}{\text{H}}_{14}{\text{N}}_2{\text{H}}^{\text{+}}$	${\text{OH}}^{\text{-}}$
Initial concentration	0.0103	0	0
Change in concentration	-x	+x	+x
Equilibrium concentration	0.0103-x	x	x

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

  ${\text{K}}_{\text{b}}\text{=}\frac{\text{x×x}}{\text{0}\text{.0103-x}}$

Nicotine ${\text{K}}_{\text{b}}$ value is $\text{1}{\text{.0×10}}^{\text{-6}}$ 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 0.0103-x is very small than 0.0103 then it can be negligible and as follows,

  $\begin{array}{l}\text{1}{\text{.1×10}}^{\text{-6}}\text{=}\frac{{\text{x}}^{\text{2}}}{\text{0}\text{.0103}}\\ \\ {\text{x}}^{\text{2}}\text{=}\text{0}\text{.0103×(1}{\text{.1×10}}^{\text{-6}}\text{)}\\ \\ \text{x}\text{=}\sqrt{\text{0}\text{.0103×(1}{\text{.1×10}}^{\text{-6}}\text{)}}\\ \\ \text{=}1.06{\text{×10}}^{\text{-4}}\text{(}\because \text{x}\text{=}{\text{[OH}}^{\text{-}}\text{])}\end{array}$

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

  $\begin{array}{l}\text{pOH}\text{=}{\text{-log[OH}}^{-}\text{] }\\ \\ \text{=}\text{-log(1}\text{.06}\times {10}^{-4}\text{)}\\ \\ \text{=}3.97\end{array}$

Therefore, the calculated $\text{pOH}$ value of 0.0103 M aqueous nicotine is 3.97.

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

  $\begin{array}{l}\text{pH+pOH = 14}\\ \\ \text{ pH = 14-pOH}\\ \\ \text{ = 14-3}\text{.97 }\\ \\ \text{ = 10}\text{.03}\\ \end{array}$

Therefore, the calculated $\text{pH}$ value of 0.0103 M aqueous nicotine is 10.03.

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.

  $\text{Percentage}\text{deprotonated}\text{=}\frac{\text{concentration}\text{of}{\text{[OH}}^{-}\text{]}}{\text{initial}\text{concentration}\text{of}{\text{[C}}_{10}{\text{H}}_{14}{\text{N}}_2\text{]}}\text{×100\%}$

Concentration of ${\text{[OH}}^{-}\text{]}$ is $1.06\times {10}^{-4}$ M

Initial concentration of ${\text{[C}}_{10}{\text{H}}_{14}{\text{N}}_2\text{]}$ is $\text{0}\text{.0103}\text{M}$

Substitute the obtained values in above equation

  $\begin{array}{l}\text{Percentage}\text{deprotonated =}\frac{1.06\times {10}^{-4}}{0.0103}\text{×100\%}\\ \\ \text{= 1}\text{.03\%}\end{array}$

Therefore, the percentage deprotonation of 0.0103 M aqueous nicotine is 1.03%

(c)

Interpretation Introduction

Interpretation:

The $\text{pH}$, pOH and percentage deprotonation of $\text{0}\text{.060}\text{M}$ aqueous quinine has to be calculated if the $\text{pKa}$ value of quinine is 8.52.

Concept introduction:

Refer to part (a).

Answer to Problem 6D.6E

The $\text{pH}$, pOH and percentage deprotonation of $\text{0}\text{.060}\text{M}$ aqueous quinine is 10.65, 3.35 and 0.743% respectively.

Explanation of Solution

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

  ${\text{C}}_{\text{20}}{\text{H}}_{\text{24}}{\text{N}}_{\text{2}}{\text{O}}_{\text{2}}{}_{\text{(aq)}}{\text{+H}}_{\text{2}}{\text{O}}_{\text{(l)}}\rightleftharpoons {\text{C}}_{\text{20}}{\text{H}}_{\text{24}}{\text{N}}_{\text{2}}{\text{O}}_{\text{2}}{\text{H}}^{\text{+}}{}_{\text{(aq)}}{\text{+OH}}^{\text{-}}{}_{\text{(aq)}}$

The equilibrium expression for the above reaction is given below.

  ${\text{K}}_{\text{b}}\text{=}\frac{{\text{[C}}_{\text{20}}{\text{H}}_{\text{24}}{\text{N}}_{\text{2}}{\text{O}}_{\text{2}}{\text{H}}^{\text{+}}\text{]}{\text{[OH}}^{\text{-}}\text{]}}{{\text{[C}}_{\text{20}}{\text{H}}_{\text{24}}{\text{N}}_{\text{2}}{\text{O}}_{\text{2}}\text{]}}$

	${\text{C}}_{\text{20}}{\text{H}}_{\text{24}}{\text{N}}_{\text{2}}{\text{O}}_{\text{2}}$	${\text{C}}_{\text{20}}{\text{H}}_{\text{24}}{\text{N}}_{\text{2}}{\text{O}}_{\text{2}}{\text{H}}^{\text{+}}$	${\text{OH}}^{\text{-}}$
Initial concentration	0.060	0	0
Change in concentration	-x	+x	+x
Equilibrium concentration	0.060-x	x	x

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

  ${\text{K}}_{\text{b}}\text{=}\frac{\text{x×x}}{\text{0}\text{.060-x}}$

The ${\text{pK}}_{\text{a}}$ value of quinine is 8.52 using this we can calculate the ${\text{K}}_{\text{a}}{\text{ and K}}_{\text{b}}$ values and the following equation is given below,

  $\begin{array}{l}{\text{pK}}_{\text{a}}\text{=}{\text{-log(k}}_{\text{a}}\text{)}\\ \\ {\text{K}}_{\text{a}}\text{=}{\text{10}}^{{\text{-pK}}_{\text{a}}}\\ \\ \text{=}{\text{10}}^{\text{-8}\text{.52}}\\ \\ \text{=}\text{3}{\text{.02×10}}^{\text{-9}}\end{array}$

Therefore, the ${\text{K}}_{\text{a}}$ value of 0.060 M aqueous quinine is $\text{3}{\text{.02×10}}^{\text{-9}}$.

The ${\text{K}}_{\text{b}}$ value is calculated using the given formula,

  $\begin{array}{l}{\text{K}}_{\text{a}}{\text{+K}}_{\text{b}}\text{=}{\text{K}}_{\text{w}}\\ \\ {\text{K}}_{\text{b}}=\frac{{\text{K}}_{\text{w}}}{{\text{K}}_{\text{a}}}\\ \\ =\frac{1.0\times {10}^{-14}}{3.02\times {10}^{-9}}\\ \\ =3.31\times {10}^{-6}\end{array}$

Hence, the ${\text{K}}_{\text{b}}$ value of 0.060 M aqueous quinine is $\text{3}{\text{.31×10}}^{\text{-6}}$.

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

  ${\text{K}}_{\text{b}}\text{=}\frac{\text{x×x}}{\text{0}\text{.060-x}}$

The obtained quinine ${\text{K}}_{\text{b}}$ ($\text{3}{\text{.31×10}}^{\text{-6}}$) value is substituted in the above equation and then the value of x is calculated.  Assume that the x present in 0.060-x is very small than 0.060 then it can be negligible and as follows,

  $\begin{array}{l}\text{3}{\text{.31×10}}^{\text{-6}}\text{=}\frac{{\text{x}}^{\text{2}}}{\text{0}\text{.060}}\\ \\ {\text{x}}^{\text{2}}\text{=}\text{0}\text{.060×(3}{\text{.31×10}}^{\text{-6}}\text{)}\\ \\ \text{x}\text{=}\sqrt{\text{0}\text{.060×(3}{\text{.31×10}}^{\text{-6}}\text{)}}\\ \\ \text{=}4.46{\text{×10}}^{\text{-4}}\text{(}\because \text{x}\text{=}{\text{[OH}}^{\text{-}}\text{])}\end{array}$

Therefore, the concentration of ${\text{OH}}^{\text{-}}$ ion is $\text{4}{\text{.46×10}}^{\text{-4}}$.

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

  $\begin{array}{l}\text{pOH}\text{=}{\text{-log[OH}}^{-}\text{] }\\ \\ \text{=}\text{-log(4}\text{.46}\times {10}^{-4}\text{)}\\ \\ \text{=}3.35\end{array}$

Therefore, the calculated $\text{pOH}$ value of 0.060 M aqueous quinine is 3.35.

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

  $\begin{array}{l}\text{pH+pOH = 14}\\ \\ \text{ pH = 14-pOH}\\ \\ \text{ = 14-3}\text{.35 }\\ \\ \text{ = 10}\text{.65}\\ \end{array}$

Therefore, the calculated $\text{pH}$ value of 0.060 M aqueous quinine is 10.65.

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.

  $\text{Percentage}\text{deprotonated}\text{=}\frac{\text{concentration}\text{of}{\text{[OH}}^{-}\text{]}}{\text{initial}\text{concentration}\text{of}{\text{[C}}_{\text{20}}{\text{H}}_{\text{24}}{\text{N}}_{\text{2}}{\text{O}}_{\text{2}}\text{]}}\text{×100\%}$

Concentration of ${\text{[OH}}^{-}\text{]}$ is $4.46\times {10}^{-4}$ M

Initial concentration of ${\text{[NH}}_2\text{OH]}$ is $\text{0}\text{.060 M}$

Substitute the obtained values in above equation

  $\begin{array}{l}\text{Percentage}\text{deprotonated =}\frac{4.46\times {10}^{-4}}{0.060}\text{×100\%}\\ \\ \text{= 0}\text{.743\%}\end{array}$

Therefore, the percentage deprotonation of 0.060 M aqueous quinine is 0.743%

(d)

Interpretation Introduction

Interpretation:

The $\text{pH}$, pOH and percentage deprotonation of $\text{0}\text{.045}\text{M}$ aqueous strychnine has to be calculated using the ${\text{K}}_{\text{a}}$ value of conjugate acid is $\text{5}{\text{.49×10}}^{\text{-9}}$.

Concept introduction:

Refer to part (a).

Answer to Problem 6D.6E

The $\text{pH}$, pOH and percentage deprotonation of $\text{0}\text{.045}\text{M}$ aqueous strychnine is 10.46, 3.54 and 0.636% respectively.

Explanation of Solution

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

  ${\text{C}}_{21}{\text{H}}_{22}{\text{N}}_2{\text{O}}_2{}_{\text{(aq)}}{\text{+H}}_{\text{2}}{\text{O}}_{\text{(l)}}\rightleftharpoons {\text{C}}_{21}{\text{H}}_{22}{\text{N}}_2{\text{O}}_2{\text{H}}^{+}{}_{\text{(aq)}}{\text{+OH}}^{-}{}_{\text{(aq)}}$

The equilibrium expression for the above reaction is given below.

  ${\text{K}}_{\text{b}}\text{=}\frac{{\text{[C}}_{\text{21}}{\text{H}}_{\text{22}}{\text{N}}_{\text{2}}{\text{O}}_{\text{2}}{\text{H}}^{\text{+}}\text{]}{\text{[OH}}^{\text{-}}\text{]}}{{\text{[C}}_{\text{21}}{\text{H}}_{\text{22}}{\text{N}}_{\text{2}}{\text{O}}_{\text{2}}\text{]}}$

	${\text{C}}_{21}{\text{H}}_{22}{\text{N}}_2{\text{O}}_2$	${\text{C}}_{\text{21}}{\text{H}}_{\text{22}}{\text{N}}_{\text{2}}{\text{O}}_{\text{2}}{\text{H}}^{\text{+}}$	${\text{OH}}^{\text{-}}$
Initial concentration	0.045	0	0
Change in concentration	-x	+x	+x
Equilibrium concentration	0.045-x	x	x

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

  ${\text{K}}_{\text{b}}\text{=}\frac{\text{x×x}}{\text{0}\text{.045-x}}$

Strychnine ${\text{K}}_{\text{a}}$ value is $\text{5}{\text{.49×10}}^{\text{-9}}$ using this we can calculate the ${\text{K}}_{\text{b}}$ value of strychnine and the following equation is given below,

  $\begin{array}{l}{\text{K}}_{\text{a}}{\text{+K}}_{\text{b}}\text{=}{\text{K}}_{\text{w}}\\ \\ {\text{K}}_{\text{b}}=\frac{{\text{K}}_{\text{w}}}{{\text{K}}_{\text{a}}}\\ \\ =\frac{1.0\times {10}^{-14}}{5.49\times {10}^{-9}}\\ \\ =1.82\times {10}^{-6}\end{array}$

Hence, the ${\text{K}}_{\text{b}}$ value of 0.045 M aqueous strychnine is $\text{1}{\text{.82×10}}^{\text{-6}}$.

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

  ${\text{K}}_{\text{b}}\text{=}\frac{\text{x×x}}{\text{0}\text{.045-x}}$

The obtained quinine ${\text{K}}_{\text{b}}$ ($\text{1}{\text{.82×10}}^{\text{-6}}$) value is substituted in the above equation and then the value of x is calculated.  Assume that the x present in 0.045-x is very small than 0.045 then it can be negligible and as follows,

  $\begin{array}{l}\text{1}{\text{.82×10}}^{\text{-6}}\text{=}\frac{{\text{x}}^{\text{2}}}{\text{0}\text{.045}}\\ \\ {\text{x}}^{\text{2}}\text{=}\text{0}\text{.045×(1}{\text{.82×10}}^{\text{-6}}\text{)}\\ \\ \text{x}\text{=}\sqrt{\text{0}\text{.045×(1}{\text{.82×10}}^{\text{-6}}\text{)}}\\ \\ \text{=}2.86{\text{×10}}^{\text{-4}}\text{(}\because \text{x}\text{=}{\text{[OH}}^{\text{-}}\text{])}\end{array}$

Therefore, the concentration of ${\text{OH}}^{\text{-}}$ ion is $\text{2}{\text{.86×10}}^{\text{-4}}$.

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

  $\begin{array}{l}\text{pOH}\text{=}{\text{-log[OH}}^{-}\text{] }\\ \\ \text{=}\text{-log(2}\text{.86}\times {10}^{-4}\text{)}\\ \\ \text{=}3.54\end{array}$

Therefore, the calculated $\text{pOH}$ value of 0.045 M aqueous strychnine is 3.54.

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

  $\begin{array}{l}\text{pH+pOH = 14}\\ \\ \text{ pH = 14-pOH}\\ \\ \text{ = 14-3}\text{.54 }\\ \\ \text{ = 10}\text{.46}\\ \end{array}$

Therefore, the calculated $\text{pH}$ value of 0.045 M aqueous strychnine is 10.46.

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.

  $\text{Percentage}\text{deprotonated}\text{=}\frac{\text{concentration}\text{of}{\text{[OH}}^{-}\text{]}}{\text{initial}\text{concentration}\text{of}{\text{[C}}_{21}{\text{H}}_{22}{\text{N}}_2{\text{O}}_2\text{]}}\text{×100\%}$

Concentration of ${\text{[OH}}^{-}\text{]}$ is $\text{2}{\text{.86×10}}^{\text{-4}}\text{M}$

Initial concentration of ${\text{[C}}_{21}{\text{H}}_{22}{\text{N}}_2{\text{O}}_2\text{]}$ is $\text{0}\text{.045}\text{M}$

Substitute the obtained values in above equation

  $\begin{array}{l}\text{Percentage}\text{deprotonated =}\frac{2.86\times {10}^{-4}}{0.045}\text{×100\%}\\ \\ \text{= 0}\text{.636\%}\end{array}$

Therefore, the percentage deprotonation of 0.045 M aqueous codeine is 0.636%