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CHXI07:EQUILIBRIUM

314865 The degree of dissociation of \(\mathrm{PCl}_{5}\) at 1 atm and 298 K is 0.2 . The value of \({{\rm{K}}_{\rm{p}}}\) is

1 0.42 atm

2 0.24 atm

3 0.48 atm

4 0.042 atm

CHXI07:EQUILIBRIUM

314866 Dissociation constant of a weak acid is \(1 \times 10^{-6}\) at \(25^{\circ} \mathrm{C}\). Find the pOH of 0.01 M of its aqueous solution.

1 4

2 3

3 10

4 12

Explanation:

\(\,{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{HA}}{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \rightleftharpoons {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \,{\mkern 1mu} {\mkern 1mu} {{\text{H}}^ + } + \,\,\,{{\text{A}}^ - }\)
\({\rm{Initial}}{\mkern 1mu} {\mkern 1mu} {\rm{conc}}{\rm{.}}{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} 0.01{\rm{M}}\quad\,\,\, {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \,\,\,\,0\quad {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \,\,\,\,\,\,0\)
\({\rm{Eq}}{\mkern 1mu} {\rm{.}}{\mkern 1mu} {\rm{conc}}{\rm{.}}{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} 0.01 - 0.01{\rm{\alpha }}{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} 0.01{\rm{\alpha }}{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} 0.01{\rm{\alpha }}{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \)
\({{\rm{K}}_{\rm{a}}} = \frac{{\left[ {{{\rm{H}}^{\rm{ + }}}} \right]\left[ {{{\rm{A}}^ - }} \right]}}{{{\rm{[HA]}}}}\)
\(1 \times {10^{ - 6}} = \frac{{0.01{\rm{\alpha }} \times 0.01{\rm{\alpha }}}}{{0.01 - 0.01{\rm{\alpha }}}}\)
\( = \frac{{0.01{{\rm{\alpha }}^{\rm{2}}}}}{{1 - {\rm{\alpha }}}} = 0.01\,{{\rm{\alpha }}^{\rm{2}}}\)
\(1 \times {10^{ - 4}} = {{\rm{\alpha }}^{\rm{2}}};\)
\(\therefore {\rm{\alpha }} = 1 \times {10^{ - 2}}\quad (\because {\rm{\alpha }} < < < 1)\)
\[\begin{array}{l}
\left[ {{{\rm{H}}^{\rm{ + }}}} \right] = 0.01 \times {\rm{\alpha }} = 0.01 \times 1 \times {10^{ - 2}}\\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 1 \times {10^{ - 4}}
\end{array}\]
\(\therefore \quad {\rm{pH}} = - {\rm{log}}\left[ {{{\rm{H}}^{\rm{ + }}}} \right] = - {\rm{log}}\,{10^{ - 4}} = 4\)
\({\rm{pH}} + {\rm{pOH}} = 14\)
\(\therefore {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\rm{pOH}} = 14 - 4 = 10{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \)

CHXI07:EQUILIBRIUM

314867 Calculate the pH of \(0.1\,{\rm{M C}}{{\rm{H}}_{\rm{3}}}{\rm{COOH}}\) solution. Dissociation constant of acetic acid is \(1.8 \times 10^{-5} \mathrm{M}\).

1 2.87

2 3.52

3 2.62

4 6.54

Explanation:

\(\,{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{C}}{{\text{H}}_{\text{3}}}{\text{COOH}} \rightleftharpoons {\text{C}}{{\text{H}}_{\text{3}}}{\text{CO}}{{\text{O}}^ - } + {{\text{H}}^{\text{ + }}}\)
\({\rm{Initial}}{\mkern 1mu} \,{\rm{conc}}{\rm{.}}{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu}\,\,\, {\rm{0}}{\rm{.1M}}{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\rm{0}}{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \,\,\,\,\,\,\,\,{\mkern 1mu} {\rm{0}}\)
\({\rm{Eq}}{\mkern 1mu} {\rm{.}}{\mkern 1mu} {\rm{Conc}}{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\rm{0}}{\rm{.1}} - {\rm{0}}.{\rm{1\alpha }}{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\mkern 1mu} {\mkern 1mu} {\rm{0}}.{\rm{1\alpha }}{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu}\,\,\,\,\,\,\,{\rm{0}}.{\rm{1}}\alpha {\mkern 1mu} \)
\({{\rm{K}}_{\rm{a}}} = \frac{{[{\rm{C}}{{\rm{H}}_{\rm{3}}}{\rm{CO}}{{\rm{O}}^ - }][{{\rm{H}}^{\rm{ + }}}]}}{{[{\rm{C}}{{\rm{H}}_{\rm{3}}}{\rm{COOH}}]}}\)
\(1.8 \times {10^{ - 5}} = \frac{{{\rm{0}}.{\rm{1\alpha }}{\mkern 1mu} \, \times {\mkern 1mu} \,{\rm{0}}.{\rm{1\alpha }}}}{{0.1(1 - {\rm{\alpha }})}}\)
\(1.8 \times {10^{ - 5}} = \frac{{{\rm{0}}{\rm{.1}}{{\rm{\alpha }}^{\rm{2}}}}}{{1 - {\rm{\alpha }}}}\)
\(1.8 \times {10^{ - 5}} = 0.1\,{{\rm{\alpha }}^{\rm{2}}}{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \)\(\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(\because {\rm{\alpha }} < < < 1)\)
\[{{\rm{\alpha }}^{\rm{2}}} = 1.8 \times {10^{ - 4}}
{\rm{\alpha }} = 1.34 \times {10^{ - 2}}\]
\(\therefore \quad \left[ {{{\rm{H}}^{\rm{ + }}}} \right] = 0.1 \times 1.34 \times {10^{ - 2}} = 1.34 \times {10^{ - 3}}\)
\({\rm{pH}} = - {\rm{log}}\left[ {{{\rm{H}}^{\rm{ + }}}} \right] = - {\rm{log}}\left[ {1.34 \times {{10}^{ - 3}}} \right] = 2.87\)

CHXI07:EQUILIBRIUM

314868 The \(\mathrm{K}_{\mathrm{a}}\) of a weak acid is \(10^{-5}, \mathrm{pK}_{\mathrm{b}}\) value of its conjugate base is

1 7

2 5

3 9

4 6

CHXI07:EQUILIBRIUM

314869 What will be the \(\mathrm{pH}\) of \(0.1\,{\rm{M}}\,{\rm{C}}{{\rm{H}}_{\rm{3}}}{\rm{COON}}{{\rm{H}}_{\rm{4}}}\) ?
\({{\text{K}}_{\text{a}}} = 1.8 \times {10^{ - 5}};\;\;\;{\mkern 1mu} {\kern 1pt} {{\text{K}}_{\text{b}}} = 1.8 \times {10^{ - 5}}\)

1 7

2 6.9

3 7.1

4 14

CHXI07:EQUILIBRIUM

314865 The degree of dissociation of \(\mathrm{PCl}_{5}\) at 1 atm and 298 K is 0.2 . The value of \({{\rm{K}}_{\rm{p}}}\) is

1 0.42 atm

2 0.24 atm

3 0.48 atm

4 0.042 atm

CHXI07:EQUILIBRIUM

314866 Dissociation constant of a weak acid is \(1 \times 10^{-6}\) at \(25^{\circ} \mathrm{C}\). Find the pOH of 0.01 M of its aqueous solution.

1 4

2 3

3 10

4 12

Explanation:

\(\,{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{HA}}{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \rightleftharpoons {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \,{\mkern 1mu} {\mkern 1mu} {{\text{H}}^ + } + \,\,\,{{\text{A}}^ - }\)
\({\rm{Initial}}{\mkern 1mu} {\mkern 1mu} {\rm{conc}}{\rm{.}}{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} 0.01{\rm{M}}\quad\,\,\, {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \,\,\,\,0\quad {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \,\,\,\,\,\,0\)
\({\rm{Eq}}{\mkern 1mu} {\rm{.}}{\mkern 1mu} {\rm{conc}}{\rm{.}}{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} 0.01 - 0.01{\rm{\alpha }}{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} 0.01{\rm{\alpha }}{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} 0.01{\rm{\alpha }}{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \)
\({{\rm{K}}_{\rm{a}}} = \frac{{\left[ {{{\rm{H}}^{\rm{ + }}}} \right]\left[ {{{\rm{A}}^ - }} \right]}}{{{\rm{[HA]}}}}\)
\(1 \times {10^{ - 6}} = \frac{{0.01{\rm{\alpha }} \times 0.01{\rm{\alpha }}}}{{0.01 - 0.01{\rm{\alpha }}}}\)
\( = \frac{{0.01{{\rm{\alpha }}^{\rm{2}}}}}{{1 - {\rm{\alpha }}}} = 0.01\,{{\rm{\alpha }}^{\rm{2}}}\)
\(1 \times {10^{ - 4}} = {{\rm{\alpha }}^{\rm{2}}};\)
\(\therefore {\rm{\alpha }} = 1 \times {10^{ - 2}}\quad (\because {\rm{\alpha }} < < < 1)\)
\[\begin{array}{l}
\left[ {{{\rm{H}}^{\rm{ + }}}} \right] = 0.01 \times {\rm{\alpha }} = 0.01 \times 1 \times {10^{ - 2}}\\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 1 \times {10^{ - 4}}
\end{array}\]
\(\therefore \quad {\rm{pH}} = - {\rm{log}}\left[ {{{\rm{H}}^{\rm{ + }}}} \right] = - {\rm{log}}\,{10^{ - 4}} = 4\)
\({\rm{pH}} + {\rm{pOH}} = 14\)
\(\therefore {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\rm{pOH}} = 14 - 4 = 10{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \)

CHXI07:EQUILIBRIUM

314867 Calculate the pH of \(0.1\,{\rm{M C}}{{\rm{H}}_{\rm{3}}}{\rm{COOH}}\) solution. Dissociation constant of acetic acid is \(1.8 \times 10^{-5} \mathrm{M}\).

1 2.87

2 3.52

3 2.62

4 6.54

Explanation:

\(\,{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{C}}{{\text{H}}_{\text{3}}}{\text{COOH}} \rightleftharpoons {\text{C}}{{\text{H}}_{\text{3}}}{\text{CO}}{{\text{O}}^ - } + {{\text{H}}^{\text{ + }}}\)
\({\rm{Initial}}{\mkern 1mu} \,{\rm{conc}}{\rm{.}}{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu}\,\,\, {\rm{0}}{\rm{.1M}}{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\rm{0}}{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \,\,\,\,\,\,\,\,{\mkern 1mu} {\rm{0}}\)
\({\rm{Eq}}{\mkern 1mu} {\rm{.}}{\mkern 1mu} {\rm{Conc}}{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\rm{0}}{\rm{.1}} - {\rm{0}}.{\rm{1\alpha }}{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\mkern 1mu} {\mkern 1mu} {\rm{0}}.{\rm{1\alpha }}{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu}\,\,\,\,\,\,\,{\rm{0}}.{\rm{1}}\alpha {\mkern 1mu} \)
\({{\rm{K}}_{\rm{a}}} = \frac{{[{\rm{C}}{{\rm{H}}_{\rm{3}}}{\rm{CO}}{{\rm{O}}^ - }][{{\rm{H}}^{\rm{ + }}}]}}{{[{\rm{C}}{{\rm{H}}_{\rm{3}}}{\rm{COOH}}]}}\)
\(1.8 \times {10^{ - 5}} = \frac{{{\rm{0}}.{\rm{1\alpha }}{\mkern 1mu} \, \times {\mkern 1mu} \,{\rm{0}}.{\rm{1\alpha }}}}{{0.1(1 - {\rm{\alpha }})}}\)
\(1.8 \times {10^{ - 5}} = \frac{{{\rm{0}}{\rm{.1}}{{\rm{\alpha }}^{\rm{2}}}}}{{1 - {\rm{\alpha }}}}\)
\(1.8 \times {10^{ - 5}} = 0.1\,{{\rm{\alpha }}^{\rm{2}}}{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \)\(\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(\because {\rm{\alpha }} < < < 1)\)
\[{{\rm{\alpha }}^{\rm{2}}} = 1.8 \times {10^{ - 4}}
{\rm{\alpha }} = 1.34 \times {10^{ - 2}}\]
\(\therefore \quad \left[ {{{\rm{H}}^{\rm{ + }}}} \right] = 0.1 \times 1.34 \times {10^{ - 2}} = 1.34 \times {10^{ - 3}}\)
\({\rm{pH}} = - {\rm{log}}\left[ {{{\rm{H}}^{\rm{ + }}}} \right] = - {\rm{log}}\left[ {1.34 \times {{10}^{ - 3}}} \right] = 2.87\)

CHXI07:EQUILIBRIUM

314868 The \(\mathrm{K}_{\mathrm{a}}\) of a weak acid is \(10^{-5}, \mathrm{pK}_{\mathrm{b}}\) value of its conjugate base is

1 7

2 5

3 9

4 6

CHXI07:EQUILIBRIUM

314869 What will be the \(\mathrm{pH}\) of \(0.1\,{\rm{M}}\,{\rm{C}}{{\rm{H}}_{\rm{3}}}{\rm{COON}}{{\rm{H}}_{\rm{4}}}\) ?
\({{\text{K}}_{\text{a}}} = 1.8 \times {10^{ - 5}};\;\;\;{\mkern 1mu} {\kern 1pt} {{\text{K}}_{\text{b}}} = 1.8 \times {10^{ - 5}}\)

1 7

2 6.9

3 7.1

4 14

CHXI07:EQUILIBRIUM

314865 The degree of dissociation of \(\mathrm{PCl}_{5}\) at 1 atm and 298 K is 0.2 . The value of \({{\rm{K}}_{\rm{p}}}\) is

1 0.42 atm

2 0.24 atm

3 0.48 atm

4 0.042 atm

CHXI07:EQUILIBRIUM

314866 Dissociation constant of a weak acid is \(1 \times 10^{-6}\) at \(25^{\circ} \mathrm{C}\). Find the pOH of 0.01 M of its aqueous solution.

1 4

2 3

3 10

4 12

Explanation:

\(\,{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{HA}}{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \rightleftharpoons {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \,{\mkern 1mu} {\mkern 1mu} {{\text{H}}^ + } + \,\,\,{{\text{A}}^ - }\)
\({\rm{Initial}}{\mkern 1mu} {\mkern 1mu} {\rm{conc}}{\rm{.}}{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} 0.01{\rm{M}}\quad\,\,\, {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \,\,\,\,0\quad {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \,\,\,\,\,\,0\)
\({\rm{Eq}}{\mkern 1mu} {\rm{.}}{\mkern 1mu} {\rm{conc}}{\rm{.}}{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} 0.01 - 0.01{\rm{\alpha }}{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} 0.01{\rm{\alpha }}{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} 0.01{\rm{\alpha }}{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \)
\({{\rm{K}}_{\rm{a}}} = \frac{{\left[ {{{\rm{H}}^{\rm{ + }}}} \right]\left[ {{{\rm{A}}^ - }} \right]}}{{{\rm{[HA]}}}}\)
\(1 \times {10^{ - 6}} = \frac{{0.01{\rm{\alpha }} \times 0.01{\rm{\alpha }}}}{{0.01 - 0.01{\rm{\alpha }}}}\)
\( = \frac{{0.01{{\rm{\alpha }}^{\rm{2}}}}}{{1 - {\rm{\alpha }}}} = 0.01\,{{\rm{\alpha }}^{\rm{2}}}\)
\(1 \times {10^{ - 4}} = {{\rm{\alpha }}^{\rm{2}}};\)
\(\therefore {\rm{\alpha }} = 1 \times {10^{ - 2}}\quad (\because {\rm{\alpha }} < < < 1)\)
\[\begin{array}{l}
\left[ {{{\rm{H}}^{\rm{ + }}}} \right] = 0.01 \times {\rm{\alpha }} = 0.01 \times 1 \times {10^{ - 2}}\\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 1 \times {10^{ - 4}}
\end{array}\]
\(\therefore \quad {\rm{pH}} = - {\rm{log}}\left[ {{{\rm{H}}^{\rm{ + }}}} \right] = - {\rm{log}}\,{10^{ - 4}} = 4\)
\({\rm{pH}} + {\rm{pOH}} = 14\)
\(\therefore {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\rm{pOH}} = 14 - 4 = 10{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \)

CHXI07:EQUILIBRIUM

314867 Calculate the pH of \(0.1\,{\rm{M C}}{{\rm{H}}_{\rm{3}}}{\rm{COOH}}\) solution. Dissociation constant of acetic acid is \(1.8 \times 10^{-5} \mathrm{M}\).

1 2.87

2 3.52

3 2.62

4 6.54

Explanation:

\(\,{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{C}}{{\text{H}}_{\text{3}}}{\text{COOH}} \rightleftharpoons {\text{C}}{{\text{H}}_{\text{3}}}{\text{CO}}{{\text{O}}^ - } + {{\text{H}}^{\text{ + }}}\)
\({\rm{Initial}}{\mkern 1mu} \,{\rm{conc}}{\rm{.}}{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu}\,\,\, {\rm{0}}{\rm{.1M}}{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\rm{0}}{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \,\,\,\,\,\,\,\,{\mkern 1mu} {\rm{0}}\)
\({\rm{Eq}}{\mkern 1mu} {\rm{.}}{\mkern 1mu} {\rm{Conc}}{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\rm{0}}{\rm{.1}} - {\rm{0}}.{\rm{1\alpha }}{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\mkern 1mu} {\mkern 1mu} {\rm{0}}.{\rm{1\alpha }}{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu}\,\,\,\,\,\,\,{\rm{0}}.{\rm{1}}\alpha {\mkern 1mu} \)
\({{\rm{K}}_{\rm{a}}} = \frac{{[{\rm{C}}{{\rm{H}}_{\rm{3}}}{\rm{CO}}{{\rm{O}}^ - }][{{\rm{H}}^{\rm{ + }}}]}}{{[{\rm{C}}{{\rm{H}}_{\rm{3}}}{\rm{COOH}}]}}\)
\(1.8 \times {10^{ - 5}} = \frac{{{\rm{0}}.{\rm{1\alpha }}{\mkern 1mu} \, \times {\mkern 1mu} \,{\rm{0}}.{\rm{1\alpha }}}}{{0.1(1 - {\rm{\alpha }})}}\)
\(1.8 \times {10^{ - 5}} = \frac{{{\rm{0}}{\rm{.1}}{{\rm{\alpha }}^{\rm{2}}}}}{{1 - {\rm{\alpha }}}}\)
\(1.8 \times {10^{ - 5}} = 0.1\,{{\rm{\alpha }}^{\rm{2}}}{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \)\(\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(\because {\rm{\alpha }} < < < 1)\)
\[{{\rm{\alpha }}^{\rm{2}}} = 1.8 \times {10^{ - 4}}
{\rm{\alpha }} = 1.34 \times {10^{ - 2}}\]
\(\therefore \quad \left[ {{{\rm{H}}^{\rm{ + }}}} \right] = 0.1 \times 1.34 \times {10^{ - 2}} = 1.34 \times {10^{ - 3}}\)
\({\rm{pH}} = - {\rm{log}}\left[ {{{\rm{H}}^{\rm{ + }}}} \right] = - {\rm{log}}\left[ {1.34 \times {{10}^{ - 3}}} \right] = 2.87\)

CHXI07:EQUILIBRIUM

314868 The \(\mathrm{K}_{\mathrm{a}}\) of a weak acid is \(10^{-5}, \mathrm{pK}_{\mathrm{b}}\) value of its conjugate base is

1 7

2 5

3 9

4 6

CHXI07:EQUILIBRIUM

314869 What will be the \(\mathrm{pH}\) of \(0.1\,{\rm{M}}\,{\rm{C}}{{\rm{H}}_{\rm{3}}}{\rm{COON}}{{\rm{H}}_{\rm{4}}}\) ?
\({{\text{K}}_{\text{a}}} = 1.8 \times {10^{ - 5}};\;\;\;{\mkern 1mu} {\kern 1pt} {{\text{K}}_{\text{b}}} = 1.8 \times {10^{ - 5}}\)

1 7

2 6.9

3 7.1

4 14

CHXI07:EQUILIBRIUM

314865 The degree of dissociation of \(\mathrm{PCl}_{5}\) at 1 atm and 298 K is 0.2 . The value of \({{\rm{K}}_{\rm{p}}}\) is

1 0.42 atm

2 0.24 atm

3 0.48 atm

4 0.042 atm

CHXI07:EQUILIBRIUM

314866 Dissociation constant of a weak acid is \(1 \times 10^{-6}\) at \(25^{\circ} \mathrm{C}\). Find the pOH of 0.01 M of its aqueous solution.

1 4

2 3

3 10

4 12

Explanation:

\(\,{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{HA}}{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \rightleftharpoons {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \,{\mkern 1mu} {\mkern 1mu} {{\text{H}}^ + } + \,\,\,{{\text{A}}^ - }\)
\({\rm{Initial}}{\mkern 1mu} {\mkern 1mu} {\rm{conc}}{\rm{.}}{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} 0.01{\rm{M}}\quad\,\,\, {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \,\,\,\,0\quad {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \,\,\,\,\,\,0\)
\({\rm{Eq}}{\mkern 1mu} {\rm{.}}{\mkern 1mu} {\rm{conc}}{\rm{.}}{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} 0.01 - 0.01{\rm{\alpha }}{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} 0.01{\rm{\alpha }}{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} 0.01{\rm{\alpha }}{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \)
\({{\rm{K}}_{\rm{a}}} = \frac{{\left[ {{{\rm{H}}^{\rm{ + }}}} \right]\left[ {{{\rm{A}}^ - }} \right]}}{{{\rm{[HA]}}}}\)
\(1 \times {10^{ - 6}} = \frac{{0.01{\rm{\alpha }} \times 0.01{\rm{\alpha }}}}{{0.01 - 0.01{\rm{\alpha }}}}\)
\( = \frac{{0.01{{\rm{\alpha }}^{\rm{2}}}}}{{1 - {\rm{\alpha }}}} = 0.01\,{{\rm{\alpha }}^{\rm{2}}}\)
\(1 \times {10^{ - 4}} = {{\rm{\alpha }}^{\rm{2}}};\)
\(\therefore {\rm{\alpha }} = 1 \times {10^{ - 2}}\quad (\because {\rm{\alpha }} < < < 1)\)
\[\begin{array}{l}
\left[ {{{\rm{H}}^{\rm{ + }}}} \right] = 0.01 \times {\rm{\alpha }} = 0.01 \times 1 \times {10^{ - 2}}\\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 1 \times {10^{ - 4}}
\end{array}\]
\(\therefore \quad {\rm{pH}} = - {\rm{log}}\left[ {{{\rm{H}}^{\rm{ + }}}} \right] = - {\rm{log}}\,{10^{ - 4}} = 4\)
\({\rm{pH}} + {\rm{pOH}} = 14\)
\(\therefore {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\rm{pOH}} = 14 - 4 = 10{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \)

CHXI07:EQUILIBRIUM

314867 Calculate the pH of \(0.1\,{\rm{M C}}{{\rm{H}}_{\rm{3}}}{\rm{COOH}}\) solution. Dissociation constant of acetic acid is \(1.8 \times 10^{-5} \mathrm{M}\).

1 2.87

2 3.52

3 2.62

4 6.54

Explanation:

\(\,{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{C}}{{\text{H}}_{\text{3}}}{\text{COOH}} \rightleftharpoons {\text{C}}{{\text{H}}_{\text{3}}}{\text{CO}}{{\text{O}}^ - } + {{\text{H}}^{\text{ + }}}\)
\({\rm{Initial}}{\mkern 1mu} \,{\rm{conc}}{\rm{.}}{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu}\,\,\, {\rm{0}}{\rm{.1M}}{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\rm{0}}{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \,\,\,\,\,\,\,\,{\mkern 1mu} {\rm{0}}\)
\({\rm{Eq}}{\mkern 1mu} {\rm{.}}{\mkern 1mu} {\rm{Conc}}{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\rm{0}}{\rm{.1}} - {\rm{0}}.{\rm{1\alpha }}{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\mkern 1mu} {\mkern 1mu} {\rm{0}}.{\rm{1\alpha }}{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu}\,\,\,\,\,\,\,{\rm{0}}.{\rm{1}}\alpha {\mkern 1mu} \)
\({{\rm{K}}_{\rm{a}}} = \frac{{[{\rm{C}}{{\rm{H}}_{\rm{3}}}{\rm{CO}}{{\rm{O}}^ - }][{{\rm{H}}^{\rm{ + }}}]}}{{[{\rm{C}}{{\rm{H}}_{\rm{3}}}{\rm{COOH}}]}}\)
\(1.8 \times {10^{ - 5}} = \frac{{{\rm{0}}.{\rm{1\alpha }}{\mkern 1mu} \, \times {\mkern 1mu} \,{\rm{0}}.{\rm{1\alpha }}}}{{0.1(1 - {\rm{\alpha }})}}\)
\(1.8 \times {10^{ - 5}} = \frac{{{\rm{0}}{\rm{.1}}{{\rm{\alpha }}^{\rm{2}}}}}{{1 - {\rm{\alpha }}}}\)
\(1.8 \times {10^{ - 5}} = 0.1\,{{\rm{\alpha }}^{\rm{2}}}{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \)\(\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(\because {\rm{\alpha }} < < < 1)\)
\[{{\rm{\alpha }}^{\rm{2}}} = 1.8 \times {10^{ - 4}}
{\rm{\alpha }} = 1.34 \times {10^{ - 2}}\]
\(\therefore \quad \left[ {{{\rm{H}}^{\rm{ + }}}} \right] = 0.1 \times 1.34 \times {10^{ - 2}} = 1.34 \times {10^{ - 3}}\)
\({\rm{pH}} = - {\rm{log}}\left[ {{{\rm{H}}^{\rm{ + }}}} \right] = - {\rm{log}}\left[ {1.34 \times {{10}^{ - 3}}} \right] = 2.87\)

CHXI07:EQUILIBRIUM

314868 The \(\mathrm{K}_{\mathrm{a}}\) of a weak acid is \(10^{-5}, \mathrm{pK}_{\mathrm{b}}\) value of its conjugate base is

1 7

2 5

3 9

4 6

CHXI07:EQUILIBRIUM

314869 What will be the \(\mathrm{pH}\) of \(0.1\,{\rm{M}}\,{\rm{C}}{{\rm{H}}_{\rm{3}}}{\rm{COON}}{{\rm{H}}_{\rm{4}}}\) ?
\({{\text{K}}_{\text{a}}} = 1.8 \times {10^{ - 5}};\;\;\;{\mkern 1mu} {\kern 1pt} {{\text{K}}_{\text{b}}} = 1.8 \times {10^{ - 5}}\)

1 7

2 6.9

3 7.1

4 14

CHXI07:EQUILIBRIUM

314865 The degree of dissociation of \(\mathrm{PCl}_{5}\) at 1 atm and 298 K is 0.2 . The value of \({{\rm{K}}_{\rm{p}}}\) is

1 0.42 atm

2 0.24 atm

3 0.48 atm

4 0.042 atm

CHXI07:EQUILIBRIUM

314866 Dissociation constant of a weak acid is \(1 \times 10^{-6}\) at \(25^{\circ} \mathrm{C}\). Find the pOH of 0.01 M of its aqueous solution.

1 4

2 3

3 10

4 12

Explanation:

\(\,{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{HA}}{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \rightleftharpoons {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \,{\mkern 1mu} {\mkern 1mu} {{\text{H}}^ + } + \,\,\,{{\text{A}}^ - }\)
\({\rm{Initial}}{\mkern 1mu} {\mkern 1mu} {\rm{conc}}{\rm{.}}{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} 0.01{\rm{M}}\quad\,\,\, {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \,\,\,\,0\quad {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \,\,\,\,\,\,0\)
\({\rm{Eq}}{\mkern 1mu} {\rm{.}}{\mkern 1mu} {\rm{conc}}{\rm{.}}{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} 0.01 - 0.01{\rm{\alpha }}{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} 0.01{\rm{\alpha }}{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} 0.01{\rm{\alpha }}{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \)
\({{\rm{K}}_{\rm{a}}} = \frac{{\left[ {{{\rm{H}}^{\rm{ + }}}} \right]\left[ {{{\rm{A}}^ - }} \right]}}{{{\rm{[HA]}}}}\)
\(1 \times {10^{ - 6}} = \frac{{0.01{\rm{\alpha }} \times 0.01{\rm{\alpha }}}}{{0.01 - 0.01{\rm{\alpha }}}}\)
\( = \frac{{0.01{{\rm{\alpha }}^{\rm{2}}}}}{{1 - {\rm{\alpha }}}} = 0.01\,{{\rm{\alpha }}^{\rm{2}}}\)
\(1 \times {10^{ - 4}} = {{\rm{\alpha }}^{\rm{2}}};\)
\(\therefore {\rm{\alpha }} = 1 \times {10^{ - 2}}\quad (\because {\rm{\alpha }} < < < 1)\)
\[\begin{array}{l}
\left[ {{{\rm{H}}^{\rm{ + }}}} \right] = 0.01 \times {\rm{\alpha }} = 0.01 \times 1 \times {10^{ - 2}}\\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 1 \times {10^{ - 4}}
\end{array}\]
\(\therefore \quad {\rm{pH}} = - {\rm{log}}\left[ {{{\rm{H}}^{\rm{ + }}}} \right] = - {\rm{log}}\,{10^{ - 4}} = 4\)
\({\rm{pH}} + {\rm{pOH}} = 14\)
\(\therefore {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\rm{pOH}} = 14 - 4 = 10{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \)

CHXI07:EQUILIBRIUM

314867 Calculate the pH of \(0.1\,{\rm{M C}}{{\rm{H}}_{\rm{3}}}{\rm{COOH}}\) solution. Dissociation constant of acetic acid is \(1.8 \times 10^{-5} \mathrm{M}\).

1 2.87

2 3.52

3 2.62

4 6.54

Explanation:

\(\,{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{C}}{{\text{H}}_{\text{3}}}{\text{COOH}} \rightleftharpoons {\text{C}}{{\text{H}}_{\text{3}}}{\text{CO}}{{\text{O}}^ - } + {{\text{H}}^{\text{ + }}}\)
\({\rm{Initial}}{\mkern 1mu} \,{\rm{conc}}{\rm{.}}{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu}\,\,\, {\rm{0}}{\rm{.1M}}{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\rm{0}}{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \,\,\,\,\,\,\,\,{\mkern 1mu} {\rm{0}}\)
\({\rm{Eq}}{\mkern 1mu} {\rm{.}}{\mkern 1mu} {\rm{Conc}}{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\rm{0}}{\rm{.1}} - {\rm{0}}.{\rm{1\alpha }}{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\mkern 1mu} {\mkern 1mu} {\rm{0}}.{\rm{1\alpha }}{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu}\,\,\,\,\,\,\,{\rm{0}}.{\rm{1}}\alpha {\mkern 1mu} \)
\({{\rm{K}}_{\rm{a}}} = \frac{{[{\rm{C}}{{\rm{H}}_{\rm{3}}}{\rm{CO}}{{\rm{O}}^ - }][{{\rm{H}}^{\rm{ + }}}]}}{{[{\rm{C}}{{\rm{H}}_{\rm{3}}}{\rm{COOH}}]}}\)
\(1.8 \times {10^{ - 5}} = \frac{{{\rm{0}}.{\rm{1\alpha }}{\mkern 1mu} \, \times {\mkern 1mu} \,{\rm{0}}.{\rm{1\alpha }}}}{{0.1(1 - {\rm{\alpha }})}}\)
\(1.8 \times {10^{ - 5}} = \frac{{{\rm{0}}{\rm{.1}}{{\rm{\alpha }}^{\rm{2}}}}}{{1 - {\rm{\alpha }}}}\)
\(1.8 \times {10^{ - 5}} = 0.1\,{{\rm{\alpha }}^{\rm{2}}}{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \)\(\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(\because {\rm{\alpha }} < < < 1)\)
\[{{\rm{\alpha }}^{\rm{2}}} = 1.8 \times {10^{ - 4}}
{\rm{\alpha }} = 1.34 \times {10^{ - 2}}\]
\(\therefore \quad \left[ {{{\rm{H}}^{\rm{ + }}}} \right] = 0.1 \times 1.34 \times {10^{ - 2}} = 1.34 \times {10^{ - 3}}\)
\({\rm{pH}} = - {\rm{log}}\left[ {{{\rm{H}}^{\rm{ + }}}} \right] = - {\rm{log}}\left[ {1.34 \times {{10}^{ - 3}}} \right] = 2.87\)

CHXI07:EQUILIBRIUM

314868 The \(\mathrm{K}_{\mathrm{a}}\) of a weak acid is \(10^{-5}, \mathrm{pK}_{\mathrm{b}}\) value of its conjugate base is

1 7

2 5

3 9

4 6

CHXI07:EQUILIBRIUM

314869 What will be the \(\mathrm{pH}\) of \(0.1\,{\rm{M}}\,{\rm{C}}{{\rm{H}}_{\rm{3}}}{\rm{COON}}{{\rm{H}}_{\rm{4}}}\) ?
\({{\text{K}}_{\text{a}}} = 1.8 \times {10^{ - 5}};\;\;\;{\mkern 1mu} {\kern 1pt} {{\text{K}}_{\text{b}}} = 1.8 \times {10^{ - 5}}\)

1 7

2 6.9

3 7.1

4 14