QBManager

Ionic Equilibrium

229808 The dissociation constant of a weak base is $1 \times 10^{- 5}$ at $25^{\circ} C$ . The $pH$ of its $0.1 M$ solution at the same temperature will be

1 11

2 3

3 6

4 13

Explanation:

Given,
dissociation constant of weak base $= 1 \times 10^{- 5}$ concentration $= 0.1 M$
By Ostwald dilution law,
$[{OH}^{-}]$ in weak base $= \sqrt{K_{b} C}$
$= \sqrt{1 \times 10^{- 5} \times 0.1} = 1 \times 10^{- 3} M$
$\begin{aligned} pOH = - \log [{OH}^{-}] \\ = - \log (1 \times 10^{- 3}) \\ pOH = 3 \end{aligned}$
We know that-
$\begin{aligned} \Rightarrow pH + pOH = 14 \\ \Rightarrow pH = 14 - 3 = 11 \end{aligned}$

COMEDK-2011

Ionic Equilibrium

229809 The $pH$ of a solution of $NaOH$ is 12. The mass of $NaOH$ present in $500 mL$ of the solution is

1

400 mg

2

200 mg

3

20 mg

4

40 mg

Explanation:

Given that,
$pH = 12$
$∵ pH + pOH = 14$
$pOH = 14 - 12 = 2$
Hence, $NaOH$ concentration $= 10^{- 2} M = 10^{- 2} N$
We know that, $N = \frac{w \times 1000}{Eq.wt \times Volume (mL)}$
$\Rightarrow 10^{- 2} = \frac{w \times 1000}{40 \times 500}$
$\Rightarrow w = \frac{10^{- 2} \times 40 \times 500}{1000} = 10^{- 2} \times 20 = 0.2 g = 200 mg$

COMEDK-2011

Ionic Equilibrium

229810 The $pOH$ of $0.0005 M$ sulphuric acid is

1 5

2 3

3 11

4 12

Explanation:

The equation for ionization is-
$H_{2} {SO}_{4} ◻ 2 H^{+} + {SO}_{4}^{2 -}$
Each molecule of acid gives 2 molecules of $H^{+}$
So, $[H^{+}] = 2 \times 0.0005 = 1 \times 10^{- 3} M$
$pH = - \log [H^{+}] = - \log (1 \times 10^{- 3}) = 3$
We know that-
$\begin{aligned} ∵ pH + pOH = 14 \\ 3 + pOH = 14 \\ pOH = 14 - 3 \\ pOH = 11 \end{aligned}$

COMEDK-2011

Ionic Equilibrium

229811 The $pH$ of $HCl$ is 5 . If $10 mL$ of this solution is diluted to $100 mL$ , the $pH$ of the resultant solution is

1 5.1

2 6.9

3 11

4 12

Explanation:

Given that, $pH$ of $HCl = 5$
So, Concentration of $[H^{+}] = 10^{- 5} mol / L$
Concentration of $H^{+}$ ions in $10 ml$ of the solution
$\begin{aligned} = \frac{10^{- 5} \times 10}{1000} \\ = 10^{- 7} mol \end{aligned}$
We know that, solution is diluted to $100 ml$ i.e. 10 times
No. of $H^{+}$ ions $= \frac{10^{- 7}}{10} = 10^{- 8}$
Total $[H^{+}] = 10^{- 8} + 10^{- 7}$
$([H^{+}]$ from water cannot be neglected.)
$\begin{aligned} = 10^{- 8} (1 + 10) = 11 \times 10^{- 8} M \\ pH = - \log (11 \times 10^{- 8}) = - (1.0413 - 8) = 6.95 \end{aligned}$

COMEDK-2011

Ionic Equilibrium

229812 The $pH$ of a mixture of $10 ml$ of $0.1 M H_{2} {SO}_{4}, 5$ $ml$ of $0.2 M HCl$ and $5 ml$ of $0.1 M Ca (OH)_{2}$ is

1 1

2 0.5

3 0

4 1.5

Explanation:

{l}
|Exp: Molarity of mixture of $Double subscripts: use braces to clarify$ |
|---|
| $= 1$ millimole of $Double subscripts: use braces to clarify$ |
| $= 2$ millimole of $^{-}$ ions |
| $5$ of $0.2 = 1$ millimole of $^{+}$ ions |
| $5$ of $0.1 ()_{2} = 0.5$ millimole of $()_{2}$ |
| $= 1$ millimole of $^{-}$ ions |
|1 millimole of $^{-}$ ions will neutralize the 1 millimole |
|of $^{+}$ ions. |
|So, 2 millimoles of $^{+}$ ions will remain in solution. |
|{ $[^{+}] 20 = 0.1 = 10^{- 1}$ mol/ } |
|Hence, $= - [^{+}] = - (10^{- 1}) = 1$

COMEDK-2012

Ionic Equilibrium

229808 The dissociation constant of a weak base is $1 \times 10^{- 5}$ at $25^{\circ} C$ . The $pH$ of its $0.1 M$ solution at the same temperature will be

1 11

2 3

3 6

4 13

Explanation:

Given,
dissociation constant of weak base $= 1 \times 10^{- 5}$ concentration $= 0.1 M$
By Ostwald dilution law,
$[{OH}^{-}]$ in weak base $= \sqrt{K_{b} C}$
$= \sqrt{1 \times 10^{- 5} \times 0.1} = 1 \times 10^{- 3} M$
$\begin{aligned} pOH = - \log [{OH}^{-}] \\ = - \log (1 \times 10^{- 3}) \\ pOH = 3 \end{aligned}$
We know that-
$\begin{aligned} \Rightarrow pH + pOH = 14 \\ \Rightarrow pH = 14 - 3 = 11 \end{aligned}$

COMEDK-2011

Ionic Equilibrium

229809 The $pH$ of a solution of $NaOH$ is 12. The mass of $NaOH$ present in $500 mL$ of the solution is

1

400 mg

2

200 mg

3

20 mg

4

40 mg

Explanation:

Given that,
$pH = 12$
$∵ pH + pOH = 14$
$pOH = 14 - 12 = 2$
Hence, $NaOH$ concentration $= 10^{- 2} M = 10^{- 2} N$
We know that, $N = \frac{w \times 1000}{Eq.wt \times Volume (mL)}$
$\Rightarrow 10^{- 2} = \frac{w \times 1000}{40 \times 500}$
$\Rightarrow w = \frac{10^{- 2} \times 40 \times 500}{1000} = 10^{- 2} \times 20 = 0.2 g = 200 mg$

COMEDK-2011

Ionic Equilibrium

229810 The $pOH$ of $0.0005 M$ sulphuric acid is

1 5

2 3

3 11

4 12

Explanation:

The equation for ionization is-
$H_{2} {SO}_{4} ◻ 2 H^{+} + {SO}_{4}^{2 -}$
Each molecule of acid gives 2 molecules of $H^{+}$
So, $[H^{+}] = 2 \times 0.0005 = 1 \times 10^{- 3} M$
$pH = - \log [H^{+}] = - \log (1 \times 10^{- 3}) = 3$
We know that-
$\begin{aligned} ∵ pH + pOH = 14 \\ 3 + pOH = 14 \\ pOH = 14 - 3 \\ pOH = 11 \end{aligned}$

COMEDK-2011

Ionic Equilibrium

229811 The $pH$ of $HCl$ is 5 . If $10 mL$ of this solution is diluted to $100 mL$ , the $pH$ of the resultant solution is

1 5.1

2 6.9

3 11

4 12

Explanation:

Given that, $pH$ of $HCl = 5$
So, Concentration of $[H^{+}] = 10^{- 5} mol / L$
Concentration of $H^{+}$ ions in $10 ml$ of the solution
$\begin{aligned} = \frac{10^{- 5} \times 10}{1000} \\ = 10^{- 7} mol \end{aligned}$
We know that, solution is diluted to $100 ml$ i.e. 10 times
No. of $H^{+}$ ions $= \frac{10^{- 7}}{10} = 10^{- 8}$
Total $[H^{+}] = 10^{- 8} + 10^{- 7}$
$([H^{+}]$ from water cannot be neglected.)
$\begin{aligned} = 10^{- 8} (1 + 10) = 11 \times 10^{- 8} M \\ pH = - \log (11 \times 10^{- 8}) = - (1.0413 - 8) = 6.95 \end{aligned}$

COMEDK-2011

Ionic Equilibrium

229812 The $pH$ of a mixture of $10 ml$ of $0.1 M H_{2} {SO}_{4}, 5$ $ml$ of $0.2 M HCl$ and $5 ml$ of $0.1 M Ca (OH)_{2}$ is

1 1

2 0.5

3 0

4 1.5

Explanation:

{l}
|Exp: Molarity of mixture of $Double subscripts: use braces to clarify$ |
|---|
| $= 1$ millimole of $Double subscripts: use braces to clarify$ |
| $= 2$ millimole of $^{-}$ ions |
| $5$ of $0.2 = 1$ millimole of $^{+}$ ions |
| $5$ of $0.1 ()_{2} = 0.5$ millimole of $()_{2}$ |
| $= 1$ millimole of $^{-}$ ions |
|1 millimole of $^{-}$ ions will neutralize the 1 millimole |
|of $^{+}$ ions. |
|So, 2 millimoles of $^{+}$ ions will remain in solution. |
|{ $[^{+}] 20 = 0.1 = 10^{- 1}$ mol/ } |
|Hence, $= - [^{+}] = - (10^{- 1}) = 1$

COMEDK-2012

Ionic Equilibrium

229808 The dissociation constant of a weak base is $1 \times 10^{- 5}$ at $25^{\circ} C$ . The $pH$ of its $0.1 M$ solution at the same temperature will be

1 11

2 3

3 6

4 13

Explanation:

Given,
dissociation constant of weak base $= 1 \times 10^{- 5}$ concentration $= 0.1 M$
By Ostwald dilution law,
$[{OH}^{-}]$ in weak base $= \sqrt{K_{b} C}$
$= \sqrt{1 \times 10^{- 5} \times 0.1} = 1 \times 10^{- 3} M$
$\begin{aligned} pOH = - \log [{OH}^{-}] \\ = - \log (1 \times 10^{- 3}) \\ pOH = 3 \end{aligned}$
We know that-
$\begin{aligned} \Rightarrow pH + pOH = 14 \\ \Rightarrow pH = 14 - 3 = 11 \end{aligned}$

COMEDK-2011

Ionic Equilibrium

229809 The $pH$ of a solution of $NaOH$ is 12. The mass of $NaOH$ present in $500 mL$ of the solution is

1

400 mg

2

200 mg

3

20 mg

4

40 mg

Explanation:

Given that,
$pH = 12$
$∵ pH + pOH = 14$
$pOH = 14 - 12 = 2$
Hence, $NaOH$ concentration $= 10^{- 2} M = 10^{- 2} N$
We know that, $N = \frac{w \times 1000}{Eq.wt \times Volume (mL)}$
$\Rightarrow 10^{- 2} = \frac{w \times 1000}{40 \times 500}$
$\Rightarrow w = \frac{10^{- 2} \times 40 \times 500}{1000} = 10^{- 2} \times 20 = 0.2 g = 200 mg$

COMEDK-2011

Ionic Equilibrium

229810 The $pOH$ of $0.0005 M$ sulphuric acid is

1 5

2 3

3 11

4 12

Explanation:

The equation for ionization is-
$H_{2} {SO}_{4} ◻ 2 H^{+} + {SO}_{4}^{2 -}$
Each molecule of acid gives 2 molecules of $H^{+}$
So, $[H^{+}] = 2 \times 0.0005 = 1 \times 10^{- 3} M$
$pH = - \log [H^{+}] = - \log (1 \times 10^{- 3}) = 3$
We know that-
$\begin{aligned} ∵ pH + pOH = 14 \\ 3 + pOH = 14 \\ pOH = 14 - 3 \\ pOH = 11 \end{aligned}$

COMEDK-2011

Ionic Equilibrium

229811 The $pH$ of $HCl$ is 5 . If $10 mL$ of this solution is diluted to $100 mL$ , the $pH$ of the resultant solution is

1 5.1

2 6.9

3 11

4 12

Explanation:

Given that, $pH$ of $HCl = 5$
So, Concentration of $[H^{+}] = 10^{- 5} mol / L$
Concentration of $H^{+}$ ions in $10 ml$ of the solution
$\begin{aligned} = \frac{10^{- 5} \times 10}{1000} \\ = 10^{- 7} mol \end{aligned}$
We know that, solution is diluted to $100 ml$ i.e. 10 times
No. of $H^{+}$ ions $= \frac{10^{- 7}}{10} = 10^{- 8}$
Total $[H^{+}] = 10^{- 8} + 10^{- 7}$
$([H^{+}]$ from water cannot be neglected.)
$\begin{aligned} = 10^{- 8} (1 + 10) = 11 \times 10^{- 8} M \\ pH = - \log (11 \times 10^{- 8}) = - (1.0413 - 8) = 6.95 \end{aligned}$

COMEDK-2011

Ionic Equilibrium

229812 The $pH$ of a mixture of $10 ml$ of $0.1 M H_{2} {SO}_{4}, 5$ $ml$ of $0.2 M HCl$ and $5 ml$ of $0.1 M Ca (OH)_{2}$ is

1 1

2 0.5

3 0

4 1.5

Explanation:

{l}
|Exp: Molarity of mixture of $Double subscripts: use braces to clarify$ |
|---|
| $= 1$ millimole of $Double subscripts: use braces to clarify$ |
| $= 2$ millimole of $^{-}$ ions |
| $5$ of $0.2 = 1$ millimole of $^{+}$ ions |
| $5$ of $0.1 ()_{2} = 0.5$ millimole of $()_{2}$ |
| $= 1$ millimole of $^{-}$ ions |
|1 millimole of $^{-}$ ions will neutralize the 1 millimole |
|of $^{+}$ ions. |
|So, 2 millimoles of $^{+}$ ions will remain in solution. |
|{ $[^{+}] 20 = 0.1 = 10^{- 1}$ mol/ } |
|Hence, $= - [^{+}] = - (10^{- 1}) = 1$

COMEDK-2012

Ionic Equilibrium

229808 The dissociation constant of a weak base is $1 \times 10^{- 5}$ at $25^{\circ} C$ . The $pH$ of its $0.1 M$ solution at the same temperature will be

1 11

2 3

3 6

4 13

Explanation:

Given,
dissociation constant of weak base $= 1 \times 10^{- 5}$ concentration $= 0.1 M$
By Ostwald dilution law,
$[{OH}^{-}]$ in weak base $= \sqrt{K_{b} C}$
$= \sqrt{1 \times 10^{- 5} \times 0.1} = 1 \times 10^{- 3} M$
$\begin{aligned} pOH = - \log [{OH}^{-}] \\ = - \log (1 \times 10^{- 3}) \\ pOH = 3 \end{aligned}$
We know that-
$\begin{aligned} \Rightarrow pH + pOH = 14 \\ \Rightarrow pH = 14 - 3 = 11 \end{aligned}$

COMEDK-2011

Ionic Equilibrium

229809 The $pH$ of a solution of $NaOH$ is 12. The mass of $NaOH$ present in $500 mL$ of the solution is

1

400 mg

2

200 mg

3

20 mg

4

40 mg

Explanation:

Given that,
$pH = 12$
$∵ pH + pOH = 14$
$pOH = 14 - 12 = 2$
Hence, $NaOH$ concentration $= 10^{- 2} M = 10^{- 2} N$
We know that, $N = \frac{w \times 1000}{Eq.wt \times Volume (mL)}$
$\Rightarrow 10^{- 2} = \frac{w \times 1000}{40 \times 500}$
$\Rightarrow w = \frac{10^{- 2} \times 40 \times 500}{1000} = 10^{- 2} \times 20 = 0.2 g = 200 mg$

COMEDK-2011

Ionic Equilibrium

229810 The $pOH$ of $0.0005 M$ sulphuric acid is

1 5

2 3

3 11

4 12

Explanation:

The equation for ionization is-
$H_{2} {SO}_{4} ◻ 2 H^{+} + {SO}_{4}^{2 -}$
Each molecule of acid gives 2 molecules of $H^{+}$
So, $[H^{+}] = 2 \times 0.0005 = 1 \times 10^{- 3} M$
$pH = - \log [H^{+}] = - \log (1 \times 10^{- 3}) = 3$
We know that-
$\begin{aligned} ∵ pH + pOH = 14 \\ 3 + pOH = 14 \\ pOH = 14 - 3 \\ pOH = 11 \end{aligned}$

COMEDK-2011

Ionic Equilibrium

229811 The $pH$ of $HCl$ is 5 . If $10 mL$ of this solution is diluted to $100 mL$ , the $pH$ of the resultant solution is

1 5.1

2 6.9

3 11

4 12

Explanation:

Given that, $pH$ of $HCl = 5$
So, Concentration of $[H^{+}] = 10^{- 5} mol / L$
Concentration of $H^{+}$ ions in $10 ml$ of the solution
$\begin{aligned} = \frac{10^{- 5} \times 10}{1000} \\ = 10^{- 7} mol \end{aligned}$
We know that, solution is diluted to $100 ml$ i.e. 10 times
No. of $H^{+}$ ions $= \frac{10^{- 7}}{10} = 10^{- 8}$
Total $[H^{+}] = 10^{- 8} + 10^{- 7}$
$([H^{+}]$ from water cannot be neglected.)
$\begin{aligned} = 10^{- 8} (1 + 10) = 11 \times 10^{- 8} M \\ pH = - \log (11 \times 10^{- 8}) = - (1.0413 - 8) = 6.95 \end{aligned}$

COMEDK-2011

Ionic Equilibrium

229812 The $pH$ of a mixture of $10 ml$ of $0.1 M H_{2} {SO}_{4}, 5$ $ml$ of $0.2 M HCl$ and $5 ml$ of $0.1 M Ca (OH)_{2}$ is

1 1

2 0.5

3 0

4 1.5

Explanation:

{l}
|Exp: Molarity of mixture of $Double subscripts: use braces to clarify$ |
|---|
| $= 1$ millimole of $Double subscripts: use braces to clarify$ |
| $= 2$ millimole of $^{-}$ ions |
| $5$ of $0.2 = 1$ millimole of $^{+}$ ions |
| $5$ of $0.1 ()_{2} = 0.5$ millimole of $()_{2}$ |
| $= 1$ millimole of $^{-}$ ions |
|1 millimole of $^{-}$ ions will neutralize the 1 millimole |
|of $^{+}$ ions. |
|So, 2 millimoles of $^{+}$ ions will remain in solution. |
|{ $[^{+}] 20 = 0.1 = 10^{- 1}$ mol/ } |
|Hence, $= - [^{+}] = - (10^{- 1}) = 1$

COMEDK-2012

Ionic Equilibrium

229808 The dissociation constant of a weak base is $1 \times 10^{- 5}$ at $25^{\circ} C$ . The $pH$ of its $0.1 M$ solution at the same temperature will be

1 11

2 3

3 6

4 13

Explanation:

Given,
dissociation constant of weak base $= 1 \times 10^{- 5}$ concentration $= 0.1 M$
By Ostwald dilution law,
$[{OH}^{-}]$ in weak base $= \sqrt{K_{b} C}$
$= \sqrt{1 \times 10^{- 5} \times 0.1} = 1 \times 10^{- 3} M$
$\begin{aligned} pOH = - \log [{OH}^{-}] \\ = - \log (1 \times 10^{- 3}) \\ pOH = 3 \end{aligned}$
We know that-
$\begin{aligned} \Rightarrow pH + pOH = 14 \\ \Rightarrow pH = 14 - 3 = 11 \end{aligned}$

COMEDK-2011

Ionic Equilibrium

229809 The $pH$ of a solution of $NaOH$ is 12. The mass of $NaOH$ present in $500 mL$ of the solution is

1

400 mg

2

200 mg

3

20 mg

4

40 mg

Explanation:

Given that,
$pH = 12$
$∵ pH + pOH = 14$
$pOH = 14 - 12 = 2$
Hence, $NaOH$ concentration $= 10^{- 2} M = 10^{- 2} N$
We know that, $N = \frac{w \times 1000}{Eq.wt \times Volume (mL)}$
$\Rightarrow 10^{- 2} = \frac{w \times 1000}{40 \times 500}$
$\Rightarrow w = \frac{10^{- 2} \times 40 \times 500}{1000} = 10^{- 2} \times 20 = 0.2 g = 200 mg$

COMEDK-2011

Ionic Equilibrium

229810 The $pOH$ of $0.0005 M$ sulphuric acid is

1 5

2 3

3 11

4 12

Explanation:

The equation for ionization is-
$H_{2} {SO}_{4} ◻ 2 H^{+} + {SO}_{4}^{2 -}$
Each molecule of acid gives 2 molecules of $H^{+}$
So, $[H^{+}] = 2 \times 0.0005 = 1 \times 10^{- 3} M$
$pH = - \log [H^{+}] = - \log (1 \times 10^{- 3}) = 3$
We know that-
$\begin{aligned} ∵ pH + pOH = 14 \\ 3 + pOH = 14 \\ pOH = 14 - 3 \\ pOH = 11 \end{aligned}$

COMEDK-2011

Ionic Equilibrium

229811 The $pH$ of $HCl$ is 5 . If $10 mL$ of this solution is diluted to $100 mL$ , the $pH$ of the resultant solution is

1 5.1

2 6.9

3 11

4 12

Explanation:

Given that, $pH$ of $HCl = 5$
So, Concentration of $[H^{+}] = 10^{- 5} mol / L$
Concentration of $H^{+}$ ions in $10 ml$ of the solution
$\begin{aligned} = \frac{10^{- 5} \times 10}{1000} \\ = 10^{- 7} mol \end{aligned}$
We know that, solution is diluted to $100 ml$ i.e. 10 times
No. of $H^{+}$ ions $= \frac{10^{- 7}}{10} = 10^{- 8}$
Total $[H^{+}] = 10^{- 8} + 10^{- 7}$
$([H^{+}]$ from water cannot be neglected.)
$\begin{aligned} = 10^{- 8} (1 + 10) = 11 \times 10^{- 8} M \\ pH = - \log (11 \times 10^{- 8}) = - (1.0413 - 8) = 6.95 \end{aligned}$

COMEDK-2011

Ionic Equilibrium

229812 The $pH$ of a mixture of $10 ml$ of $0.1 M H_{2} {SO}_{4}, 5$ $ml$ of $0.2 M HCl$ and $5 ml$ of $0.1 M Ca (OH)_{2}$ is

1 1

2 0.5

3 0

4 1.5

Explanation:

{l}
|Exp: Molarity of mixture of $Double subscripts: use braces to clarify$ |
|---|
| $= 1$ millimole of $Double subscripts: use braces to clarify$ |
| $= 2$ millimole of $^{-}$ ions |
| $5$ of $0.2 = 1$ millimole of $^{+}$ ions |
| $5$ of $0.1 ()_{2} = 0.5$ millimole of $()_{2}$ |
| $= 1$ millimole of $^{-}$ ions |
|1 millimole of $^{-}$ ions will neutralize the 1 millimole |
|of $^{+}$ ions. |
|So, 2 millimoles of $^{+}$ ions will remain in solution. |
|{ $[^{+}] 20 = 0.1 = 10^{- 1}$ mol/ } |
|Hence, $= - [^{+}] = - (10^{- 1}) = 1$

COMEDK-2012

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