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CHXI07:EQUILIBRIUM

314996 HCOOH and ${CH}_{3} COOH$ solutions have equal pH . If $K_{1} / K_{2}$ is 4 , their molar concentration ratio will be

1 2

2 0.5

3 4

4 0.25

Explanation:

$pH = - \log [H^{+}]$
and $[H^{+}] = C α = \sqrt{K_{a} C}$
Given pH of two acids as same, so
$\sqrt{K_{a_{1}} C_{1}} = \sqrt{K_{a_{2}} C_{2}}$
$\frac{K_{a_{1}}}{K_{a_{2}}} = \frac{C_{2}}{C_{1}}$
$4 = \frac{C_{2}}{C_{1}}$
or $\frac{C_{1}}{C_{2}} = \frac{1}{4} = 0.25$

CHXI07:EQUILIBRIUM

314997 When $50 mL 0 .6 M HCl$ and $50 mL 0 .3 M NaOH$ are mixed, the species with maximum concentration in the resulting solution is

1

H^{+}

2

{Na}^{+}

3

{Cl}^{-}

4

{OH}^{-}

Explanation:

$\begin{matrix} HCl + & NaOH \to & NaCl + & H_{2} O \\ Intial mili moles & 30 & 15 & 0 & 0 \\ At time't' & 15 & 0 & 15 & 15 \end{matrix}$
$[H^{+}] = 0.15 M [{Cl}^{-}] = 0.30 M$
$[{Na}^{+}] = 0.15 M [{OH}^{-}] = 10^{- 13} M$

CHXI07:EQUILIBRIUM

314998 Equal volumes of $0.1 M$ potassium hydroxide and $0.1 M$ sulphuric acid are mixed. The $pH$ of resulting solution is

1 7

2 0

3 Less than 7

4 Greater than 7

Explanation:

m.eq. of acid $= 0.1 \times 2 \times V$ m.eq. of base $= 0.1 \times V$
Since number of milli equivalents of acid is greater than the base the solution is acidic in nature.

CHXI07:EQUILIBRIUM

314999 $5 mL$ of $0 .4 N N a O H$ is mxied with $20 mL$ of $0 .1 N H C l$ The $pH$ of the resulting soluiton will be

1 6

2 7

3 8

4 5

Explanation:

Millimoles of $NaOH = 5 \times 0.4 = 2$
Millimoles of $HCl = 20 \times 0.1 = 2$
So, 2 millimoles of $NaOH$ and 2 millimoles of $HCl$ completely neutralise each other, resulting into a neutral solution $(pH = 7)$ .

CHXI07:EQUILIBRIUM

315000 Determine the $pH$ of the solution that results from the addition of $20.00 mL$ of $0.01 M Ca (OH)_{2}$ to $30.00 mL$ of $0.01 M HCl$ .

1 11.30

2 10.55

3 2.70

4 83.35

Explanation:

Millimoles of $H^{+} = 30 \times 0.01 = 0.3$
Millimole of ${OH}^{-} = 20 \times 0.01 \times 2 = 0.4$
Remaining millimoles of ${OH}^{-}$
$\begin{aligned} = & 0.4 - 0.3 = 0.1 \\ \Rightarrow [{OH}^{-}] & = \frac{0.1}{50} or 2 \times 10^{- 3} \end{aligned}$
So, $pOH = 2.7$
$pH = 14 - 2.7 = 11.30$

CHXI07:EQUILIBRIUM

314996 HCOOH and ${CH}_{3} COOH$ solutions have equal pH . If $K_{1} / K_{2}$ is 4 , their molar concentration ratio will be

1 2

2 0.5

3 4

4 0.25

Explanation:

$pH = - \log [H^{+}]$
and $[H^{+}] = C α = \sqrt{K_{a} C}$
Given pH of two acids as same, so
$\sqrt{K_{a_{1}} C_{1}} = \sqrt{K_{a_{2}} C_{2}}$
$\frac{K_{a_{1}}}{K_{a_{2}}} = \frac{C_{2}}{C_{1}}$
$4 = \frac{C_{2}}{C_{1}}$
or $\frac{C_{1}}{C_{2}} = \frac{1}{4} = 0.25$

CHXI07:EQUILIBRIUM

314997 When $50 mL 0 .6 M HCl$ and $50 mL 0 .3 M NaOH$ are mixed, the species with maximum concentration in the resulting solution is

1

H^{+}

2

{Na}^{+}

3

{Cl}^{-}

4

{OH}^{-}

Explanation:

$\begin{matrix} HCl + & NaOH \to & NaCl + & H_{2} O \\ Intial mili moles & 30 & 15 & 0 & 0 \\ At time't' & 15 & 0 & 15 & 15 \end{matrix}$
$[H^{+}] = 0.15 M [{Cl}^{-}] = 0.30 M$
$[{Na}^{+}] = 0.15 M [{OH}^{-}] = 10^{- 13} M$

CHXI07:EQUILIBRIUM

314998 Equal volumes of $0.1 M$ potassium hydroxide and $0.1 M$ sulphuric acid are mixed. The $pH$ of resulting solution is

1 7

2 0

3 Less than 7

4 Greater than 7

Explanation:

m.eq. of acid $= 0.1 \times 2 \times V$ m.eq. of base $= 0.1 \times V$
Since number of milli equivalents of acid is greater than the base the solution is acidic in nature.

CHXI07:EQUILIBRIUM

314999 $5 mL$ of $0 .4 N N a O H$ is mxied with $20 mL$ of $0 .1 N H C l$ The $pH$ of the resulting soluiton will be

1 6

2 7

3 8

4 5

Explanation:

Millimoles of $NaOH = 5 \times 0.4 = 2$
Millimoles of $HCl = 20 \times 0.1 = 2$
So, 2 millimoles of $NaOH$ and 2 millimoles of $HCl$ completely neutralise each other, resulting into a neutral solution $(pH = 7)$ .

CHXI07:EQUILIBRIUM

315000 Determine the $pH$ of the solution that results from the addition of $20.00 mL$ of $0.01 M Ca (OH)_{2}$ to $30.00 mL$ of $0.01 M HCl$ .

1 11.30

2 10.55

3 2.70

4 83.35

Explanation:

Millimoles of $H^{+} = 30 \times 0.01 = 0.3$
Millimole of ${OH}^{-} = 20 \times 0.01 \times 2 = 0.4$
Remaining millimoles of ${OH}^{-}$
$\begin{aligned} = & 0.4 - 0.3 = 0.1 \\ \Rightarrow [{OH}^{-}] & = \frac{0.1}{50} or 2 \times 10^{- 3} \end{aligned}$
So, $pOH = 2.7$
$pH = 14 - 2.7 = 11.30$

CHXI07:EQUILIBRIUM

314996 HCOOH and ${CH}_{3} COOH$ solutions have equal pH . If $K_{1} / K_{2}$ is 4 , their molar concentration ratio will be

1 2

2 0.5

3 4

4 0.25

Explanation:

$pH = - \log [H^{+}]$
and $[H^{+}] = C α = \sqrt{K_{a} C}$
Given pH of two acids as same, so
$\sqrt{K_{a_{1}} C_{1}} = \sqrt{K_{a_{2}} C_{2}}$
$\frac{K_{a_{1}}}{K_{a_{2}}} = \frac{C_{2}}{C_{1}}$
$4 = \frac{C_{2}}{C_{1}}$
or $\frac{C_{1}}{C_{2}} = \frac{1}{4} = 0.25$

CHXI07:EQUILIBRIUM

314997 When $50 mL 0 .6 M HCl$ and $50 mL 0 .3 M NaOH$ are mixed, the species with maximum concentration in the resulting solution is

1

H^{+}

2

{Na}^{+}

3

{Cl}^{-}

4

{OH}^{-}

Explanation:

$\begin{matrix} HCl + & NaOH \to & NaCl + & H_{2} O \\ Intial mili moles & 30 & 15 & 0 & 0 \\ At time't' & 15 & 0 & 15 & 15 \end{matrix}$
$[H^{+}] = 0.15 M [{Cl}^{-}] = 0.30 M$
$[{Na}^{+}] = 0.15 M [{OH}^{-}] = 10^{- 13} M$

CHXI07:EQUILIBRIUM

314998 Equal volumes of $0.1 M$ potassium hydroxide and $0.1 M$ sulphuric acid are mixed. The $pH$ of resulting solution is

1 7

2 0

3 Less than 7

4 Greater than 7

Explanation:

m.eq. of acid $= 0.1 \times 2 \times V$ m.eq. of base $= 0.1 \times V$
Since number of milli equivalents of acid is greater than the base the solution is acidic in nature.

CHXI07:EQUILIBRIUM

314999 $5 mL$ of $0 .4 N N a O H$ is mxied with $20 mL$ of $0 .1 N H C l$ The $pH$ of the resulting soluiton will be

1 6

2 7

3 8

4 5

Explanation:

Millimoles of $NaOH = 5 \times 0.4 = 2$
Millimoles of $HCl = 20 \times 0.1 = 2$
So, 2 millimoles of $NaOH$ and 2 millimoles of $HCl$ completely neutralise each other, resulting into a neutral solution $(pH = 7)$ .

CHXI07:EQUILIBRIUM

315000 Determine the $pH$ of the solution that results from the addition of $20.00 mL$ of $0.01 M Ca (OH)_{2}$ to $30.00 mL$ of $0.01 M HCl$ .

1 11.30

2 10.55

3 2.70

4 83.35

Explanation:

Millimoles of $H^{+} = 30 \times 0.01 = 0.3$
Millimole of ${OH}^{-} = 20 \times 0.01 \times 2 = 0.4$
Remaining millimoles of ${OH}^{-}$
$\begin{aligned} = & 0.4 - 0.3 = 0.1 \\ \Rightarrow [{OH}^{-}] & = \frac{0.1}{50} or 2 \times 10^{- 3} \end{aligned}$
So, $pOH = 2.7$
$pH = 14 - 2.7 = 11.30$

CHXI07:EQUILIBRIUM

314996 HCOOH and ${CH}_{3} COOH$ solutions have equal pH . If $K_{1} / K_{2}$ is 4 , their molar concentration ratio will be

1 2

2 0.5

3 4

4 0.25

Explanation:

$pH = - \log [H^{+}]$
and $[H^{+}] = C α = \sqrt{K_{a} C}$
Given pH of two acids as same, so
$\sqrt{K_{a_{1}} C_{1}} = \sqrt{K_{a_{2}} C_{2}}$
$\frac{K_{a_{1}}}{K_{a_{2}}} = \frac{C_{2}}{C_{1}}$
$4 = \frac{C_{2}}{C_{1}}$
or $\frac{C_{1}}{C_{2}} = \frac{1}{4} = 0.25$

CHXI07:EQUILIBRIUM

314997 When $50 mL 0 .6 M HCl$ and $50 mL 0 .3 M NaOH$ are mixed, the species with maximum concentration in the resulting solution is

1

H^{+}

2

{Na}^{+}

3

{Cl}^{-}

4

{OH}^{-}

Explanation:

$\begin{matrix} HCl + & NaOH \to & NaCl + & H_{2} O \\ Intial mili moles & 30 & 15 & 0 & 0 \\ At time't' & 15 & 0 & 15 & 15 \end{matrix}$
$[H^{+}] = 0.15 M [{Cl}^{-}] = 0.30 M$
$[{Na}^{+}] = 0.15 M [{OH}^{-}] = 10^{- 13} M$

CHXI07:EQUILIBRIUM

314998 Equal volumes of $0.1 M$ potassium hydroxide and $0.1 M$ sulphuric acid are mixed. The $pH$ of resulting solution is

1 7

2 0

3 Less than 7

4 Greater than 7

Explanation:

m.eq. of acid $= 0.1 \times 2 \times V$ m.eq. of base $= 0.1 \times V$
Since number of milli equivalents of acid is greater than the base the solution is acidic in nature.

CHXI07:EQUILIBRIUM

314999 $5 mL$ of $0 .4 N N a O H$ is mxied with $20 mL$ of $0 .1 N H C l$ The $pH$ of the resulting soluiton will be

1 6

2 7

3 8

4 5

Explanation:

Millimoles of $NaOH = 5 \times 0.4 = 2$
Millimoles of $HCl = 20 \times 0.1 = 2$
So, 2 millimoles of $NaOH$ and 2 millimoles of $HCl$ completely neutralise each other, resulting into a neutral solution $(pH = 7)$ .

CHXI07:EQUILIBRIUM

315000 Determine the $pH$ of the solution that results from the addition of $20.00 mL$ of $0.01 M Ca (OH)_{2}$ to $30.00 mL$ of $0.01 M HCl$ .

1 11.30

2 10.55

3 2.70

4 83.35

Explanation:

Millimoles of $H^{+} = 30 \times 0.01 = 0.3$
Millimole of ${OH}^{-} = 20 \times 0.01 \times 2 = 0.4$
Remaining millimoles of ${OH}^{-}$
$\begin{aligned} = & 0.4 - 0.3 = 0.1 \\ \Rightarrow [{OH}^{-}] & = \frac{0.1}{50} or 2 \times 10^{- 3} \end{aligned}$
So, $pOH = 2.7$
$pH = 14 - 2.7 = 11.30$

CHXI07:EQUILIBRIUM

314996 HCOOH and ${CH}_{3} COOH$ solutions have equal pH . If $K_{1} / K_{2}$ is 4 , their molar concentration ratio will be

1 2

2 0.5

3 4

4 0.25

Explanation:

$pH = - \log [H^{+}]$
and $[H^{+}] = C α = \sqrt{K_{a} C}$
Given pH of two acids as same, so
$\sqrt{K_{a_{1}} C_{1}} = \sqrt{K_{a_{2}} C_{2}}$
$\frac{K_{a_{1}}}{K_{a_{2}}} = \frac{C_{2}}{C_{1}}$
$4 = \frac{C_{2}}{C_{1}}$
or $\frac{C_{1}}{C_{2}} = \frac{1}{4} = 0.25$

CHXI07:EQUILIBRIUM

314997 When $50 mL 0 .6 M HCl$ and $50 mL 0 .3 M NaOH$ are mixed, the species with maximum concentration in the resulting solution is

1

H^{+}

2

{Na}^{+}

3

{Cl}^{-}

4

{OH}^{-}

Explanation:

$\begin{matrix} HCl + & NaOH \to & NaCl + & H_{2} O \\ Intial mili moles & 30 & 15 & 0 & 0 \\ At time't' & 15 & 0 & 15 & 15 \end{matrix}$
$[H^{+}] = 0.15 M [{Cl}^{-}] = 0.30 M$
$[{Na}^{+}] = 0.15 M [{OH}^{-}] = 10^{- 13} M$

CHXI07:EQUILIBRIUM

314998 Equal volumes of $0.1 M$ potassium hydroxide and $0.1 M$ sulphuric acid are mixed. The $pH$ of resulting solution is

1 7

2 0

3 Less than 7

4 Greater than 7

Explanation:

m.eq. of acid $= 0.1 \times 2 \times V$ m.eq. of base $= 0.1 \times V$
Since number of milli equivalents of acid is greater than the base the solution is acidic in nature.

CHXI07:EQUILIBRIUM

314999 $5 mL$ of $0 .4 N N a O H$ is mxied with $20 mL$ of $0 .1 N H C l$ The $pH$ of the resulting soluiton will be

1 6

2 7

3 8

4 5

Explanation:

Millimoles of $NaOH = 5 \times 0.4 = 2$
Millimoles of $HCl = 20 \times 0.1 = 2$
So, 2 millimoles of $NaOH$ and 2 millimoles of $HCl$ completely neutralise each other, resulting into a neutral solution $(pH = 7)$ .

CHXI07:EQUILIBRIUM

315000 Determine the $pH$ of the solution that results from the addition of $20.00 mL$ of $0.01 M Ca (OH)_{2}$ to $30.00 mL$ of $0.01 M HCl$ .

1 11.30

2 10.55

3 2.70

4 83.35

Explanation:

Millimoles of $H^{+} = 30 \times 0.01 = 0.3$
Millimole of ${OH}^{-} = 20 \times 0.01 \times 2 = 0.4$
Remaining millimoles of ${OH}^{-}$
$\begin{aligned} = & 0.4 - 0.3 = 0.1 \\ \Rightarrow [{OH}^{-}] & = \frac{0.1}{50} or 2 \times 10^{- 3} \end{aligned}$
So, $pOH = 2.7$
$pH = 14 - 2.7 = 11.30$

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