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CHXI07:EQUILIBRIUM

315009 If $50 mL$ of $0.1 HBr$ is mixed with $50 m L 0.2 M N a O H$ , find $pH$ of resulting mixture

1 2.7

2 12.7

3 10.7

4 1.3

Explanation:

$\begin{array}{l} N_{Mix} V_{Mix} = N_{1} V_{1} - N_{2} V_{2} (1) \\ (Base) (Acid) \end{array}$
Given, $M_{1} = 0.2 M, N_{1} = 0.2 N$
$\begin{aligned} M_{2} = 0.1 M, N_{2} = 0.1 N \\ V_{1} = 50 mL, V_{2} = 50 mL, V_{mix} = 100 mL \end{aligned}$
On substituting the given values in equation (1)
$N_{mix} \times 100 = 50 \times 0.2 - 50 \times 0.1$
$N_{m i x} = \frac{5}{100} = 5 \times 10^{- 2}$ $∴ [{OH}^{-}] = 5 \times 10^{- 2}$
Now, pOH $= - \log [{OH}^{-}] = - \log [5 \times 10^{- 2}] $ $ = 2 - \log 5 = 1.3$
Also, $pH + pOH = 14 $ $ pH = 14 - 1.3 = 12.7$

CHXI07:EQUILIBRIUM

315010 Following solutions were prepared by mixing different volumes of $NaOH$ and $HCl$ of different concentrations :
(A) $60 mL \frac{M}{10} HCl + 40 mL \frac{M}{10} NaOH$
(B) $55 mL \frac{M}{10} HCl + 45 mL \frac{M}{10} NaOH$
(C) $75 mL \frac{M}{5} HCl + 25 mL \frac{M}{5} NaOH$
(D) $100 mL \frac{M}{10} HCl + 100 mL \frac{M}{10} NaOH$
$pH$ of which one of them will be equal to 1?

1 D

2 C

3 A

4 B

NEET - 2018

CHXI07:EQUILIBRIUM

315011 The $pH$ of $0 .005$ $M$ $Ba(OH)_{2}$ is

1 2.301

2 11.699

3 12

4 7

Explanation:

$[{OH}^{-}] = 2 \times 0.005 = 0.01$ or $10^{- 2}$
$pOH = - \log [{OH}^{-}] = 2 pH = 14 - pOH = 14 - 2 = 12$

CHXI07:EQUILIBRIUM

315012 What is the $p H o f 0 .02 M N a O H$ solution?

1 10.3

2 11.3

3 11.7

4 12.3

Explanation:

$[{OH}^{-}] = 2 \times 10^{- 2}$
$P^{OH} = - \log_{10} [{OH}^{-}]$
$= - \log 2 \times 10^{- 2}$
$= 2 - \log 2$
$= 2 - 0.3$
$P^{OH} = 1.7$
and $P^{H} = 14 - P^{OH} = 14 - 1.7$
$P^{H} = 12.3$

MHTCET - 2021

CHXI07:EQUILIBRIUM

315013 Which of the following has highest $pH$ ?

1

\frac{M}{4} K O H

2

\frac{M}{4} N a O H

3

\frac{M}{4} N H_{4} O H

4

\frac{M}{4} C a (O H)_{2}

Explanation:

$pH$ of solution $\propto [{OH}^{-}]$
$\frac{M}{4} Ca (OH)_{2}$ solution will give highest concentration of $O H^{-}$
Hence, it has highest $pH$ .

CHXI07:EQUILIBRIUM

315009 If $50 mL$ of $0.1 HBr$ is mixed with $50 m L 0.2 M N a O H$ , find $pH$ of resulting mixture

1 2.7

2 12.7

3 10.7

4 1.3

Explanation:

$\begin{array}{l} N_{Mix} V_{Mix} = N_{1} V_{1} - N_{2} V_{2} (1) \\ (Base) (Acid) \end{array}$
Given, $M_{1} = 0.2 M, N_{1} = 0.2 N$
$\begin{aligned} M_{2} = 0.1 M, N_{2} = 0.1 N \\ V_{1} = 50 mL, V_{2} = 50 mL, V_{mix} = 100 mL \end{aligned}$
On substituting the given values in equation (1)
$N_{mix} \times 100 = 50 \times 0.2 - 50 \times 0.1$
$N_{m i x} = \frac{5}{100} = 5 \times 10^{- 2}$ $∴ [{OH}^{-}] = 5 \times 10^{- 2}$
Now, pOH $= - \log [{OH}^{-}] = - \log [5 \times 10^{- 2}] $ $ = 2 - \log 5 = 1.3$
Also, $pH + pOH = 14 $ $ pH = 14 - 1.3 = 12.7$

CHXI07:EQUILIBRIUM

315010 Following solutions were prepared by mixing different volumes of $NaOH$ and $HCl$ of different concentrations :
(A) $60 mL \frac{M}{10} HCl + 40 mL \frac{M}{10} NaOH$
(B) $55 mL \frac{M}{10} HCl + 45 mL \frac{M}{10} NaOH$
(C) $75 mL \frac{M}{5} HCl + 25 mL \frac{M}{5} NaOH$
(D) $100 mL \frac{M}{10} HCl + 100 mL \frac{M}{10} NaOH$
$pH$ of which one of them will be equal to 1?

1 D

2 C

3 A

4 B

NEET - 2018

CHXI07:EQUILIBRIUM

315011 The $pH$ of $0 .005$ $M$ $Ba(OH)_{2}$ is

1 2.301

2 11.699

3 12

4 7

Explanation:

$[{OH}^{-}] = 2 \times 0.005 = 0.01$ or $10^{- 2}$
$pOH = - \log [{OH}^{-}] = 2 pH = 14 - pOH = 14 - 2 = 12$

CHXI07:EQUILIBRIUM

315012 What is the $p H o f 0 .02 M N a O H$ solution?

1 10.3

2 11.3

3 11.7

4 12.3

Explanation:

$[{OH}^{-}] = 2 \times 10^{- 2}$
$P^{OH} = - \log_{10} [{OH}^{-}]$
$= - \log 2 \times 10^{- 2}$
$= 2 - \log 2$
$= 2 - 0.3$
$P^{OH} = 1.7$
and $P^{H} = 14 - P^{OH} = 14 - 1.7$
$P^{H} = 12.3$

MHTCET - 2021

CHXI07:EQUILIBRIUM

315013 Which of the following has highest $pH$ ?

1

\frac{M}{4} K O H

2

\frac{M}{4} N a O H

3

\frac{M}{4} N H_{4} O H

4

\frac{M}{4} C a (O H)_{2}

Explanation:

$pH$ of solution $\propto [{OH}^{-}]$
$\frac{M}{4} Ca (OH)_{2}$ solution will give highest concentration of $O H^{-}$
Hence, it has highest $pH$ .

CHXI07:EQUILIBRIUM

315009 If $50 mL$ of $0.1 HBr$ is mixed with $50 m L 0.2 M N a O H$ , find $pH$ of resulting mixture

1 2.7

2 12.7

3 10.7

4 1.3

Explanation:

$\begin{array}{l} N_{Mix} V_{Mix} = N_{1} V_{1} - N_{2} V_{2} (1) \\ (Base) (Acid) \end{array}$
Given, $M_{1} = 0.2 M, N_{1} = 0.2 N$
$\begin{aligned} M_{2} = 0.1 M, N_{2} = 0.1 N \\ V_{1} = 50 mL, V_{2} = 50 mL, V_{mix} = 100 mL \end{aligned}$
On substituting the given values in equation (1)
$N_{mix} \times 100 = 50 \times 0.2 - 50 \times 0.1$
$N_{m i x} = \frac{5}{100} = 5 \times 10^{- 2}$ $∴ [{OH}^{-}] = 5 \times 10^{- 2}$
Now, pOH $= - \log [{OH}^{-}] = - \log [5 \times 10^{- 2}] $ $ = 2 - \log 5 = 1.3$
Also, $pH + pOH = 14 $ $ pH = 14 - 1.3 = 12.7$

CHXI07:EQUILIBRIUM

315010 Following solutions were prepared by mixing different volumes of $NaOH$ and $HCl$ of different concentrations :
(A) $60 mL \frac{M}{10} HCl + 40 mL \frac{M}{10} NaOH$
(B) $55 mL \frac{M}{10} HCl + 45 mL \frac{M}{10} NaOH$
(C) $75 mL \frac{M}{5} HCl + 25 mL \frac{M}{5} NaOH$
(D) $100 mL \frac{M}{10} HCl + 100 mL \frac{M}{10} NaOH$
$pH$ of which one of them will be equal to 1?

1 D

2 C

3 A

4 B

NEET - 2018

CHXI07:EQUILIBRIUM

315011 The $pH$ of $0 .005$ $M$ $Ba(OH)_{2}$ is

1 2.301

2 11.699

3 12

4 7

Explanation:

$[{OH}^{-}] = 2 \times 0.005 = 0.01$ or $10^{- 2}$
$pOH = - \log [{OH}^{-}] = 2 pH = 14 - pOH = 14 - 2 = 12$

CHXI07:EQUILIBRIUM

315012 What is the $p H o f 0 .02 M N a O H$ solution?

1 10.3

2 11.3

3 11.7

4 12.3

Explanation:

$[{OH}^{-}] = 2 \times 10^{- 2}$
$P^{OH} = - \log_{10} [{OH}^{-}]$
$= - \log 2 \times 10^{- 2}$
$= 2 - \log 2$
$= 2 - 0.3$
$P^{OH} = 1.7$
and $P^{H} = 14 - P^{OH} = 14 - 1.7$
$P^{H} = 12.3$

MHTCET - 2021

CHXI07:EQUILIBRIUM

315013 Which of the following has highest $pH$ ?

1

\frac{M}{4} K O H

2

\frac{M}{4} N a O H

3

\frac{M}{4} N H_{4} O H

4

\frac{M}{4} C a (O H)_{2}

Explanation:

$pH$ of solution $\propto [{OH}^{-}]$
$\frac{M}{4} Ca (OH)_{2}$ solution will give highest concentration of $O H^{-}$
Hence, it has highest $pH$ .

CHXI07:EQUILIBRIUM

315009 If $50 mL$ of $0.1 HBr$ is mixed with $50 m L 0.2 M N a O H$ , find $pH$ of resulting mixture

1 2.7

2 12.7

3 10.7

4 1.3

Explanation:

$\begin{array}{l} N_{Mix} V_{Mix} = N_{1} V_{1} - N_{2} V_{2} (1) \\ (Base) (Acid) \end{array}$
Given, $M_{1} = 0.2 M, N_{1} = 0.2 N$
$\begin{aligned} M_{2} = 0.1 M, N_{2} = 0.1 N \\ V_{1} = 50 mL, V_{2} = 50 mL, V_{mix} = 100 mL \end{aligned}$
On substituting the given values in equation (1)
$N_{mix} \times 100 = 50 \times 0.2 - 50 \times 0.1$
$N_{m i x} = \frac{5}{100} = 5 \times 10^{- 2}$ $∴ [{OH}^{-}] = 5 \times 10^{- 2}$
Now, pOH $= - \log [{OH}^{-}] = - \log [5 \times 10^{- 2}] $ $ = 2 - \log 5 = 1.3$
Also, $pH + pOH = 14 $ $ pH = 14 - 1.3 = 12.7$

CHXI07:EQUILIBRIUM

315010 Following solutions were prepared by mixing different volumes of $NaOH$ and $HCl$ of different concentrations :
(A) $60 mL \frac{M}{10} HCl + 40 mL \frac{M}{10} NaOH$
(B) $55 mL \frac{M}{10} HCl + 45 mL \frac{M}{10} NaOH$
(C) $75 mL \frac{M}{5} HCl + 25 mL \frac{M}{5} NaOH$
(D) $100 mL \frac{M}{10} HCl + 100 mL \frac{M}{10} NaOH$
$pH$ of which one of them will be equal to 1?

1 D

2 C

3 A

4 B

NEET - 2018

CHXI07:EQUILIBRIUM

315011 The $pH$ of $0 .005$ $M$ $Ba(OH)_{2}$ is

1 2.301

2 11.699

3 12

4 7

Explanation:

$[{OH}^{-}] = 2 \times 0.005 = 0.01$ or $10^{- 2}$
$pOH = - \log [{OH}^{-}] = 2 pH = 14 - pOH = 14 - 2 = 12$

CHXI07:EQUILIBRIUM

315012 What is the $p H o f 0 .02 M N a O H$ solution?

1 10.3

2 11.3

3 11.7

4 12.3

Explanation:

$[{OH}^{-}] = 2 \times 10^{- 2}$
$P^{OH} = - \log_{10} [{OH}^{-}]$
$= - \log 2 \times 10^{- 2}$
$= 2 - \log 2$
$= 2 - 0.3$
$P^{OH} = 1.7$
and $P^{H} = 14 - P^{OH} = 14 - 1.7$
$P^{H} = 12.3$

MHTCET - 2021

CHXI07:EQUILIBRIUM

315013 Which of the following has highest $pH$ ?

1

\frac{M}{4} K O H

2

\frac{M}{4} N a O H

3

\frac{M}{4} N H_{4} O H

4

\frac{M}{4} C a (O H)_{2}

Explanation:

$pH$ of solution $\propto [{OH}^{-}]$
$\frac{M}{4} Ca (OH)_{2}$ solution will give highest concentration of $O H^{-}$
Hence, it has highest $pH$ .

CHXI07:EQUILIBRIUM

315009 If $50 mL$ of $0.1 HBr$ is mixed with $50 m L 0.2 M N a O H$ , find $pH$ of resulting mixture

1 2.7

2 12.7

3 10.7

4 1.3

Explanation:

$\begin{array}{l} N_{Mix} V_{Mix} = N_{1} V_{1} - N_{2} V_{2} (1) \\ (Base) (Acid) \end{array}$
Given, $M_{1} = 0.2 M, N_{1} = 0.2 N$
$\begin{aligned} M_{2} = 0.1 M, N_{2} = 0.1 N \\ V_{1} = 50 mL, V_{2} = 50 mL, V_{mix} = 100 mL \end{aligned}$
On substituting the given values in equation (1)
$N_{mix} \times 100 = 50 \times 0.2 - 50 \times 0.1$
$N_{m i x} = \frac{5}{100} = 5 \times 10^{- 2}$ $∴ [{OH}^{-}] = 5 \times 10^{- 2}$
Now, pOH $= - \log [{OH}^{-}] = - \log [5 \times 10^{- 2}] $ $ = 2 - \log 5 = 1.3$
Also, $pH + pOH = 14 $ $ pH = 14 - 1.3 = 12.7$

CHXI07:EQUILIBRIUM

315010 Following solutions were prepared by mixing different volumes of $NaOH$ and $HCl$ of different concentrations :
(A) $60 mL \frac{M}{10} HCl + 40 mL \frac{M}{10} NaOH$
(B) $55 mL \frac{M}{10} HCl + 45 mL \frac{M}{10} NaOH$
(C) $75 mL \frac{M}{5} HCl + 25 mL \frac{M}{5} NaOH$
(D) $100 mL \frac{M}{10} HCl + 100 mL \frac{M}{10} NaOH$
$pH$ of which one of them will be equal to 1?

1 D

2 C

3 A

4 B

NEET - 2018

CHXI07:EQUILIBRIUM

315011 The $pH$ of $0 .005$ $M$ $Ba(OH)_{2}$ is

1 2.301

2 11.699

3 12

4 7

Explanation:

$[{OH}^{-}] = 2 \times 0.005 = 0.01$ or $10^{- 2}$
$pOH = - \log [{OH}^{-}] = 2 pH = 14 - pOH = 14 - 2 = 12$

CHXI07:EQUILIBRIUM

315012 What is the $p H o f 0 .02 M N a O H$ solution?

1 10.3

2 11.3

3 11.7

4 12.3

Explanation:

$[{OH}^{-}] = 2 \times 10^{- 2}$
$P^{OH} = - \log_{10} [{OH}^{-}]$
$= - \log 2 \times 10^{- 2}$
$= 2 - \log 2$
$= 2 - 0.3$
$P^{OH} = 1.7$
and $P^{H} = 14 - P^{OH} = 14 - 1.7$
$P^{H} = 12.3$

MHTCET - 2021

CHXI07:EQUILIBRIUM

315013 Which of the following has highest $pH$ ?

1

\frac{M}{4} K O H

2

\frac{M}{4} N a O H

3

\frac{M}{4} N H_{4} O H

4

\frac{M}{4} C a (O H)_{2}

Explanation:

$pH$ of solution $\propto [{OH}^{-}]$
$\frac{M}{4} Ca (OH)_{2}$ solution will give highest concentration of $O H^{-}$
Hence, it has highest $pH$ .