QBManager

Ionic Equilibrium

229837 The $pH$ value of $N / 10 NaOH$ is:

1 9

2 10

3 12

4 13

Explanation:

$NaOH \to {Na}^{+} + {OH}^{-}$
Normality of $NaOH =$ molarity of $NaOH = 0.1 M$
$\begin{array}{ll} [OH] = 10^{- 1} M \\ [H^{+}] [{OH}^{-}] = 10^{- 14} \\ ∴ & [H^{+}] = \frac{10^{- 14}}{10^{- 1}} = 10^{- 13} m \\ pH = - \log [H^{+}] \\ ∴ & pH = - \log 10^{- 13} \\ ∴ & pH = 13 \end{array}$

AP-EAMCET-(2001)

Ionic Equilibrium

229839 The $pH$ of a solution containing $0.10 M$ sodium acetate and $0.03 M$ acetic acid is $({pK}_{n}$ for ${CH}_{3} COOH = 4.57)$

1 4.09

2 6.09

3 5.09

4 7.09

Explanation:

According to Henderson's equation
$\begin{aligned} pH & = {pK}_{a} + \log (\frac{salt}{acid}) \\ {CH}_{3} COOH & + NaOH \to {CH}_{3} COONa + H_{2} O \end{aligned}$
Given value,
${pK}_{a} = 4.57$
Sodium acetate concentration $= 0.10 M$ acetic acid concentration $= 0.03 M$
Putting the value we get
$pH = 4.57 + \log \frac{0.10}{0.03}$
$pH = 5.09$

AP-EAMCET-(2000)

Ionic Equilibrium

229840 Calculate the $pOH$ of $0.10 M HCl$ solution.

1 13

2 1

3 7

4 14

Explanation:

Given,
$\begin{aligned} [H^{+}] = 0.10 M \\ pOH = ? \end{aligned}$
Formula $pH = - \log [H^{+}]$
$\begin{aligned} pH & = - \log [\frac{0.10}{100}] \\ pH & = - \log 1 + \log 10 \\ pH & = 1 \\ ∵ pH + pOH & = 14 \Rightarrow 1 + pOH = 14 \Rightarrow pOH = 13 \end{aligned}$

Shift-II

Ionic Equilibrium

229841 One litre of an aqueous solution has $3.65 g$ of $HCl$ . If it is desired to increase the $pH$ of the solution to 2 then $H_{3} O^{+}$ ion concentration present initially should be

1 also doubled

2 reduced to half

3 increased by 10 times

4 reduced by 10 times.

Explanation:

Given - $amount of aqueous solution = 1 L$
$amount of HCl = 3.65 g$
$Molar mass of HCl = 36.5$
$∴ Molarity = \frac{3.65}{36.5 \times 1} = \frac{1}{10} M$
$pH = - \log [\frac{1}{10}] = 1$
$Given - pH = 2$
$= - \log_{10} [H^{+}] = 2$
$\frac{1}{[H]^{+}} = 100$
$or [H^{+}] = \frac{1}{10 Ω} M$
From equation (i) and (ii), it is clear that initially concentration reduced by 10 times.

SRM JEEE - 2012

Ionic Equilibrium

229837 The $pH$ value of $N / 10 NaOH$ is:

1 9

2 10

3 12

4 13

Explanation:

$NaOH \to {Na}^{+} + {OH}^{-}$
Normality of $NaOH =$ molarity of $NaOH = 0.1 M$
$\begin{array}{ll} [OH] = 10^{- 1} M \\ [H^{+}] [{OH}^{-}] = 10^{- 14} \\ ∴ & [H^{+}] = \frac{10^{- 14}}{10^{- 1}} = 10^{- 13} m \\ pH = - \log [H^{+}] \\ ∴ & pH = - \log 10^{- 13} \\ ∴ & pH = 13 \end{array}$

AP-EAMCET-(2001)

Ionic Equilibrium

229839 The $pH$ of a solution containing $0.10 M$ sodium acetate and $0.03 M$ acetic acid is $({pK}_{n}$ for ${CH}_{3} COOH = 4.57)$

1 4.09

2 6.09

3 5.09

4 7.09

Explanation:

According to Henderson's equation
$\begin{aligned} pH & = {pK}_{a} + \log (\frac{salt}{acid}) \\ {CH}_{3} COOH & + NaOH \to {CH}_{3} COONa + H_{2} O \end{aligned}$
Given value,
${pK}_{a} = 4.57$
Sodium acetate concentration $= 0.10 M$ acetic acid concentration $= 0.03 M$
Putting the value we get
$pH = 4.57 + \log \frac{0.10}{0.03}$
$pH = 5.09$

AP-EAMCET-(2000)

Ionic Equilibrium

229840 Calculate the $pOH$ of $0.10 M HCl$ solution.

1 13

2 1

3 7

4 14

Explanation:

Given,
$\begin{aligned} [H^{+}] = 0.10 M \\ pOH = ? \end{aligned}$
Formula $pH = - \log [H^{+}]$
$\begin{aligned} pH & = - \log [\frac{0.10}{100}] \\ pH & = - \log 1 + \log 10 \\ pH & = 1 \\ ∵ pH + pOH & = 14 \Rightarrow 1 + pOH = 14 \Rightarrow pOH = 13 \end{aligned}$

Shift-II

Ionic Equilibrium

229841 One litre of an aqueous solution has $3.65 g$ of $HCl$ . If it is desired to increase the $pH$ of the solution to 2 then $H_{3} O^{+}$ ion concentration present initially should be

1 also doubled

2 reduced to half

3 increased by 10 times

4 reduced by 10 times.

Explanation:

Given - $amount of aqueous solution = 1 L$
$amount of HCl = 3.65 g$
$Molar mass of HCl = 36.5$
$∴ Molarity = \frac{3.65}{36.5 \times 1} = \frac{1}{10} M$
$pH = - \log [\frac{1}{10}] = 1$
$Given - pH = 2$
$= - \log_{10} [H^{+}] = 2$
$\frac{1}{[H]^{+}} = 100$
$or [H^{+}] = \frac{1}{10 Ω} M$
From equation (i) and (ii), it is clear that initially concentration reduced by 10 times.

SRM JEEE - 2012

Ionic Equilibrium

229837 The $pH$ value of $N / 10 NaOH$ is:

1 9

2 10

3 12

4 13

Explanation:

$NaOH \to {Na}^{+} + {OH}^{-}$
Normality of $NaOH =$ molarity of $NaOH = 0.1 M$
$\begin{array}{ll} [OH] = 10^{- 1} M \\ [H^{+}] [{OH}^{-}] = 10^{- 14} \\ ∴ & [H^{+}] = \frac{10^{- 14}}{10^{- 1}} = 10^{- 13} m \\ pH = - \log [H^{+}] \\ ∴ & pH = - \log 10^{- 13} \\ ∴ & pH = 13 \end{array}$

AP-EAMCET-(2001)

Ionic Equilibrium

229839 The $pH$ of a solution containing $0.10 M$ sodium acetate and $0.03 M$ acetic acid is $({pK}_{n}$ for ${CH}_{3} COOH = 4.57)$

1 4.09

2 6.09

3 5.09

4 7.09

Explanation:

According to Henderson's equation
$\begin{aligned} pH & = {pK}_{a} + \log (\frac{salt}{acid}) \\ {CH}_{3} COOH & + NaOH \to {CH}_{3} COONa + H_{2} O \end{aligned}$
Given value,
${pK}_{a} = 4.57$
Sodium acetate concentration $= 0.10 M$ acetic acid concentration $= 0.03 M$
Putting the value we get
$pH = 4.57 + \log \frac{0.10}{0.03}$
$pH = 5.09$

AP-EAMCET-(2000)

Ionic Equilibrium

229840 Calculate the $pOH$ of $0.10 M HCl$ solution.

1 13

2 1

3 7

4 14

Explanation:

Given,
$\begin{aligned} [H^{+}] = 0.10 M \\ pOH = ? \end{aligned}$
Formula $pH = - \log [H^{+}]$
$\begin{aligned} pH & = - \log [\frac{0.10}{100}] \\ pH & = - \log 1 + \log 10 \\ pH & = 1 \\ ∵ pH + pOH & = 14 \Rightarrow 1 + pOH = 14 \Rightarrow pOH = 13 \end{aligned}$

Shift-II

Ionic Equilibrium

229841 One litre of an aqueous solution has $3.65 g$ of $HCl$ . If it is desired to increase the $pH$ of the solution to 2 then $H_{3} O^{+}$ ion concentration present initially should be

1 also doubled

2 reduced to half

3 increased by 10 times

4 reduced by 10 times.

Explanation:

Given - $amount of aqueous solution = 1 L$
$amount of HCl = 3.65 g$
$Molar mass of HCl = 36.5$
$∴ Molarity = \frac{3.65}{36.5 \times 1} = \frac{1}{10} M$
$pH = - \log [\frac{1}{10}] = 1$
$Given - pH = 2$
$= - \log_{10} [H^{+}] = 2$
$\frac{1}{[H]^{+}} = 100$
$or [H^{+}] = \frac{1}{10 Ω} M$
From equation (i) and (ii), it is clear that initially concentration reduced by 10 times.

SRM JEEE - 2012

Ionic Equilibrium

229837 The $pH$ value of $N / 10 NaOH$ is:

1 9

2 10

3 12

4 13

Explanation:

$NaOH \to {Na}^{+} + {OH}^{-}$
Normality of $NaOH =$ molarity of $NaOH = 0.1 M$
$\begin{array}{ll} [OH] = 10^{- 1} M \\ [H^{+}] [{OH}^{-}] = 10^{- 14} \\ ∴ & [H^{+}] = \frac{10^{- 14}}{10^{- 1}} = 10^{- 13} m \\ pH = - \log [H^{+}] \\ ∴ & pH = - \log 10^{- 13} \\ ∴ & pH = 13 \end{array}$

AP-EAMCET-(2001)

Ionic Equilibrium

229839 The $pH$ of a solution containing $0.10 M$ sodium acetate and $0.03 M$ acetic acid is $({pK}_{n}$ for ${CH}_{3} COOH = 4.57)$

1 4.09

2 6.09

3 5.09

4 7.09

Explanation:

According to Henderson's equation
$\begin{aligned} pH & = {pK}_{a} + \log (\frac{salt}{acid}) \\ {CH}_{3} COOH & + NaOH \to {CH}_{3} COONa + H_{2} O \end{aligned}$
Given value,
${pK}_{a} = 4.57$
Sodium acetate concentration $= 0.10 M$ acetic acid concentration $= 0.03 M$
Putting the value we get
$pH = 4.57 + \log \frac{0.10}{0.03}$
$pH = 5.09$

AP-EAMCET-(2000)

Ionic Equilibrium

229840 Calculate the $pOH$ of $0.10 M HCl$ solution.

1 13

2 1

3 7

4 14

Explanation:

Given,
$\begin{aligned} [H^{+}] = 0.10 M \\ pOH = ? \end{aligned}$
Formula $pH = - \log [H^{+}]$
$\begin{aligned} pH & = - \log [\frac{0.10}{100}] \\ pH & = - \log 1 + \log 10 \\ pH & = 1 \\ ∵ pH + pOH & = 14 \Rightarrow 1 + pOH = 14 \Rightarrow pOH = 13 \end{aligned}$

Shift-II

Ionic Equilibrium

229841 One litre of an aqueous solution has $3.65 g$ of $HCl$ . If it is desired to increase the $pH$ of the solution to 2 then $H_{3} O^{+}$ ion concentration present initially should be

1 also doubled

2 reduced to half

3 increased by 10 times

4 reduced by 10 times.

Explanation:

Given - $amount of aqueous solution = 1 L$
$amount of HCl = 3.65 g$
$Molar mass of HCl = 36.5$
$∴ Molarity = \frac{3.65}{36.5 \times 1} = \frac{1}{10} M$
$pH = - \log [\frac{1}{10}] = 1$
$Given - pH = 2$
$= - \log_{10} [H^{+}] = 2$
$\frac{1}{[H]^{+}} = 100$
$or [H^{+}] = \frac{1}{10 Ω} M$
From equation (i) and (ii), it is clear that initially concentration reduced by 10 times.

SRM JEEE - 2012