QBManager

Ionic Equilibrium

229824 $pH$ of a salt solution of weak acid $({pK}_{n} = 4)$ \& weak base $({pK}_{b} = 5)$ at $25^{\circ} C$ is:

1 6.5

2 6

3 7

4 7.5

Explanation:

Given that,
${pK}_{a} = 4, {pK}_{b} = 5$
then,
$\begin{aligned} pH & = \frac{1}{2} ({pK}_{π} + {pK}_{a} - {pK}_{b}) \\ = \frac{1}{2} (14 + 4 - 5) \\ = 6.5 \end{aligned}$

AIIMS 26 May 2019(Evening)

Ionic Equilibrium

229825 The aqueous solution of which of the following salt will have the lowest $pH$ ?

1

{NaClO}_{3}

2

NaClO

3

{NaClO}_{4}

4

{NaClO}_{2}

Explanation:

Salt of SA + SB : $pH$ will remain neutral at 7 Salt of $SA + WB : pH < 7$ (acidic)
Salt of WA $+ SB : pH > 7$ (basic)
${NaClO}_{4} \Rightarrow$ Neutral $(pH = 7)$
${HClO}_{4} + NaOH \to {NaClO}_{4} + H_{2} O$
$(SA) (SB)$
Where,
$\begin{aligned} SA & = strong acid \\ SB & = strong base \\ WB & = weak base \\ WA & = weak acid \end{aligned}$
${NaClO}_{4}$ , on hydrolysis gives strongest acid ${HClO}_{4}$ as compared to other salt. This is strongly ionized and gives highest concentration of $H^{+}$ therefore, aqueous solution of ${NaCl}_{4}$ will have lowest $pH$ value.

AIIMS-1996

Ionic Equilibrium

229827 The $pH$ of $0.001 (N)$ acetic acid solution, which is $10 %$ dissociated, is

1 3

2 1

3 4

4 2

Explanation:

Given that,
Concentration of acetic acid $= 0.001 N$ for $10 %$ dissociation, the effective concentration of $H^{+}$ is $1 / 10^{th}$ of total concentration.
Hence, effective concentration
$\begin{aligned} = \frac{1}{10} \times 0.001 = 0.0001 N = 1 \times 10^{- 4} N \\ pH = - \log [H^{+}] \\ pH = - \log [1 \times 10^{- 4}] \\ = 4 \end{aligned}$

AIIMS-1996

Ionic Equilibrium

229828 Assertion : In a titration of weak acid and $NaOH$ , the $pH$ at half equivalence point is ${pK}_{a}$ .
**Reason :** At half equivalence point, it forms an acidic buffer and the buffer capacity is maximum where [acid] $=$ [salt $]$

1 If both Assertion and Reason are correct and the Reason is the correct explanation of Assertion.

2 If both assertion and Reason are correct, but Reason is not the correct explanation of assertion.

3 If Assertion if correct but Reason is incorrect.

4 If both the Assertion and Reason are incorrect.

Explanation:

$pH = {pK}_{a} + \log \frac{[A^{-}]}{[HA]}$
At the half-equivalence point say we have 10 moles of weak acid, and so there will be 5 moles of strong base as the name suggests (half equivalence).
$HA + {OH}^{-} \to A^{-} + H_{2} O$
for the half equivalent point, $[H . A] = [A^{-}]$
So, $pH = {pK}_{a}$

AIIMS-2009

Ionic Equilibrium

229824 $pH$ of a salt solution of weak acid $({pK}_{n} = 4)$ \& weak base $({pK}_{b} = 5)$ at $25^{\circ} C$ is:

1 6.5

2 6

3 7

4 7.5

Explanation:

Given that,
${pK}_{a} = 4, {pK}_{b} = 5$
then,
$\begin{aligned} pH & = \frac{1}{2} ({pK}_{π} + {pK}_{a} - {pK}_{b}) \\ = \frac{1}{2} (14 + 4 - 5) \\ = 6.5 \end{aligned}$

AIIMS 26 May 2019(Evening)

Ionic Equilibrium

229825 The aqueous solution of which of the following salt will have the lowest $pH$ ?

1

{NaClO}_{3}

2

NaClO

3

{NaClO}_{4}

4

{NaClO}_{2}

Explanation:

Salt of SA + SB : $pH$ will remain neutral at 7 Salt of $SA + WB : pH < 7$ (acidic)
Salt of WA $+ SB : pH > 7$ (basic)
${NaClO}_{4} \Rightarrow$ Neutral $(pH = 7)$
${HClO}_{4} + NaOH \to {NaClO}_{4} + H_{2} O$
$(SA) (SB)$
Where,
$\begin{aligned} SA & = strong acid \\ SB & = strong base \\ WB & = weak base \\ WA & = weak acid \end{aligned}$
${NaClO}_{4}$ , on hydrolysis gives strongest acid ${HClO}_{4}$ as compared to other salt. This is strongly ionized and gives highest concentration of $H^{+}$ therefore, aqueous solution of ${NaCl}_{4}$ will have lowest $pH$ value.

AIIMS-1996

Ionic Equilibrium

229827 The $pH$ of $0.001 (N)$ acetic acid solution, which is $10 %$ dissociated, is

1 3

2 1

3 4

4 2

Explanation:

Given that,
Concentration of acetic acid $= 0.001 N$ for $10 %$ dissociation, the effective concentration of $H^{+}$ is $1 / 10^{th}$ of total concentration.
Hence, effective concentration
$\begin{aligned} = \frac{1}{10} \times 0.001 = 0.0001 N = 1 \times 10^{- 4} N \\ pH = - \log [H^{+}] \\ pH = - \log [1 \times 10^{- 4}] \\ = 4 \end{aligned}$

AIIMS-1996

Ionic Equilibrium

229828 Assertion : In a titration of weak acid and $NaOH$ , the $pH$ at half equivalence point is ${pK}_{a}$ .
**Reason :** At half equivalence point, it forms an acidic buffer and the buffer capacity is maximum where [acid] $=$ [salt $]$

1 If both Assertion and Reason are correct and the Reason is the correct explanation of Assertion.

2 If both assertion and Reason are correct, but Reason is not the correct explanation of assertion.

3 If Assertion if correct but Reason is incorrect.

4 If both the Assertion and Reason are incorrect.

Explanation:

$pH = {pK}_{a} + \log \frac{[A^{-}]}{[HA]}$
At the half-equivalence point say we have 10 moles of weak acid, and so there will be 5 moles of strong base as the name suggests (half equivalence).
$HA + {OH}^{-} \to A^{-} + H_{2} O$
for the half equivalent point, $[H . A] = [A^{-}]$
So, $pH = {pK}_{a}$

AIIMS-2009

Ionic Equilibrium

229824 $pH$ of a salt solution of weak acid $({pK}_{n} = 4)$ \& weak base $({pK}_{b} = 5)$ at $25^{\circ} C$ is:

1 6.5

2 6

3 7

4 7.5

Explanation:

Given that,
${pK}_{a} = 4, {pK}_{b} = 5$
then,
$\begin{aligned} pH & = \frac{1}{2} ({pK}_{π} + {pK}_{a} - {pK}_{b}) \\ = \frac{1}{2} (14 + 4 - 5) \\ = 6.5 \end{aligned}$

AIIMS 26 May 2019(Evening)

Ionic Equilibrium

229825 The aqueous solution of which of the following salt will have the lowest $pH$ ?

1

{NaClO}_{3}

2

NaClO

3

{NaClO}_{4}

4

{NaClO}_{2}

Explanation:

Salt of SA + SB : $pH$ will remain neutral at 7 Salt of $SA + WB : pH < 7$ (acidic)
Salt of WA $+ SB : pH > 7$ (basic)
${NaClO}_{4} \Rightarrow$ Neutral $(pH = 7)$
${HClO}_{4} + NaOH \to {NaClO}_{4} + H_{2} O$
$(SA) (SB)$
Where,
$\begin{aligned} SA & = strong acid \\ SB & = strong base \\ WB & = weak base \\ WA & = weak acid \end{aligned}$
${NaClO}_{4}$ , on hydrolysis gives strongest acid ${HClO}_{4}$ as compared to other salt. This is strongly ionized and gives highest concentration of $H^{+}$ therefore, aqueous solution of ${NaCl}_{4}$ will have lowest $pH$ value.

AIIMS-1996

Ionic Equilibrium

229827 The $pH$ of $0.001 (N)$ acetic acid solution, which is $10 %$ dissociated, is

1 3

2 1

3 4

4 2

Explanation:

Given that,
Concentration of acetic acid $= 0.001 N$ for $10 %$ dissociation, the effective concentration of $H^{+}$ is $1 / 10^{th}$ of total concentration.
Hence, effective concentration
$\begin{aligned} = \frac{1}{10} \times 0.001 = 0.0001 N = 1 \times 10^{- 4} N \\ pH = - \log [H^{+}] \\ pH = - \log [1 \times 10^{- 4}] \\ = 4 \end{aligned}$

AIIMS-1996

Ionic Equilibrium

229828 Assertion : In a titration of weak acid and $NaOH$ , the $pH$ at half equivalence point is ${pK}_{a}$ .
**Reason :** At half equivalence point, it forms an acidic buffer and the buffer capacity is maximum where [acid] $=$ [salt $]$

1 If both Assertion and Reason are correct and the Reason is the correct explanation of Assertion.

2 If both assertion and Reason are correct, but Reason is not the correct explanation of assertion.

3 If Assertion if correct but Reason is incorrect.

4 If both the Assertion and Reason are incorrect.

Explanation:

$pH = {pK}_{a} + \log \frac{[A^{-}]}{[HA]}$
At the half-equivalence point say we have 10 moles of weak acid, and so there will be 5 moles of strong base as the name suggests (half equivalence).
$HA + {OH}^{-} \to A^{-} + H_{2} O$
for the half equivalent point, $[H . A] = [A^{-}]$
So, $pH = {pK}_{a}$

AIIMS-2009

Ionic Equilibrium

229824 $pH$ of a salt solution of weak acid $({pK}_{n} = 4)$ \& weak base $({pK}_{b} = 5)$ at $25^{\circ} C$ is:

1 6.5

2 6

3 7

4 7.5

Explanation:

Given that,
${pK}_{a} = 4, {pK}_{b} = 5$
then,
$\begin{aligned} pH & = \frac{1}{2} ({pK}_{π} + {pK}_{a} - {pK}_{b}) \\ = \frac{1}{2} (14 + 4 - 5) \\ = 6.5 \end{aligned}$

AIIMS 26 May 2019(Evening)

Ionic Equilibrium

229825 The aqueous solution of which of the following salt will have the lowest $pH$ ?

1

{NaClO}_{3}

2

NaClO

3

{NaClO}_{4}

4

{NaClO}_{2}

Explanation:

Salt of SA + SB : $pH$ will remain neutral at 7 Salt of $SA + WB : pH < 7$ (acidic)
Salt of WA $+ SB : pH > 7$ (basic)
${NaClO}_{4} \Rightarrow$ Neutral $(pH = 7)$
${HClO}_{4} + NaOH \to {NaClO}_{4} + H_{2} O$
$(SA) (SB)$
Where,
$\begin{aligned} SA & = strong acid \\ SB & = strong base \\ WB & = weak base \\ WA & = weak acid \end{aligned}$
${NaClO}_{4}$ , on hydrolysis gives strongest acid ${HClO}_{4}$ as compared to other salt. This is strongly ionized and gives highest concentration of $H^{+}$ therefore, aqueous solution of ${NaCl}_{4}$ will have lowest $pH$ value.

AIIMS-1996

Ionic Equilibrium

229827 The $pH$ of $0.001 (N)$ acetic acid solution, which is $10 %$ dissociated, is

1 3

2 1

3 4

4 2

Explanation:

Given that,
Concentration of acetic acid $= 0.001 N$ for $10 %$ dissociation, the effective concentration of $H^{+}$ is $1 / 10^{th}$ of total concentration.
Hence, effective concentration
$\begin{aligned} = \frac{1}{10} \times 0.001 = 0.0001 N = 1 \times 10^{- 4} N \\ pH = - \log [H^{+}] \\ pH = - \log [1 \times 10^{- 4}] \\ = 4 \end{aligned}$

AIIMS-1996

Ionic Equilibrium

229828 Assertion : In a titration of weak acid and $NaOH$ , the $pH$ at half equivalence point is ${pK}_{a}$ .
**Reason :** At half equivalence point, it forms an acidic buffer and the buffer capacity is maximum where [acid] $=$ [salt $]$

1 If both Assertion and Reason are correct and the Reason is the correct explanation of Assertion.

2 If both assertion and Reason are correct, but Reason is not the correct explanation of assertion.

3 If Assertion if correct but Reason is incorrect.

4 If both the Assertion and Reason are incorrect.

Explanation:

$pH = {pK}_{a} + \log \frac{[A^{-}]}{[HA]}$
At the half-equivalence point say we have 10 moles of weak acid, and so there will be 5 moles of strong base as the name suggests (half equivalence).
$HA + {OH}^{-} \to A^{-} + H_{2} O$
for the half equivalent point, $[H . A] = [A^{-}]$
So, $pH = {pK}_{a}$

AIIMS-2009

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