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CHXI07:EQUILIBRIUM

314434 How many of the following combinations of reactants will react less than $2 %$ of theoretically possible extent?
$\begin{array}{l} (a) C H_{3} C O O H + H_{2} O \\ (b) C H_{3} C O O^{⊖} + H_{2} O \\ (c) C H_{3} C O O^{⊖} + H_{3} O^{\oplus} \\ (d) C H_{3} C O O H + K O H \\ (e) C H_{3} C O O K + H C l (a q) \\ (f) H C l (g) + H_{2} O \\ (g) C l^{-} + H_{3} O^{\oplus} (h) C l^{⊖} + H_{2} O \\ (i) \overset{\oplus}{N H} H_{4} + K O H (j) \overset{\oplus}{N} H_{4} + \overset{⊖}{O} H \\ (k) N H_{3} + H_{2} O (l) N H_{3} + H_{3} O^{\oplus} \\ (m) N H_{3} + H C l (a q) (n) K^{\oplus} + \overset{⊖}{O} H \end{array}$

1 6

2 2

3 5

4 3

Explanation:

(a), (b), (g), (h) and (k)
$\begin{array}{l} (a) C H_{3} COOH + H_{2} O \to H_{3} O^{\oplus} + C H_{3} CO O^{⊖} (< 2 %) \\ W_{A} \end{array}$
(b) $C H_{3} C O O^{⊖} + H_{2} O \to C H_{3} C O O H + \overset{⊖}{O} H (< 2 %)$
$S a l t o f W_{A}$
$\begin{array}{l} (g) C l^{⊖} + H_{3} O^{\oplus} \to N o r e a c t i o n \\ (h) C l^{⊖} + H_{2} O \to N o r e a c t i o n \\ (k) N H_{3} + H_{2} O \to \overset{\oplus}{N} H_{4} + \overset{⊖}{O} H \\ (n) K^{\oplus} + \overset{⊖}{O} H \to N o r e a c t i o n . \end{array}$

CHXI07:EQUILIBRIUM

314435 Which of the following species acts as Bronsted base but not as acid ?

1

{CH}_{3} COOH

2

{HCO}_{3}^{-}

3

H_{2} {PO}_{2}^{-}

4

H_{2} {PO}_{3}^{-}

Explanation:

$H_{2} {PO}_{2}^{-}$ ion does not contain any $H^{+}$ ions to donate.

CHXI07:EQUILIBRIUM

314436 In ${HS}^{-}, I^{-}, {RNH}_{2}$ and ${NH}_{3}$ order of proton accepting tendency will be

1

I^{-} > N H_{3} > RN H_{2} > H S^{-}

2

{NH}_{3} > {RNH}_{2} > {HS}^{-} > I^{-}

3

{RNH}_{2} > {NH}_{3} > {HS}^{-} > I^{-}

4

{HS}^{-} > {RNH}_{2} > {NH}_{3} > I^{-}

Explanation:

${RNH}_{2}$ is a stronger base than ${NH}_{3}$ . And $HI$ is a strong acid; Thus, $I^{-}$ is a weak base.

CHXI07:EQUILIBRIUM

314437 Ammonia gas dissolves in water to form ${NH}_{4} OH$ . In this reaction, water acts as

1 Conjugate base

2 A non - polar molecule

3 An acid

4 A base

Explanation:

$H_{2} O + {NH}_{3} ⇌ {NH}_{4}^{+} + {OH}^{-}$
In this reaction, $H_{2} O$ acts as an acid, as it donates a proton.

AIIMS - 2012

CHXI07:EQUILIBRIUM

314434 How many of the following combinations of reactants will react less than $2 %$ of theoretically possible extent?
$\begin{array}{l} (a) C H_{3} C O O H + H_{2} O \\ (b) C H_{3} C O O^{⊖} + H_{2} O \\ (c) C H_{3} C O O^{⊖} + H_{3} O^{\oplus} \\ (d) C H_{3} C O O H + K O H \\ (e) C H_{3} C O O K + H C l (a q) \\ (f) H C l (g) + H_{2} O \\ (g) C l^{-} + H_{3} O^{\oplus} (h) C l^{⊖} + H_{2} O \\ (i) \overset{\oplus}{N H} H_{4} + K O H (j) \overset{\oplus}{N} H_{4} + \overset{⊖}{O} H \\ (k) N H_{3} + H_{2} O (l) N H_{3} + H_{3} O^{\oplus} \\ (m) N H_{3} + H C l (a q) (n) K^{\oplus} + \overset{⊖}{O} H \end{array}$

1 6

2 2

3 5

4 3

Explanation:

(a), (b), (g), (h) and (k)
$\begin{array}{l} (a) C H_{3} COOH + H_{2} O \to H_{3} O^{\oplus} + C H_{3} CO O^{⊖} (< 2 %) \\ W_{A} \end{array}$
(b) $C H_{3} C O O^{⊖} + H_{2} O \to C H_{3} C O O H + \overset{⊖}{O} H (< 2 %)$
$S a l t o f W_{A}$
$\begin{array}{l} (g) C l^{⊖} + H_{3} O^{\oplus} \to N o r e a c t i o n \\ (h) C l^{⊖} + H_{2} O \to N o r e a c t i o n \\ (k) N H_{3} + H_{2} O \to \overset{\oplus}{N} H_{4} + \overset{⊖}{O} H \\ (n) K^{\oplus} + \overset{⊖}{O} H \to N o r e a c t i o n . \end{array}$

CHXI07:EQUILIBRIUM

314435 Which of the following species acts as Bronsted base but not as acid ?

1

{CH}_{3} COOH

2

{HCO}_{3}^{-}

3

H_{2} {PO}_{2}^{-}

4

H_{2} {PO}_{3}^{-}

Explanation:

$H_{2} {PO}_{2}^{-}$ ion does not contain any $H^{+}$ ions to donate.

CHXI07:EQUILIBRIUM

314436 In ${HS}^{-}, I^{-}, {RNH}_{2}$ and ${NH}_{3}$ order of proton accepting tendency will be

1

I^{-} > N H_{3} > RN H_{2} > H S^{-}

2

{NH}_{3} > {RNH}_{2} > {HS}^{-} > I^{-}

3

{RNH}_{2} > {NH}_{3} > {HS}^{-} > I^{-}

4

{HS}^{-} > {RNH}_{2} > {NH}_{3} > I^{-}

Explanation:

${RNH}_{2}$ is a stronger base than ${NH}_{3}$ . And $HI$ is a strong acid; Thus, $I^{-}$ is a weak base.

CHXI07:EQUILIBRIUM

314437 Ammonia gas dissolves in water to form ${NH}_{4} OH$ . In this reaction, water acts as

1 Conjugate base

2 A non - polar molecule

3 An acid

4 A base

Explanation:

$H_{2} O + {NH}_{3} ⇌ {NH}_{4}^{+} + {OH}^{-}$
In this reaction, $H_{2} O$ acts as an acid, as it donates a proton.

AIIMS - 2012

CHXI07:EQUILIBRIUM

314434 How many of the following combinations of reactants will react less than $2 %$ of theoretically possible extent?
$\begin{array}{l} (a) C H_{3} C O O H + H_{2} O \\ (b) C H_{3} C O O^{⊖} + H_{2} O \\ (c) C H_{3} C O O^{⊖} + H_{3} O^{\oplus} \\ (d) C H_{3} C O O H + K O H \\ (e) C H_{3} C O O K + H C l (a q) \\ (f) H C l (g) + H_{2} O \\ (g) C l^{-} + H_{3} O^{\oplus} (h) C l^{⊖} + H_{2} O \\ (i) \overset{\oplus}{N H} H_{4} + K O H (j) \overset{\oplus}{N} H_{4} + \overset{⊖}{O} H \\ (k) N H_{3} + H_{2} O (l) N H_{3} + H_{3} O^{\oplus} \\ (m) N H_{3} + H C l (a q) (n) K^{\oplus} + \overset{⊖}{O} H \end{array}$

1 6

2 2

3 5

4 3

Explanation:

(a), (b), (g), (h) and (k)
$\begin{array}{l} (a) C H_{3} COOH + H_{2} O \to H_{3} O^{\oplus} + C H_{3} CO O^{⊖} (< 2 %) \\ W_{A} \end{array}$
(b) $C H_{3} C O O^{⊖} + H_{2} O \to C H_{3} C O O H + \overset{⊖}{O} H (< 2 %)$
$S a l t o f W_{A}$
$\begin{array}{l} (g) C l^{⊖} + H_{3} O^{\oplus} \to N o r e a c t i o n \\ (h) C l^{⊖} + H_{2} O \to N o r e a c t i o n \\ (k) N H_{3} + H_{2} O \to \overset{\oplus}{N} H_{4} + \overset{⊖}{O} H \\ (n) K^{\oplus} + \overset{⊖}{O} H \to N o r e a c t i o n . \end{array}$

CHXI07:EQUILIBRIUM

314435 Which of the following species acts as Bronsted base but not as acid ?

1

{CH}_{3} COOH

2

{HCO}_{3}^{-}

3

H_{2} {PO}_{2}^{-}

4

H_{2} {PO}_{3}^{-}

Explanation:

$H_{2} {PO}_{2}^{-}$ ion does not contain any $H^{+}$ ions to donate.

CHXI07:EQUILIBRIUM

314436 In ${HS}^{-}, I^{-}, {RNH}_{2}$ and ${NH}_{3}$ order of proton accepting tendency will be

1

I^{-} > N H_{3} > RN H_{2} > H S^{-}

2

{NH}_{3} > {RNH}_{2} > {HS}^{-} > I^{-}

3

{RNH}_{2} > {NH}_{3} > {HS}^{-} > I^{-}

4

{HS}^{-} > {RNH}_{2} > {NH}_{3} > I^{-}

Explanation:

${RNH}_{2}$ is a stronger base than ${NH}_{3}$ . And $HI$ is a strong acid; Thus, $I^{-}$ is a weak base.

CHXI07:EQUILIBRIUM

314437 Ammonia gas dissolves in water to form ${NH}_{4} OH$ . In this reaction, water acts as

1 Conjugate base

2 A non - polar molecule

3 An acid

4 A base

Explanation:

$H_{2} O + {NH}_{3} ⇌ {NH}_{4}^{+} + {OH}^{-}$
In this reaction, $H_{2} O$ acts as an acid, as it donates a proton.

AIIMS - 2012

CHXI07:EQUILIBRIUM

314434 How many of the following combinations of reactants will react less than $2 %$ of theoretically possible extent?
$\begin{array}{l} (a) C H_{3} C O O H + H_{2} O \\ (b) C H_{3} C O O^{⊖} + H_{2} O \\ (c) C H_{3} C O O^{⊖} + H_{3} O^{\oplus} \\ (d) C H_{3} C O O H + K O H \\ (e) C H_{3} C O O K + H C l (a q) \\ (f) H C l (g) + H_{2} O \\ (g) C l^{-} + H_{3} O^{\oplus} (h) C l^{⊖} + H_{2} O \\ (i) \overset{\oplus}{N H} H_{4} + K O H (j) \overset{\oplus}{N} H_{4} + \overset{⊖}{O} H \\ (k) N H_{3} + H_{2} O (l) N H_{3} + H_{3} O^{\oplus} \\ (m) N H_{3} + H C l (a q) (n) K^{\oplus} + \overset{⊖}{O} H \end{array}$

1 6

2 2

3 5

4 3

Explanation:

(a), (b), (g), (h) and (k)
$\begin{array}{l} (a) C H_{3} COOH + H_{2} O \to H_{3} O^{\oplus} + C H_{3} CO O^{⊖} (< 2 %) \\ W_{A} \end{array}$
(b) $C H_{3} C O O^{⊖} + H_{2} O \to C H_{3} C O O H + \overset{⊖}{O} H (< 2 %)$
$S a l t o f W_{A}$
$\begin{array}{l} (g) C l^{⊖} + H_{3} O^{\oplus} \to N o r e a c t i o n \\ (h) C l^{⊖} + H_{2} O \to N o r e a c t i o n \\ (k) N H_{3} + H_{2} O \to \overset{\oplus}{N} H_{4} + \overset{⊖}{O} H \\ (n) K^{\oplus} + \overset{⊖}{O} H \to N o r e a c t i o n . \end{array}$

CHXI07:EQUILIBRIUM

314435 Which of the following species acts as Bronsted base but not as acid ?

1

{CH}_{3} COOH

2

{HCO}_{3}^{-}

3

H_{2} {PO}_{2}^{-}

4

H_{2} {PO}_{3}^{-}

Explanation:

$H_{2} {PO}_{2}^{-}$ ion does not contain any $H^{+}$ ions to donate.

CHXI07:EQUILIBRIUM

314436 In ${HS}^{-}, I^{-}, {RNH}_{2}$ and ${NH}_{3}$ order of proton accepting tendency will be

1

I^{-} > N H_{3} > RN H_{2} > H S^{-}

2

{NH}_{3} > {RNH}_{2} > {HS}^{-} > I^{-}

3

{RNH}_{2} > {NH}_{3} > {HS}^{-} > I^{-}

4

{HS}^{-} > {RNH}_{2} > {NH}_{3} > I^{-}

Explanation:

${RNH}_{2}$ is a stronger base than ${NH}_{3}$ . And $HI$ is a strong acid; Thus, $I^{-}$ is a weak base.

CHXI07:EQUILIBRIUM

314437 Ammonia gas dissolves in water to form ${NH}_{4} OH$ . In this reaction, water acts as

1 Conjugate base

2 A non - polar molecule

3 An acid

4 A base

Explanation:

$H_{2} O + {NH}_{3} ⇌ {NH}_{4}^{+} + {OH}^{-}$
In this reaction, $H_{2} O$ acts as an acid, as it donates a proton.

AIIMS - 2012

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