1
MCQ (Single Correct Answer)

AIPMT 2009

Which of the following molecules acts as a Lewis acid?
A
(CH3)2O
B
(CH3)3P
C
(CH3)3N
D
(CH3)3B

Explanation

CH3)3 B – is an electron deficient compound due to inclete octate of B, thus behave as a lewis acid.
2
MCQ (Single Correct Answer)

AIPMT 2009

The ionization constant of ammonium hydroxide is 1.77 $$ \times $$ 10$$-$$5 at 298 K. Hydrolysis constant of ammonium chloride is
A
6.50 $$ \times $$ 10$$-$$12
B
5.65 $$ \times $$ 10$$-$$13
C
5.65 $$ \times $$ 10$$-$$12
D
5.65 $$ \times $$ 10$$-$$10

Explanation

Ammonium chloride is a salt of weak base and strong acid. In this case hydrolysis constant Kh can be calculated as

Kh = $${{{K_w}} \over {{K_b}}}$$ = $${{{{10}^{ - 14}}} \over {1.77 \times {{10}^{ - 5}}}}$$ = 5.65 $$ \times $$ 10$$-$$10
3
MCQ (Single Correct Answer)

AIPMT 2009

What is the [OH$$-$$] in the final solution prepared by mixing 20.0 mL of 0.050 M HCl with 30.0 mL of 0.10 M Ba(OH)2?
A
0.40 M
B
0.0050 M
C
0.12 M
D
0.10 M

Explanation

Number of equivalents of H+ = 20.0 × 0.050 milliequivalents

= 1.0 milliequivalents

Number of equivalents of OH = 2 × 30.0 × 0.10

= 6.0 milliequivalents

$$ \therefore $$ Equivalents of OH- left after neutralization

= 6 – 1 = 5 milliequivalents

Total volume after neutralization

= 20.0 + 30.0 mL

= 50 mL

[OH-] = $${5 \over {50}}$$ = 0.1 M
4
MCQ (Single Correct Answer)

AIPMT 2009

The dissociation constants for acetic acid and HCN at 25oC are 1.5 $$ \times $$ 10$$-$$5 and 4.5 $$ \times $$ 10$$-$$10 respectively. The equilibrium constant for the equilibrium
CN$$-$$ + CH3COOH $$\rightleftharpoons$$ HCN + CH3COO$$-$$ would be
A
3.0 $$ \times $$ 10$$-$$5
B
3.0 $$ \times $$ 10$$-$$4
C
3.0 $$ \times $$ 104
D
3.0 $$ \times $$ 105

Explanation

CH3COOH $$\rightleftharpoons$$ CH3COO + H+,   K1 = 1.5 $$ \times $$ 10$$-$$5

$$ \Rightarrow $$ K1 = $${{\left[ {C{H_3}CO{O^ - }} \right]\left[ {{H^ + }} \right]} \over {\left[ {C{H_3}COOH} \right]}}$$ = 1.5 $$ \times $$ 10$$-$$5

HCN $$\rightleftharpoons$$ CN + H+,     K2 = 4.5 × 10–10

K2 = $${{\left[ {C{N^ - }} \right]\left[ {{H^ + }} \right]} \over {\left[ {HCN} \right]}}$$ = 4.5 × 10–10

CN$$-$$ + CH3COOH $$\rightleftharpoons$$ HCN + CH3COO$$-$$

K = $${{\left[ {HCN} \right]\left[ {C{H_3}CO{O^ - }} \right]} \over {\left[ {C{N^ - }} \right]\left[ {C{H_3}COOH} \right]}}$$

$$ \Rightarrow $$ K = $${{{K_1}} \over {{K_2}}}$$ = $${{1.5 \times {{10}^{ - 5}}} \over {4.5 \times {{10}^{ - 10}}}}$$

= 3.33 × 104

$$ \simeq $$ 3.0 × 104

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