1

NEET 2021

MCQ (Single Correct Answer)
The pKb of dimethyl amine and pKa of acetic acid are 3.27 and 4.77 respectively at T (K). The correct option for the pH of dimethyl ammonium acetate solution is :
A
6.25
B
8.50
C
5.50
D
7.75

Explanation

Dimethylammonium acetate is a salt of weak acid and weak base whose pH can be calculated as

pH = 7 + $${1 \over 2}$$(pKa $$-$$ pKb)

pKa of acetic acid = 4.77

pKb of dimethyl amine = 3.27

pH = 7 + $${1 \over 2}$$(4.77 $$-$$ 3.27)

= 7.75
2

NEET 2020 Phase 1

MCQ (Single Correct Answer)
Find out the solubility of Ni(OH)2 in 0.1M NaOH. Given that the ionic product of Ni(OH)2 is 2 $$ \times $$ 10-15.
A
2 $$ \times $$ 10-8 M
B
1 $$ \times $$ 10-13 M
C
1 $$ \times $$ 108 M
D
2 $$ \times $$ 10-13 M

Explanation



Total [OH- ] = 2s + 0.1 $$ \simeq $$ 0.1

Ionic product = [Ni2+][OH-]2

$$ \Rightarrow $$ 2 $$ \times $$ 10-15 = s(0.1)2

$$ \Rightarrow $$ s = 2 $$ \times $$ 10-13
3

NEET 2020 Phase 1

MCQ (Single Correct Answer)
Hydrolysis of sucrose is given by the following reaction.
Sucrose + H2O ⇌ Glucose + Fructose
If the equilibrium constant (Kc) is 2 $$ \times $$ 1013 at 800 K, the value of $$\Delta r{G^\Theta }$$ at the same temperature will be :
A
8.314 J $$mo{l^{ - 1}}{K^{ - 1}} \times 300K \times In(2 \times {10^{13}})$$
B
8.314 J $$mo{l^{ - 1}}{K^{ - 1}} \times 300K \times In(3 \times {10^{13}})$$
C
$$ - $$ 8.314 J $$mo{l^{ - 1}}{K^{ - 1}} \times 300K \times In(4 \times {10^{13}})$$
D
-8.314 J $$mo{l^{ - 1}}{K^{ - 1}} \times 300K \times In(2 \times {10^{13}})$$

Explanation

$$\Delta G$$ = $$\Delta G$$ $$^\circ $$ + RT ln Q

At equilibrium $$\Delta G$$ = 0, Q = Keq

So, $${\Delta _r}G^\circ $$ = $$ - $$ RT in Keq

$${\Delta _r}G^\circ $$ = $$ - $$ 8.314 J mol-1 K-1 $$ \times $$ 300K $$ \times $$ ln (2 $$ \times $$ 1013)
4

NEET 2019

MCQ (Single Correct Answer)
Which will make basic buffer?
A
100 mL of 0.1 M HCl + 200 mL of 0.1 M NH4OH
B
100 mL of 0.1 M HCl + 100 mL of 0.1 M NHOH
C
50 mL of 0.1 M NaOH + 25 mL of 0.1 M CH3COOH
D
100 mL of 0.1 M CH3COOH + 100 mL of 0.1 M NaOH

Explanation

Basic buffer is mixture of weak base and salt of weak base with strong acid.

milli mole of HCl = 100 × 0.1 = 10 milli mole

milli mole of NH4OH = 200 × 0.1 = 20 milli mole
HCl + NH4OH $$ \to $$ NH4Cl + H2O
10 20 0
0 20 - 10 = 10 10


Here in the final solution 10 milli mole weak base NH4OH and 10 milli mole salt of weak base and strong acid NH4Cl present. So it is a basic buffer.

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