1
MCQ (Single Correct Answer)

AIPMT 2012 Prelims

Buffer solutions have constant acidity and alkalinity because
A
these give unionised acid or base on reaction with added acid or alkali
B
acids and alkalies in these solutions are shielded from attack by other ions
C
they have large excess of H+ or OH$$-$$ ions
D
they have fixed value of pH

Explanation

For this example,

CH3COOH ⇌ CH3COO + H+ ;

CH3COONa ⇌ CH3COO + Na+

when few drops of HCl are added to this buffer, the H+ of HCl immediatly combine with CH3COO ions to form undissociated acetic acid molecules. Thus there will be no H+ ions to combine with CH3COO ions to form undissociated acetic acid molecules. Thus there will be no appreciable change in its pH value. Like wise if few drops of NaOH are added, the OH ions will combine with H+ ions to form unionised water molecule. Thus pH of solution will remain constant.
2
MCQ (Single Correct Answer)

AIPMT 2012 Prelims

Equimolar solutions of the following substances were prepared separately. Which one of these will record the highest pH value ?
A
BaCl2
B
AlCl3
C
LiCl
D
BeCl2

Explanation

All of the given salts have same anion i.e., Cl which on hydrolysis gives HCl which is a strong acid.

Now, among the salts which have cation that gives a strongest base on hydrolysis of salt have the highest pH value. As Ba form Ba(OH)2 which is a stronger base thus, it results in the highest pH value.
3
MCQ (Single Correct Answer)

AIPMT 2012 Prelims

pH of a saturated solution of Ba(OH)2 is 12. The value of solubility product (Ksp) of Ba(OH)2 is
A
3.3 $$ \times $$ 10$$-$$7
B
5.0 $$ \times $$ 10$$-$$7
C
4.0 $$ \times $$ 10$$-$$6
D
5.0$$ \times $$ 10$$-$$6

Explanation

Ba(OH)2 Ba2+ + 2OH
At equilibrium x 2x


pH = – log[H+]

12 = – log [H+]

$$ \Rightarrow $$ [H+] = 10–12

As, [H+][OH ] = 10–14

10–12 [OH ] = 10–14

$$ \Rightarrow $$ [OH ] = 10–2

As [OH ] = 2x = 10–2 then x = 5.0 × 10–3

Now, Ksp = [Ba2+][OH ]2

Ksp = (5 × 10–3) (10–2)2 = 5.0 × 10–7
4
MCQ (Single Correct Answer)

AIPMT 2011 Mains

In qualitative analysis, the metals of group I can be separated from other ions by precipitating them as chloride salts. A solution initially contains Ag+ and pb2+ at a concentration of 0.10 M. Aqueous HCl is added to this solution until the Cl$$-$$ concentration is 0.10 M. What will the concentrations of Ag+ and Pb2+ be at equilibrium ?

(Ksp for AgCl = 1.8 $$ \times $$ 10$$-$$10, Ksp for PbCl2 = 1.7 $$ \times $$ 10$$-$$5)
A
[Ag+] = 1.8 $$ \times $$ 10$$-$$7 M, [Pb2+] = 1.7 $$ \times $$ 10$$-$$6 M
B
[Ag+] = 1.8 $$ \times $$ 10$$-$$11 M, [Pb2+] = 8.5 $$ \times $$ 10$$-$$5 M
C
[Ag+] = 1.8 $$ \times $$ 10$$-$$9 M, [Pb2+] = 1.7 $$ \times $$ 10$$-$$3 M
D
[Ag+] = 1.8 $$ \times $$ 10$$-$$11 M, [Pb2+] = 1.7 $$ \times $$ 10$$-$$4 M

Explanation

Ksp[AgCl] = [Ag+][Cl-]

$$ \Rightarrow $$ [Ag+] = $${{1.8 \times {{10}^{ - 10}}} \over {{{10}^{ - 1}}}}$$ = 1.8 $$ \times $$ 10$$-$$9 M

Ksp[PbCl2] = [Pb2+][Cl-]2

$$ \Rightarrow $$ [Pb2+] = $${{1.7 \times {{10}^{ - 5}}} \over {{{10}^{ - 1}} \times {{10}^{ - 1}}}}$$ = 1.7 $$ \times $$ 10$$-$$3 M

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