1
MCQ (Single Correct Answer)

AIPMT 2010 Prelims

In a buffer solution containing equal concentration of B$$-$$ and HB, the Kb for B$$-$$ is 10$$-$$10. The pH of buffer solution is
A
10
B
7
C
6
D
4

Explanation

pOH = pKb + log $${{\left[ {Salt} \right]} \over {\left[ {Acid} \right]}}$$

$$ \Rightarrow $$ pOH = - log Kb + log $${{\left[ {Salt} \right]} \over {\left[ {Acid} \right]}}$$

$$ \Rightarrow $$ pOH = –log10–10 + log 1

[As conc. of HB and B are same]

$$ \Rightarrow $$ pOH = 10

$$ \Rightarrow $$ pH = 14 – pOH = 14 – 10 = 4
2
MCQ (Single Correct Answer)

AIPMT 2010 Prelims

What is [H+] in mol/L of a solution that is 0.20 M in CH3COONa and 0.10 M in CH3COOH? Ka for CH3COOH = 1.8 $$ \times $$ 10$$-$$5
A
3.5 $$ \times $$ 10$$-$$4
B
1.1 $$ \times $$ 10$$-$$5
C
1.8 $$ \times $$ 10$$-$$5
D
9.0 $$ \times $$ 10$$-$$6

Explanation

CH3COOH and CH3COONa constitute to form an acidic buffer.

$$ \Rightarrow $$ pH = pKa + log$${{\left[ {Salt} \right]} \over {\left[ {Acid} \right]}}$$

pH = –log(1.8 × 10–5) + log $${{\left( {0.20} \right)} \over {\left( {0.10} \right)}}$$

= 4.74 + log 2

= 4.74 + 0.3010 = 5.041

Now, pH = – log[H+]

$$ \Rightarrow $$ 5.041 = – log[H+]

$$ \Rightarrow $$ [H+] = 10–5.041 = 9.0 × 106 mol L–1
3
MCQ (Single Correct Answer)

AIPMT 2010 Prelims

If pH of a saturated solution of Ba(OH)2 is 12, the value of its Ksp is
A
4.00 $$ \times $$ 10$$-$$6 M3
B
4.00 $$ \times $$ 10$$-$$7 M3
C
5.00 $$ \times $$ 10$$-$$6 M3
D
5.00$$ \times $$ 10$$-$$7 M3

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 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

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