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1

AIPMT 2002

MCQ (Single Correct Answer)
Solubility of MX2 type electrolytes is 0.5 $$ \times $$ 10$$-$$4 mole/lit., then find out Ksp of electrolytes.
A
5 $$ \times $$ 10$$-$$12
B
25 $$ \times $$ 10$$-$$10
C
1 $$ \times $$ 10$$-$$13
D
5 $$ \times $$ 10$$-$$13

Explanation

MX2 Ag2+ + 2X-
s s 2s


Ksp = [M2+] [X]2 = (S)(2S)2 = 4S3

$$ \Rightarrow $$ Ksp = 4(0.5 × 10–4)3 = 5 × 10–13
2

AIPMT 2001

MCQ (Single Correct Answer)
Correct relation between dissociation constants of a dibasic acid is
A
Ka1 = Ka2
B
Ka1 > Ka2
C
Ka1 < Ka2
D
Ka1 = $${1 \over {{K_{{a_2}}}}}$$

Explanation

In polyprotic acids the loss of second proton occurs much less readily than the first. Usually the Ka values for successive loss of protons from these acids differ by at least a factor of 10–3.

$$ \therefore $$ Ka1 > Ka2
3

AIPMT 2001

MCQ (Single Correct Answer)
Ionisation constant of CH3COOH is 1.7 $$ \times $$ 10$$-$$5 and concentration of H+ ions is 3.4 $$ \times $$ 10$$-$$4. Then find out initial concentration of CH3COOH molecules.
A
3.4 $$ \times $$ 10$$-$$4
B
3.4 $$ \times $$ 10$$-$$3
C
6.8 $$ \times $$ 10$$-$$4
D
6.8 $$ \times $$ 10$$-$$3

Explanation

CH3COOH ⇌ CH3COO + H+

Kc = $${{{\left[ {C{H_3}CO{O^ - }} \right]\left[ {{H^ + }} \right]} \over {\left[ {C{H_3}COOH} \right]}}}$$

$$ \Rightarrow $$ [CH3COOH] = $${{{3.4 \times {{10}^{ - 4}} \times 3.4 \times {{10}^{ - 4}}} \over {1.7 \times {{10}^{ - 5}}}}}$$

= 6.8 × 10–3
4

AIPMT 2001

MCQ (Single Correct Answer)
Solubility of M2S salt is 3.5 $$ \times $$ 10$$-$$6 then find out solubility product.
A
1.7 $$ \times $$ 10$$-$$6
B
1.7 $$ \times $$ 10$$-$$16
C
1.7 $$ \times $$ 10$$-$$18
D
1.7 $$ \times $$ 10$$-$$12

Explanation

M2S ⇌ 2M+ + S2–

Ksp = [M+]2[S2–] = (2s)2(s) = 4s 3

Ksp = 4(3.5 × 10–6)3 = 1.7 × 10–16

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