1

### AIPMT 2001

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
2

### AIPMT 2001

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
3

### AIPMT 2001

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
4

### AIPMT 2001

In HS$-$, I$-$, R $-$ NH2, NH3 order of proton accepting tendency will be
A
I$-$ > NH3 > R $-$ NH2 > HS$-$
B
NH3 > R $-$ NH2 > HS$-$ > I$-$
C
R $-$ NH2 > NH3 > HS$-$ > I$-$
D
HS$-$ > R $-$ NH2 > NH3 > I$-$

## Explanation

Strong base has higher tendency to accept the proton. Increasing order of base and hence the order of accepting tendency of proton is

R $-$ NH2 > NH3 > HS$-$ > I$-$