The solubility of AgCl(s) with solubility product 1.6 $$ \times $$ 10$$-$$10 in 0.1 M NaCl solution would be
A
1.26 $$ \times $$ 10$$-$$5 M
B
1.6 $$ \times $$ 10$$-$$9 M
C
1.6 $$ \times $$ 10$$-$$11 M
D
zero
Explanation
AgCl ⇌ Ag+ + Cl– s s s + 0.1
Concentration of Cl–
is (s + 0.1) mol L–1 because
s mol L–1 from ionization of AgCl and 0.1 mol L–1
from ionization of 0.1 M NaCl.
Now, Ksp = [Ag+][Cl–
]
$$ \Rightarrow $$ 1.6 × 10–10 = s (s + 0.1)
$$ \Rightarrow $$ 1.6 × 10–10 = s (0.1) {$$ \because $$ s << 0.1}
$$ \Rightarrow $$ s = 1.6 × 10–9 M
2
NEET 2016 Phase 2
MCQ (Single Correct Answer)
Which of the following fluro-compounds is most likely to behave as a Lewis base ?
A
BF3
B
PF3
C
CF4
D
SiF4
Explanation
PF3 is lewis base because on P atom there is lone pair.
3
NEET 2016 Phase 2
MCQ (Single Correct Answer)
The percentage of pyridine (C5H5N) that forms pyridinium ion (C5H5N+H) ina 0.10 M aqueous pyridine solution (Kb for C5H5N = 1.7 $$ \times $$ 10$$-$$9) is
A
0.0060%
B
0.013%
C
0.77%
D
1.6%
Explanation
C5H5N + H2O → C5H5N+H + OH–
So, the amount of pyridine that forms pyridinium ion is $$\alpha $$.