Dilution processes of different aqueous solutions, with water, are given in LIST-I. The effects of dilution of the solution on [H + ] are given in LIST-II.
LIST-I LIST-II
(P) (10 mL of 0.1 M NaOH + 20 mL of (1) the value of [H + ] does not change
0.1 M acetic acid) diluted to 60 mL on dilution
(Q) (20 mL of 0.1 M NaOH + 20 mL of (2) the value of [H + ] changes to half of
0.1 M acetic acid) diluted to 80 mL its initial value on dilution
(R) (20 mL of 0.1 M HCl + 20 mL of (3) the value of [H + ] changes to two
0.1 M ammonia solution) diluted to 80 mL times of its initial value on dilution
(S) 10 mL saturated solution of Ni(OH) 2 in (4) the value of [H + ] changes to 
equilibrium with excess solid Ni(OH) 2 is times of its initial value on dilution
diluted to 20 mL (solid Ni(OH) 2 is still
present after dilution).
(5) the value of [H + ] changes to 
times of its initial value on dilution
Match each process given in LIST-I with one or more effect(s) in LIST-II. the correct option is
Text Solution
Verified by ExpertsD
(P → 1) ; (Q → 5); (R → 4) ; (S → 1)
Sol.
(P) NaOH + CH 3 COOH 
CH 3 COONa + H 2 O
M.Mole 1 2
Now solution contains 1 m. mole CH 3 COOH & 1 m.mole CH 3 COONa in 30 ml solution.
It is a Buffer solution
∴ [H + ] does not change with dilution.
(Q) NaOH + CH 3 COOH 
CH 3 COONa + H 2 O
M.Mole 2 2
Now solution contain 2 m.mole of CH 3 COONa in 40 ml solution (salt of weak acid strong base)
[H + ] initial = 
Now on dilution upto 80 ml, now can. Becomes 
.
∴ [H + ] new = 
= [H + ] initial × 
(R) HCl + NH 3 
NH 4 Cl
M.Mole 2 2
Now solution contain 2 m.mole of NH 4 Cl in 40 ml solution (salt of SA & WB)
[H + ] initial = 
Now on dilution upto 80 ml, new conc. becomes 
.
[H + ] new = 
= 
(S) Ni(OH) 2 (s) 
Ni 2+ + 2OH –
it is sparingly soluble salt
∴ on dilution [OH – ] conc. in saturated solution of Ni(OH) 2 remains const.
[H + ] new = [H + ] initial
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