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Dilution processes of different aqueous solution, with water, are given in LIST-I. The effects of dilution of the solutions on [H+] are given in LIST -II.
(Note: Degree of dissociation (α) of weak acid and weak base is << 1; degree of hydrolysis of salt << 1; [H+] represents the concentration of H+ ions)
LIST -ILIST-IIP. 10 mL of 0.1 M NaOH +20 mL of1. The value of [H+] does not0.1 M acetic acid diluted to 60 mLchange on dilutionQ. 20 mL of 0.1 M NaOH +20 mL of2. The value of [H+] changes to half0.1 M acetic acid diluted to 80 mLof its initial value on dilutionR. 20 mL of 0.1 M HCl + 20 mL of 0.13. The value of [H+] changes to twoM ammonia solution diluted to 80 mLtimes of its initial value on dilutionS. 10 mL saturated solution of Ni(OH)24. The value of [H+] changes to 12in equilibrium with excess solid Ni(OH)2times of its initial value on dilutionis diluted to 20 mL (solid Ni(OH)2is stillpresent after dilution).5. The value of [H+] changes to 2times 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

A
P4;Q2;R3;S1
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B
P4;Q3;R2;S3
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C
P1;Q4;R5;S3
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D
P1;Q5;R4;S1
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Solution

The correct option is D P1;Q5;R4;S1
KwKa=[CH3COO][OH][CH3COOH]
[OH]=(KwKa×C)
(P) is a buffer, so [H+] does not change on dilution, as [salt] =[acid].

(Q) contains only CH3COONa
So, CH3COO+H2OCH3COOH+OH
[OH]=Kh×C[H+] decreases by 2 times

(R) is also salt hydrolysis
So, NH+4+H2ONH4OH+H+
[H+]=KwKb×C
So, C is made 12 so ,[H+] becomes 12

(S) It is the case of solubility equilibria. So, dilution does not affect [H+] or [OH]

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