Dilution process 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 `(alpha)` of weak acid and weak base is ltlt 1, degree of hydrolysis of salt `lt lt 1 , [H^(+)]` represents the concentration of `H^(+)` ions)
`|{:(," List-I",," List -II"),((P),"(10 mL of 0.1 M NaOH+20 mL of 0.1 acetic acid)dilutes to 60 mL",(1),"The value of "[H^(+)]" does not change on dilution"),((Q),"(20 mL of 0.1 M NaOH + 20 mL of 0.1 M acetic acid)diluted to 80 mL",(2),"The value of "[H^(+)]" change to half of its initial value on dilution"),((R ),"(20 mL of 0.1 M HCl + 20 mL of 0.1 M ammonia solution)diluted to 80 mL",(3),"The value of "[H^(+)]" change to two times of its initial value on dilution"),((S),"10 mL saturated solution of "Ni(OH)_(2)" in equilibrium with excess solid "Ni(OH)_(2)" is ",(4),"The value of "[H^(+)]" changes to "(1)/(sqrt(2))" 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 "sqrt(2)" times of its initial value on dilution"):}|`

Answer

`P to 1; Q to 5; R to 4; S to 1`

Explanation

`P to 1, Q to 5, R to 4, S to 1`
P. `underset("0.1M, 20ml")(CH_(3)COOH)+underset("0.1M, 10ml")(NaOH)to CH_(3)COONa+H_(2)O`
`pH=pKa rArr [H^(+)]` will not change on dilution correct match : F - 1
Q. `{:(underset("0.1M, 20ml")(CH_(3)COOH)+underset("0.1M, 20ml")(NaOH)to CH_(3)COONa+H_(2)O),(" "- " " -" 0.05M"):}`
`[OH^(-)]=sqrt(K_(H)C)=sqrt(((k_(w))/(k_(a))C))`
`[H^(+)]_(1)=sqrt((k_(w)k_(a))/(C))`
`([H^(+)]_(2))/([H^(+)]_(1))=sqrt((C_(1))/(C_(2)))=sqrt((0.05)/(0.025))=sqrt(2)`
Correct match : Q-5
R. `underset("0.1M, 20ml")(NH_(4)OH)+underset("0.1M, 20ml")(HCl)to underset(0.05M)(NH_(4)Cl)`
`[H^(+)]=sqrt(K_(H)C)`
`([H^(+)]_(2))/([H^(+)]_(1))=sqrt((C_(2))/(C_(1)))=(1)/(sqrt(2))`
Correct match : R-4
S. Because of dilution solubility does not change so `[H^(+)]` = constant.