Pls solve Q121 Following five solution of KOH were prepare as- - Chemistry -

Question

Pls solve

Q121 Following five solution of KOH were prepare as-

First → 0 1 moles in 1 L
Second → 0 2 moles in 2 L
â€‹ â€‹ Third → 0 3 moles
in 3 L
Fourth â€‹ → 0 4 moles in 4 L
Fifth â€‹ → 0 5 moles in 5 L

The pH of resultant solution is :-

(1) 2 (2) 1 (3) 13 (4) 7

Vartika Jain · Accepted Answer

Dear Student,

We know that, Moles of final solution = sum of the moles of five solutionsi
e  Moles of final solution = 0 1+0
2+0 3+0 4+0
5                                              =1
5Now, [KOH] =MolesTotal Volume=1 515= 0
1Since KOH is a strong baseSo, Moles of KOH = Moles of OH-So, pH = 14- pOH             =14 -(- log [OH-])             =14+log(0
1)             =14-1             =13Hence, correct answer is (3)

Pls solve Q121. Following five solution of KOH were prepare as- First → 0.1 moles in 1 L Second → 0.2 moles in 2 L ​ ​ Third → 0.3 moles in 3 L Fourth ​ → 0.4 moles in 4 L Fifth ​ → 0.5 moles in 5 L The pH of resultant solution is :- (1) 2 (2) 1 (3) 13 (4) 7

Pls solve

Q121. Following five solution of KOH were prepare as-

First $\to$ 0.1 moles in 1 L
Second $\to$ 0.2 moles in 2 L
Third $\to$ 0.3 moles in 3 L
Fourth $\to$ 0.4 moles in 4 L
Fifth $\to$ 0.5 moles in 5 L

The pH of resultant solution is :-

(1) 2 (2) 1 (3) 13 (4) 7