please give a well explained solution

Q
2.5 mL of 2/5M weak mono acidic base (Kb = 1 × 10–12 at 25°C) is titrated with 2/15 M HCl in water at 25°C. The concentration of H+ at equivalence point is (Kw = 1 × 10–14 at 25°C)
(A) 3.7 × 10–13 M
(B) 3.2 × 10–7 M
(C) 3.2 × 10–2 M
(D) 2.7 × 10–2 M

20 Answers

11
Mani Pal Singh ·

its an 2008 question
but BEWARE to post the solution

I HAVE A LOT OF THINGS TO ASK

11
Mani Pal Singh ·

nahin yaar yeh to abhi net se download mara hai
mere pass doosre solution pade hain

uss mein

1
Aditya ·

yes...but wats mani's doubt?

1
vector ·

after seeing soln it s luking easy

1
Aditya ·

wat u didn't get here?

11
Mani Pal Singh ·

2.5 mL 2/5 M base is neutralized by 15/2 mL of 2/15M acid

Salt concentration = 1/10
C = 0.1
Kw/Kb = Ch2/1-h
On solving, h = 0.27
[H+] = Ch = 2.7 X 10 2 M

1
Aditya ·

Mani..cud u pls post the soln with u...may b it can help!

1
vector ·

plz explain

11
Mani Pal Singh ·

bhai thoda elaborate to kar

13
Двҥїяuρ now in medical c ·

10h2+h-1=0
h=(-1+√41)/20=0.27

ch=[H+]=0.1*0.27=2.7*10-2

1
ANKIT GOYAL ·

mani u just find out concentration of the salt at the equivalence point.
then u use the formula ph=.5(pkw-pkb-logc)

1
Aditya ·

d.

1
vector ·

wat s d ans

11
Mani Pal Singh ·

sorry boss
copy paste mein galti ho gaya
[17]

13
Двҥїяuρ now in medical c ·

[16][16]

52M....i wasted 30 min for it!!!!!

1
Aditya ·

Hey Mani,

It shud be 2/5 M and not 52M..!

11
Mani Pal Singh ·

actually this is JEE 2008 question

I HAVE THE SOLUTION WITH ME BUT STILL I AM NOT CONVINCED

11
Mani Pal Singh ·

bhai ........................
abhi the question and the options r perfect!!!!!!!!!!!!!

1
Philip Calvert ·

please explain a bit
me too want explanation

13
Двҥїяuρ now in medical c ·

is it 52M or 5.2M??

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