2.5mL of2/5M weak monoacidic base (Kb)=(1)(10)^-12 at 25*C.The concentration of H^+ at equivale points is(Kw=1(10^-14)at25*C(1)(3.7)(10^-13)M(2)(3.2)(10^-7)M(3)(3.2)(10^-2)M(4)(3.2)(10^-2)M

BOH + H₂O  <====>  B⁺ + OH^-

1000 ml solution have molarity = 2/5

1ml = 2/5 *10^-3

2.5ml = (2/5)*(2.5)*10^-3 = 10^-3M

now , concentration (OH^-) = (kC)^1/2

k = 10^-12 , C = 10^-3 so

[OH^-] = [3.33*10^-8]

[H⁺] = 10^-14/[OH^-] = 3.2*10^-7

approve if u like my ans

Dear student,

The reaction can be written as
BOH + H₂O → B⁺ + OH^-
1000 ml solution have molarity = 2/5
1ml = 2/5 *10^-3
2.5ml = (2/5)*(2.5)*10^-3 = 10^-3M
Now , concentration of (OH^-) = (kC)^1/2
k = 10^-12 , C = 10^-3
So,
[OH^-] = [3.33*10^-8]

[H+] = 10^-14/[OH^-] = 3.2*10^-7

Hope it helps.
Thanks and regards,
Kushagra