Ionic EquilibriumHard
Question
The pH of 0.1 M – N2H4 solution is (For N2H4, Kb1 = 3.6 × 10−6, Kb2 = 6.4 × 10−12, log 2 = 0.3, log 3 = 0.48)
Options
A.3.22
B.2.72
C.10.78
D.11.22
Solution
For POH, 2nd protonation may be neglected
$\left\lbrack OH^{-} \right\rbrack = \sqrt{K_{b_{1}}.C} = \sqrt{3.6 \times 10^{- 6} \times 0.1} = 6 \times 10^{- 4}\text{ M}$
∴ POH = –log(6 × 10–4) = 3.22 ⇒ PH = 10.78
Create a free account to view solution
View Solution FreeMore Ionic Equilibrium Questions
Unexposed silver halides are removed from photographic film when they react with sodium thiosulphate to form the complex...Given Ag(NH3)2+$\rightleftharpoons$ Ag+ + 2NH3, Kc = 7.2 × 10–8 and Ksp of AgCl = 1.8 × 10–10 at 298 K. If ammonia is a...The pH at which water is maximum dissociated at 25o C, is...There exist an equilibrium between solid BaSO4, Ba2+ and SO42− ions in aqueous medium. Now, if equilibrium is disturbed ...Nitration of aniline in strong acidic medium also gives m-nitroaniline because...