Ionic EquilibriumHard
Question
The degree of dissociation of pure water at 25°C is found to be 1.8 × 10–9. The dissociation constant, Kd of water, at 25°C is
Options
A.10−14
B.1.8 × 10–16
C.5.56 × 10–13
D.1.8 × 10–14
Solution
$K_{d} = \frac{\left\lbrack H^{+} \right\rbrack\left\lbrack OH^{-} \right\rbrack}{\left\lbrack H_{2}O \right\rbrack}$
$= \frac{\alpha^{2}.C}{1 - \alpha} \approx \alpha^{2}.C = \left( 1.8 \times 10^{- 9} \right)^{2} \times \frac{1000}{18} = 1.8 \times 10^{- 16}$
Create a free account to view solution
View Solution FreeMore Ionic Equilibrium Questions
The pH of which salt is independent of its concentration :1. (CH3COO)C5H5NH 2. NaH2PO4 3. Na2HPO4 4.NH4CN...pH of a saturated solution of Ba ( OH )2 is 12. The value of solubility product Ksp of Ba ( OH2 ) is...A physician wishes to prepare a buffer solution at pH = 3.58 that efficiently resists a change in pH yet contains only s...A solution is prepared in which 0.1 mole each of HCl, CH3COOH and CHCl2COOH is present in a litre. If the ionization con...The dissociation constant of a weak monoprotic acid is numerically equal to the dissociation constant of its conjugate b...