If degree of dissociation of pure water at 100° C is 1.8 × 10-8, then the dissociation constant of water will be (density of H 2 O =1 g / cc )

Question

If degree of dissociation of pure water at 100° C is 1.8 × 10-8, then the dissociation constant of water will be (density of H 2 O =1 g / cc ) (A) 1 × 10-12 (B) 1 × 10-14 (C) 1.8 × 10-12 (D) 1.8 × 10-14

Tardigrade Admin · Accepted Answer

As, molarity, =( text wt. of solute per litre of solution / text Mol. wt. of solute ) Molarity of H 2 O =(1000/18) mole / litre undersetc(1-α)H 2 O leftharpoons undersetcαH ++ undersetcαOH - Thus, Ka=(c α2/1-r)=c α2=1.8 × 10-14

Q. If degree of dissociation of pure water at $10 0^{\circ} C$ is $1.8 \times 1 0^{- 8}$ , then the dissociation constant of water will be (density of $H_{2} O = 1 g / cc$ )

$1 \times 1 0^{- 12}$

$1 \times 1 0^{- 14}$

$1.8 \times 1 0^{- 12}$

$1.8 \times 1 0^{- 14}$

As, molarity, $= \frac{wt. of solute per litre of solution}{Mol. wt. of solute}$
Molarity of $H_{2} O = \frac{1000}{18}$ mole / litre
$c (1 - α) H_{2} O ⇌ c α H^{+} + c α O H^{-}$
Thus, $K_{a} = \frac{c α ^{2}}{1 - r} = c α^{2} = 1.8 \times 1 0^{- 14}$

Q. If degree of dissociation of pure water at 100∘C is 1.8×10−8, then the dissociation constant of water will be (density of H2​O=1g/cc )

1×10−12

1×10−14

1.8×10−12

1.8×10−14