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 H₂O = 1 gm/cc)
(a) 1 × 10⁻¹²
(b) 1 × 10⁻¹⁴
(c) 1.8 × 10⁻¹²
(d) 1.8 × 10⁻¹⁴

Answer 1

Answer:

(c) 1.8 × 10^-12 ✔✔✔✔

Right answer

Explanation:

_________xd

Answer 2

Answer: d)1.8×10⁻¹⁴

Explanation:

For finding out the dissociation constant, we need the concentration of water(Molarity) and the Ionic concentration of H⁺ and OH⁻.

Here density is given, so we can easily calculate the molarity of water.

The ionic concentration of H⁺ and OH⁻ is equal to the product of the Concentration of water and the degree of dissociation.

$Degree\ of\ dissociation(100^{0}C)(\alpha )=1.8\times 10^{-8}\\Density\ of\ water=1g/cc(CGS)=1000kg/m^{3}(MKS)\\Molarity=Density/Molar mass=(1000kg/m^{3})/(18g/mol)=10^{6}/18mol/m^{3}\\We\ know\ that:1m^{3}=1000\ litres\\Molarity(Concentration'C')=(1000/18)mol/l=55.56M\\H^{+}=OH^{-}=C\alpha =55.56\times 1.8\times 10^{-8}\simeq 10^{-6}\\K_{d}=[H^{+}][OH^{-}]/[H_{2}O]=10^{-12}/55.56M=1.8\times 10^{-14}$