Equilibrium Question 830

Question: If degree of dissociation of pure water at $ 100{}^\circ C $ is $ 1.8\times {10^{-8}} $ , then the dissociation constant of water will be (density of $ H_2O=1g/cc $ )

Options:

A) $ 1\times {10^{-12}} $

B) $ 1\times {10^{-14}} $

C) $ 1.8\times {10^{-12}} $

D) $ 1.8\times {10^{-14}} $

Show Answer

Answer:

Correct Answer: D

Solution:

As, molarity, $ =\frac{Wt.ofsoluteperlitreofsolution}{Mol.wt.ofsolute} $ Molarity of $ H_2O=\frac{1000}{18}mol/L $

$ \underset{c(1-\alpha )}{\mathop{H_2O}},\rightarrow \underset{c\alpha }{\mathop{{H^{+}}}},+\underset{c\alpha }{\mathop{O{H^{-}}}}, $ Thus, $ K_{a}=\frac{c{{\alpha }^{2}}}{1-\alpha }=c{{\alpha }^{2}} $

$ =\frac{1000}{18}\times {{(1.8\times {10^{-8}})}^{2}} $

$ =1.8\times {10^{-14}} $



NCERT Chapter Video Solution

Dual Pane