Equilibrium Question 794
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}} $
BETA
