A) \[1\times {{10}^{-12}}\]
B) \[1\times {{10}^{-14}}\]
C) \[1.8\times {{10}^{-12}}\]
D) \[1.8\times {{10}^{-14}}\]
Correct Answer: D
Solution :
[d] As, molarity, \[=\frac{Wt.\text{ }of\text{ }solute\text{ }per\text{ }litre\text{ }of\text{ }solution}{Mol.\text{ }wt.\text{ }of\text{ }solute}\] Molarity of \[{{H}_{2}}O=\frac{1000}{18}mol/L\] \[\underset{c(1-\alpha )}{\mathop{{{H}_{2}}O}}\,\rightleftharpoons \underset{c\alpha }{\mathop{{{H}^{+}}}}\,+\underset{c\alpha }{\mathop{O{{H}^{-}}}}\,\] Thus, \[{{K}_{a}}=\frac{c{{\alpha }^{2}}}{1-\alpha }=c{{\alpha }^{2}}\] \[=\frac{1000}{18}\times {{(1.8\times {{10}^{-8}})}^{2}}\] \[=1.8\times {{10}^{-14}}\]
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