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19/10/2021 11:04 am
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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 g/cc)
(a) 1 × 10-12
(b) 1 × 10-14
(c) 1.8 × 10-12
(d) 1.8 × 10-14
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19/10/2021 11:10 am
Answer is : (d) 1.8 × 10-14
Explanation:
As, molarity
= \(\frac{Wt.\;of\;solute\;per\;litre\;of\;solution}{Mol.\;wt.\;of\;solute}\)
Molarity of H2O
= \(\frac{1000}{18}\)mol/L
H2O ⇌ H+ + OH-
c(1-α) cα cα
Thus,
Ka = \(\frac{c \alpha^2}{1-α}\)
= cα2
= \(\frac{1000}{18}\times(1.8 \times 10^{-8})^2\)
= 1.8 x 10-14
This post was modified 3 years ago by Mayank1444
This post was modified 3 years ago 2 times by admin