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Q.

The dissociation of water at a certain temperature is $1.9 \times 10^{- 5}$ percent and the density of water is $1.0 g / {cm}^{3}$ . The ionization constant of water is:

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a

$2 \times 10^{- 12}$

b

$1.00 \times 10^{- 14}$

c

$1.8 \times 10^{- 6}$

d

$2.00 \times 10^{- 16}$

answer is B.

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Detailed Solution

$k_{a}$ or

$k_{b} = c α^{2} k_{b} = \frac{1000}{18} \times {[1.9 \times 10^{- 7}]}^{2} k_{b} = 2 \times 10^{- 12}$

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The dissociation of water at a certain temperature is 1.9×10−5percent and the density of water is 1.0 g/cm3. The ionization constant of water is: