2H2O ⇌ H3O+ + OH-Kw = 1 × 10-14 at 250C. Hence, Ka is

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$2 H_{2} O ⇌ H_{3} O^{+} + {OH}^{-}$

$K_{w} = 1 \times 10^{- 14} at 25^{0} C .$ Hence, $Ka$ is

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$5.55 \times 10^{- 13}$

$1.8 \times 10^{- 16}$

$1 \times 10^{- 14}$

$1.00 \times 10^{- 7}$

answer is C.

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Detailed Solution

$K_{w} = [H^{+}] [{OH}^{-}] = 1 \times 10^{- 14}$

$∴ [H^{+}] = 1 \times 10^{- 7} M,$

density of water = 1 g / mL ; or 1000 g/L C=number of moles(n)/lit = 1000/18 = 55.55 M

Also in weak acid $[H^{+}] = \sqrt{K_{a} C}$

$∴ K_{a} = \frac{{[H^{+}]}^{2}}{C} = \frac{10^{- 14}}{55.55} = 1.8 \times 10^{- 16}$

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