The dissociation constant of water at 10°C is 5.31×10-17. The ionic product of water in mol2lit-2 is

Ki=Kw55.55⇒Kw=Ki×55.55⇒Kw=5.31×10-17×55.55=2.95×10-15 mole2lit-2Hence the correct answer is (A) 0.29×10-14

Ki=Kw55.55⇒Kw=Ki×55.55⇒Kw=5.31×10-17×55.55=2.95×10-15 mole2lit-2Hence the correct answer is (A) 0.29×10-14

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The dissociation constant of water at $10 ° C$ is $5.31 \times 10^{- 17}$ . The ionic product of water in ${mol}^{2} {lit}^{- 2}$ is

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$0.6 \times 10^{- 14}$

$0.29 \times 10^{- 14}$

$0.54 \times 10^{- 14}$

$0.82 \times 10^{- 14}$

answer is A.

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$K_{i} = \frac{K_{w}}{55.55} \Rightarrow K_{w} = K_{i} \times 55.55$

$\Rightarrow K_{w} = 5.31 \times 10^{- 17} \times 55.55 = 2.95 \times 10^{- 15} {mole}^{2} {lit}^{- 2}$

Hence the correct answer is (A) $0.29 \times 10^{- 14}$

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