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Question: If the degree of ionization of water be 1.8 x 10-9 at 298 K. Its ionization constant will be...

If the degree of ionization of water be 1.8 x 10-9 at 298 K. Its ionization constant will be

A

1.8×10−161.8 \times 10 ^ { - 16 }

B

1×10−141 \times 10 ^ { - 14 }

C

1×10−161 \times 10 ^ { - 16 }

D

1.67×10−141.67 \times 10 ^ { - 14 }

Answer

1.8×10−161.8 \times 10 ^ { - 16 }

Explanation

Solution

Ka=Kω[H2O]=10−14555=1.8×10−16\mathrm { K } _ { \mathrm { a } } = \frac { \mathrm { K } _ { \omega } } { \left[ \mathrm { H } _ { 2 } \mathrm { O } \right] } = \frac { 10 ^ { - 14 } } { 555 } = 1.8 \times 10 ^ { - 16 }