If the degree of ionization of water be 1-8 - 10-9 at 298K- What will be its ionization constant-

Degree of ionisation= 1.8 × 10^ 9 Molarity of water= 55.55M [H^+]=1.8× 10^ 9× 55.55 #160; #160; #160; #160; #160; =99.99 × 10 ...

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Standard XII

Chemistry

Introduction and Degree of Hydrolysis

If the degree...

Question

If the degree of ionization of water be $1.8 \times 10^{- 9}$ at $298$ K. What will be its ionization constant?

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Solution

Degree of ionisation= $1.8 \times 10^{- 9}$
Molarity of water= $55.55 M$
$[H^{+}] = 1.8 \times 10^{- 9} \times 55.55$
$= 99.99 \times 10^{- 9} = 1 \times 10^{- 7}$
$K_{a} = [H^{+}] [O H^{-}] = (10^{- 7})^{2} = 10^{- 14}$

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If the degree of ionization of water be 1.8×10−9 at 298K. What will be its ionization constant?

Degree of ionisation= 1.8×10−9Molarity of water= 55.55M[H+]=1.8×10−9×55.55 =99.99×10−9=1×10−7Ka=[H+][OH−]=(10−7)2=10−14

If the degree of ionization of water be $1.8 \times 10^{- 9}$ at $298$ K. What will be its ionization constant?

Degree of ionisation= $1.8 \times 10^{- 9}$
Molarity of water= $55.55 M$
$[H^{+}] = 1.8 \times 10^{- 9} \times 55.55$
$= 99.99 \times 10^{- 9} = 1 \times 10^{- 7}$
$K_{a} = [H^{+}] [O H^{-}] = (10^{- 7})^{2} = 10^{- 14}$