Question

the ionization constant of water at 90 C is​

Answer 1

If the degree of dissociation of water at 90 o C { 90 }^{ o }C 90oC is 1.28 × 10 − 8 1.28\times { 10 }^{ -8 } 1. 28×10−8 then the ionization constant of water at 90 o C { 90 }^{ o }C 90oC is: A .

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Answer 2

Answer:

The ionization constant of water at $90^{o}C$ is $9.07 * 10^{-15}$ .

Explanation:

The ionization constant:- An ionization constant$(K)$equilibrium ions and molecules which are not ionized in an aqueous solution. Basically, it is the ratio of the product and the reactant that leads to perfect stoichiometric power.

Degree of dissociation:- The term degree of dissociation is a phenomenon of producing free ions that are loaded with the current in a given concentration that is dissociated from the fraction of solute.

Degree of dissociation of water $(\alpha )$ $= 1.28$ ×$10^{-8}$

to water $c= \frac{1000}{18} =55.56$

then $H_{2}O$ ⇄$[H^{+}] + [OH^{-}]$

∴ $[K_{a} ]= \frac{[H^{+} ][OH^{-} ]}{[H_{2} O]}$                        ∵$[H_{2}O } ]= 55.56$

$[{H^{+}] } = [OH^{-} ]= c \alpha = 55.56$ × $1.28$ × $10^{-8}$ $= [71.12$ × $10^{-8}]$

$[K_{w} ]= [H^{+} ] .[OH^{-} ] = [71.12$ × $10^{-8} ] = 5057.6 * 10^{-16}$

∴ $[K_{d}] =$ $\frac{[H^{+} ][OH^{-} ]}{[H_{2} O]} =\frac{[K_{w}] }{[H_{2} O]} = \frac{5057.6 * 10^{-16} }{55.56}$

                                         $= 9.07 * 10^{-15}$

Hence, the ionization constant of water at $90^{o}C$ is $9.07 * 10^{-15}$.