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The degree of dissociation of water is ...

The degree of dissociation of water is `1.8 xx10^(-9)` at 298 K . Calculate the ionization constant and Ionic product of water at 298 K .

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`1xx10^(-14)`
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At 25^@C the degree of ionization of water was found to be 1.8 xx 10^(–9) . Calculate the ionization constant and ionic product of water at this temperature

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Knowledge Check

  • If the degree of ionization of water is 1.8xx10^(-9) at 298K . Its ionization constant will be

    A
    `1.8xx10^(-16)`
    B
    `1xx10^(-14)`
    C
    `1xx10^(-16)`
    D
    `1.67xx10^(-14)`

  • If the degree of ionization of water be 1.8xx10^(-9) at 298 K . Its ionization constant will be

    A
    `1.8xx10^(-16)`
    B
    `1xx10^(-14)`
    C
    `1xx10^(-16)`
    D
    `1.67xx10^(-14)`

  • If the degree of ionization of water be 1.8 xx 10^(-9) at 298K. Its ionization constant will be

    A
    `1.8 xx 10^(-16)`
    B
    `1 xx 10^(-14)`
    C
    `1 xx 10^(-16)`
    D
    `1.67 xx 10^(-14)`

    • Similar Questions

    Explore conceptually related problems

    The degree of dissociation of pure water at 18^@C is found to be 1.8 xx 10^(-9) . Find ionic product of water and its dissociation constant at 18^@C .

    Define ionic product of water. The degree of dissociation of pure water at 18^(@)C is found to be 1.8times10^(-9) Find the final ionic product of water and its dissociation constant at 18^(@)C

    A 0.01 M solution of acetic acid is 1.34 % ionised (degree of dissociation = 0.0134) at 298 K. What is the ionization constant of acetic acid ?

    The degree of dissociation of water at 25^(@)C is 1.9 xx 10^(-7)% and density of 1.0 gm^(-3) . The ionic constant for water is

    At 277 K, degree of dissociation water is 1xx10^(-7)% . The value of ionic product of water is

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