Home
Class 12
CHEMISTRY
The dissoctiation of water of 25^(@)C is...

The dissoctiation of water of `25^(@)C` is `1.9 xx 10^(-7)%` and the density of water is `1g//cm^(3)` the ionisation constant of water is

A

`1.0xx10^(-15)`

B

`1.0xx10^(-14)`

C

`1.0xx10^(-16)`

D

`1.0xx10^(-6)`

Text Solution

Verified by Experts

The correct Answer is:
C
`{:(,H_(2)OhArrH^(+)+OH^(-),,[H_(2)O]=55.5,),(,KrArr([H^(+)][OH^(-)])/([H_(2)O]),,[H^(+)]=[OH^(-)]=10^(-7),),(,,,at -25^(@)C,):}`
`K=(10^(-7)xx10^(-7))/(55.5)`
`K +1.8xx10^(-16)`
Promotional Banner

Topper's Solved these Questions

  • CHEMISTRY AT A GLANCE

    ALLEN|Exercise INORGANIC CHEMISTRY|300 Videos

  • CHEMISTRY AT A GLANCE

    ALLEN|Exercise ORGANIC CHEMISTRY|483 Videos

  • Chemical Equilibrium

    ALLEN|Exercise All Questions|24 Videos

  • ELECTROCHEMISTRY

    ALLEN|Exercise EXERCISE -05 [B]|38 Videos

Similar Questions

Explore conceptually related problems

If degree of dissociation of pure water at 100^@ C is 1.8 xx 10^(-8) , then the dissociation constant of water will be: (Density of H_2O = 1 g/cc)

The ionization constant of water is 1.1 xx 10^(-16) . What is its ionic product of water?

The K_(sp) of AgC1 at 25^(@)C is 1.6 xx 10^(-9) , find the solubility of salt in gL^(-1) in water.

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

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

12.25g of CH_(3)CH_(2)CHClCOOH ia added to 250g of water to make a solution. If the dissociation constant of above acid is 1.44 xx 10^(-3) , the depression in freezing point of water in .^(@)C is : ( K_(f) for water is 1.86 "K kg mol"^(-1) )