Question

At $25^oC$, the dissociation constant for pure water is given by:-
A ${\left ( {55.4 \times {{10} ^ {14}}} \right)^ {- 1}}$
B $1 \times {10^ {- 14}}$
C. $\dfrac {{1 \times {{10} ^ {- 14}}}} {{18}}$
D. none of these

Answer 1

Pure water contains ions. There is a self-ionization constant of water ($Kw$) which has no units. It is the product of the hydronium ions times the hydroxide ions. At $25^\circ C$ ${H_3}{O^ +}$ and $O{H^ -}$ are having equal concentration. In water dissociation they have equal concentration.
FORMULA USED
$Kw = {{ }}\left [{{H_3} {O^ +}} \right]\left [{O {H^ -}} \right]$

Complete step by step answer:
Here at $25^\circ C$ dissociation constant is equal to equilibrium constant. At $25^\circ C$ ${H_3}{O^ +}$ and $O{H^ -}$ are having equal concentration
Dissociation constant of water is
By using the above formula we can have $Kw$
$Kw = {{ }}\left [{{H_3} {O^ +}} \right]\left [{O {H^ -}} \right]$
$= ({10^ {- 7}}) ({10^ {- 7}})$
$= {10^ {- 14}}$
The reason why we choose ${10^ {- 7}}$ is $pH$ of the activity of hydrogen ions is ${10^ {- pH}}$ where at $25^\circ C$ both acid and bases are equal.
This is only for pure water at $25^\circ C$
Hence correct answer to this question is option “B” which is $1 \times {10^ {- 14}}$

Additional information:
-Dissociation constant
A quantity expressing the extent to which a specific substance in solution is dissociated into ions, adequate to the product of the concentrations of the respective ions divided by the concentration of the undissipated molecule.
Two species that differ by only a proton constitute a conjugate acid–base pair. For a solution of a weak acid, the equilibrium constant is named the acid ionization constant ($Ka$). Similarly, the constant for the reaction of a weak base with water is that the base ionization constant ($Kb$)
The concept of the equilibrium constant is applied in different fields of chemistry and pharmacology. In protein-ligand binding the equilibrium constant describes the affinity between a protein and a ligand
The smaller the equilibrium constant, the more tightly bound the ligand is, or the upper the affinity between ligand and protein

Note: Suppose when we choose only water dissociation instead of pure water dissociation at $25^\circ C$ then dissociation constant of both for water and pure water are differ because in dissociation constant for water molarity plays a role.