Answer
Verified
98.1k+ views
Hint: In case of pure water the amount of $[{H^ + }]$ and $[O{H^ - }]$ is the same. Water is a weak electrolyte and the amount of dissociation in pure water is very low. Due to this the amount of non-dissociated water molecules remains more or less constant. The product of $[{H^ + }]$ and $[O{H^ - }]$ is called an ionic product of water. This is denoted as \[{{\text{K}}_w}\].
Complete step by step solution:
Now Given,
Concentration of \[{{\text{H}}_{\text{3}}}{{\text{O}}^{\text{ + }}}{\text{ = 1}}{{\text{0}}^{{\text{ - 6}}}}{\text{mol/litre}}\]
The ionization reaction of water is,
\[{{\text{H}}_{\text{2}}}{\text{O}} \rightleftharpoons {{\text{H}}^{\text{ + }}}{\text{ + O}}{{\text{H}}^{\text{ - }}}\]
The expression for equilibrium constant is,
\[{\text{K = }}\dfrac{{\left[ {{{\text{H}}^{\text{ + }}}} \right]{\text{ }}\left[ {{\text{O}}{{\text{H}}^{\text{ - }}}} \right]}}{{\left[ {{{\text{H}}_{\text{2}}}{\text{O}}} \right]}}\]
After rearranging thisformula, we get,
\[{\text{K}} \times \left[ {{{\text{H}}_{\text{2}}}{\text{O}}} \right]{\text{ = }}\left[ {{{\text{H}}^{\text{ + }}}} \right]{\text{ }}\left[ {{\text{O}}{{\text{H}}^{\text{ - }}}} \right]\]
Due to the large amount of \[{{\text{H}}_{\text{2}}}{\text{O}}\] is large, that is why it can be taken as constant .
\[{{\text{K}}_w}{\text{ = }}\left[ {{{\text{H}}^{\text{ + }}}} \right]{\text{ }}\left[ {{\text{O}}{{\text{H}}^{\text{ - }}}} \right]\]
Where, \[{{\text{K}}_w}\] is the equilibrium constant.
Pure water means the concentration of hydronium ion and hydroxide ion are equal.
\[{\text{ }}\left[ {{{\text{H}}^{\text{ + }}}} \right]{\text{ = }}\left[ {{\text{O}}{{\text{H}}^{\text{ - }}}} \right]{\text{ = 1}}{{\text{0}}^{{\text{ - 6}}}}{\text{moles/litre}}\]
Now the values in the equation, \[{{\text{K}}_w}{\text{ = }}\left[ {{{\text{H}}^{\text{ + }}}} \right]{\text{ }}\left[ {{\text{O}}{{\text{H}}^{\text{ - }}}} \right]\]
$
{K_{_W}} = {10^{ - 6}} \times {10^{ - 6}} \\
{K_W} = {10^{ - 12}} \\
$
So, the correct option is B.
Note:
When a partly soluble salt dissolves into water it forms a dynamic equilibrium between the hydrated ions and its solid salt molecule. Due to this equilibrium the amount of solid salt is hydrolyzed remains unchanged after achieving an amount of hydration. Due to this a new term is introduced which is solubility product \[\left( {{{\text{K}}_{{\text{sp}}}}} \right)\].With increasing the temperature solubility of the salt increases as well as the concentrations of solvated ions. Therefore, with increasing temperature the value of solubility product increases and vice-versa.
Complete step by step solution:
Now Given,
Concentration of \[{{\text{H}}_{\text{3}}}{{\text{O}}^{\text{ + }}}{\text{ = 1}}{{\text{0}}^{{\text{ - 6}}}}{\text{mol/litre}}\]
The ionization reaction of water is,
\[{{\text{H}}_{\text{2}}}{\text{O}} \rightleftharpoons {{\text{H}}^{\text{ + }}}{\text{ + O}}{{\text{H}}^{\text{ - }}}\]
The expression for equilibrium constant is,
\[{\text{K = }}\dfrac{{\left[ {{{\text{H}}^{\text{ + }}}} \right]{\text{ }}\left[ {{\text{O}}{{\text{H}}^{\text{ - }}}} \right]}}{{\left[ {{{\text{H}}_{\text{2}}}{\text{O}}} \right]}}\]
After rearranging thisformula, we get,
\[{\text{K}} \times \left[ {{{\text{H}}_{\text{2}}}{\text{O}}} \right]{\text{ = }}\left[ {{{\text{H}}^{\text{ + }}}} \right]{\text{ }}\left[ {{\text{O}}{{\text{H}}^{\text{ - }}}} \right]\]
Due to the large amount of \[{{\text{H}}_{\text{2}}}{\text{O}}\] is large, that is why it can be taken as constant .
\[{{\text{K}}_w}{\text{ = }}\left[ {{{\text{H}}^{\text{ + }}}} \right]{\text{ }}\left[ {{\text{O}}{{\text{H}}^{\text{ - }}}} \right]\]
Where, \[{{\text{K}}_w}\] is the equilibrium constant.
Pure water means the concentration of hydronium ion and hydroxide ion are equal.
\[{\text{ }}\left[ {{{\text{H}}^{\text{ + }}}} \right]{\text{ = }}\left[ {{\text{O}}{{\text{H}}^{\text{ - }}}} \right]{\text{ = 1}}{{\text{0}}^{{\text{ - 6}}}}{\text{moles/litre}}\]
Now the values in the equation, \[{{\text{K}}_w}{\text{ = }}\left[ {{{\text{H}}^{\text{ + }}}} \right]{\text{ }}\left[ {{\text{O}}{{\text{H}}^{\text{ - }}}} \right]\]
$
{K_{_W}} = {10^{ - 6}} \times {10^{ - 6}} \\
{K_W} = {10^{ - 12}} \\
$
So, the correct option is B.
Note:
When a partly soluble salt dissolves into water it forms a dynamic equilibrium between the hydrated ions and its solid salt molecule. Due to this equilibrium the amount of solid salt is hydrolyzed remains unchanged after achieving an amount of hydration. Due to this a new term is introduced which is solubility product \[\left( {{{\text{K}}_{{\text{sp}}}}} \right)\].With increasing the temperature solubility of the salt increases as well as the concentrations of solvated ions. Therefore, with increasing temperature the value of solubility product increases and vice-versa.
Recently Updated Pages
Write a composition in approximately 450 500 words class 10 english JEE_Main
Arrange the sentences P Q R between S1 and S5 such class 10 english JEE_Main
Write an article on the need and importance of sports class 10 english JEE_Main
Name the scale on which the destructive energy of an class 11 physics JEE_Main
Choose the exact meaning of the given idiomphrase The class 9 english JEE_Main
Choose the one which best expresses the meaning of class 9 english JEE_Main
Other Pages
The focal length of a thin biconvex lens is 20cm When class 12 physics JEE_Main
A mosquito with 8 legs stands on the water surface class 11 physics JEE_Main
If a wire of resistance R is stretched to double of class 12 physics JEE_Main
Calculate CFSE of the following complex FeCN64 A 04Delta class 11 chemistry JEE_Main
Formula for number of images formed by two plane mirrors class 12 physics JEE_Main
Electric field due to uniformly charged sphere class 12 physics JEE_Main