
What is meant by the ionic product of water?
(A) Concentration of ${{H}^{+}}$ ions
(B) Concentration of $O{{H}^{-}}$ ions
(C) The product of concentration of $O{{H}^{-}}$ and ${{H}^{+}}$ ions
(D) None of these
Answer
586.2k+ views
Hint: The ionic product of water is the product of concentration of the ions of water at a given temperature. The water molecule itself is not included while writing the expression for ionic products.
Complete step by step solution:
Water is self-ionized but it occurs to a limited extent. When the two molecules of water collide there is a transfer of H from one molecule to another and the product formed is hydronium ion with a positive charge and a hydroxide ion with a negative charge.
\[{{H}_{2}}O(l)+{{H}_{2}}O(l)\rightleftharpoons {{H}_{3}}{{O}^{+}}(aq)+O{{H}^{-}}(aq)\]
Which is often simplified as
\[{{H}_{2}}O(l)+\rightleftharpoons {{H}^{+}}(aq)+O{{H}^{-}}(aq)\]
The equilibrium constant for the expression for self-ionization of water referred to as the ionic product of water. ${{K}_{W}}$ is used as the symbol for ionic products of water.
${{K}_{W}}=\left[ {{H}^{+}} \right]\left[ O{{H}^{-}} \right]$
The ionic product of water is the product of concentration of $O{{H}^{-}}$ and ${{H}^{+}}$ ions.
At ${{25}^{0}}C$ the value of ionic product of water = $1.0X{{10}^{-14}}$
Due to the self-ionization of water it can act as both acid and base. When an acid or base is added to the water the ionic product of the water remains constant, but the concentration of ions changes for different types of solution.
Note: The concentration of \[O{{H}^{-}}and{{H}^{+}}\] ions can be calculated using the ionic product of water.
Calculation, we know that
${{K}_{W}}=\left[ {{H}^{+}} \right]\left[ O{{H}^{-}} \right]$=$1.0X{{10}^{-14}}$
We know that pure water is neutral concentration of $O{{H}^{-}}$ =${{H}^{+}}$= X
${{K}_{W}}=\left[ x \right]\left[ x \right]={{x}^{2}}$
The value of x or the concentration of both the ions is =$1.0X{{10}^{-7}}$
Complete step by step solution:
Water is self-ionized but it occurs to a limited extent. When the two molecules of water collide there is a transfer of H from one molecule to another and the product formed is hydronium ion with a positive charge and a hydroxide ion with a negative charge.
\[{{H}_{2}}O(l)+{{H}_{2}}O(l)\rightleftharpoons {{H}_{3}}{{O}^{+}}(aq)+O{{H}^{-}}(aq)\]
Which is often simplified as
\[{{H}_{2}}O(l)+\rightleftharpoons {{H}^{+}}(aq)+O{{H}^{-}}(aq)\]
The equilibrium constant for the expression for self-ionization of water referred to as the ionic product of water. ${{K}_{W}}$ is used as the symbol for ionic products of water.
${{K}_{W}}=\left[ {{H}^{+}} \right]\left[ O{{H}^{-}} \right]$
The ionic product of water is the product of concentration of $O{{H}^{-}}$ and ${{H}^{+}}$ ions.
At ${{25}^{0}}C$ the value of ionic product of water = $1.0X{{10}^{-14}}$
Due to the self-ionization of water it can act as both acid and base. When an acid or base is added to the water the ionic product of the water remains constant, but the concentration of ions changes for different types of solution.
| Type of solution | Concentration of $O{{H}^{-}}$ and ${{H}^{+}}$ ions |
| Acidic solution | $O{{H}^{-}}$ <${{H}^{+}}$ |
| Basic solution | \[~~O{{H}^{-}}>{{H}^{+}}\] |
| Neutral solution | $O{{H}^{-}}$ =${{H}^{+}}$ |
Note: The concentration of \[O{{H}^{-}}and{{H}^{+}}\] ions can be calculated using the ionic product of water.
Calculation, we know that
${{K}_{W}}=\left[ {{H}^{+}} \right]\left[ O{{H}^{-}} \right]$=$1.0X{{10}^{-14}}$
We know that pure water is neutral concentration of $O{{H}^{-}}$ =${{H}^{+}}$= X
${{K}_{W}}=\left[ x \right]\left[ x \right]={{x}^{2}}$
The value of x or the concentration of both the ions is =$1.0X{{10}^{-7}}$
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