
The unit of ionic product of water ${{K}_{w}}$ is:
(A) $Mo{{l}^{-1}}{{L}^{-1}}$
(B) $Mo{{l}^{2}}{{L}^{-2}}$
(C) $Mo{{l}^{-2}}{{L}^{-2}}$
(D) $Mo{{l}^{2}}{{L}^{2}}$
Answer
586.5k+ views
Hint: Water molecules act as both acids and bases. Consider two water molecules, one water molecule acting as a base can accept a hydrogen ion, from a second water molecule that acts as an acid. This hydronium ion and hydrogen ion produce water again in an equilibrium reaction which impacts the ionic product of water.
Complete step by step solution:
The ionic product of water varies with temperature and is useful to determine the pH of pure water at different temperatures.
At a specified temperature, the difference between the solubility product and ionic product of water
The equilibrium between hydronium ions and hydrogen ion present in pure water,
$2{{H}_{2}}O(l)\rightleftharpoons {{H}_{3}}{{O}^{+}}(aq)+O{{H}^{-}}(aq)$
The concentration of hydronium ion in pure water will be $1X{{10}^{-7}}M/L$ , then the above equilibrium will be changed to,
${{H}_{2}}O(l)\rightleftharpoons {{H}^{+}}(aq)+O{{H}^{-}}(aq)$
${{K}_{w}}$, is the equilibrium constant for the above reaction is known as the ionic product of water.
${{K}_{w}}=[{{H}^{+}}][O{{H}^{-}}]$
Hydrogen ion and hydroxyl ion concentrations are expressed in terms of M/L,
${{K}_{w}}=[M/L][M/L]={{M}^{2}}{{L}^{-2}}$
Hence, the units of the ionic product of water = ${{M}^{2}}{{L}^{-2}}$
The correct answer is option B.
Note: According to Le Chateleir, the equilibrium moves to lower temperature, when the temperature of water increases. This means forward reaction favored for water equilibrium and more hydrogen ions, hydroxide ions will form. Due to this effect, the ionic product of water increases with temperature.
Complete step by step solution:
The ionic product of water varies with temperature and is useful to determine the pH of pure water at different temperatures.
At a specified temperature, the difference between the solubility product and ionic product of water
| Solubility product | Ionic product |
| This is the concentration of ions of electrolytes, each raised to the power of their coefficients in the balanced chemical equation in a saturated solution. | It is the product of the concentrations of ions of the electrolyte, each raised to the power of their coefficients in the balanced chemical equation in solution at any concentration |
| Applicable only to saturated solutions | Apply to all types of solutions of any concentrations |
| This value is constant for a given electrolyte at a constant temperature. | This value changing with concentration |
The equilibrium between hydronium ions and hydrogen ion present in pure water,
$2{{H}_{2}}O(l)\rightleftharpoons {{H}_{3}}{{O}^{+}}(aq)+O{{H}^{-}}(aq)$
The concentration of hydronium ion in pure water will be $1X{{10}^{-7}}M/L$ , then the above equilibrium will be changed to,
${{H}_{2}}O(l)\rightleftharpoons {{H}^{+}}(aq)+O{{H}^{-}}(aq)$
${{K}_{w}}$, is the equilibrium constant for the above reaction is known as the ionic product of water.
${{K}_{w}}=[{{H}^{+}}][O{{H}^{-}}]$
Hydrogen ion and hydroxyl ion concentrations are expressed in terms of M/L,
${{K}_{w}}=[M/L][M/L]={{M}^{2}}{{L}^{-2}}$
Hence, the units of the ionic product of water = ${{M}^{2}}{{L}^{-2}}$
The correct answer is option B.
Note: According to Le Chateleir, the equilibrium moves to lower temperature, when the temperature of water increases. This means forward reaction favored for water equilibrium and more hydrogen ions, hydroxide ions will form. Due to this effect, the ionic product of water increases with temperature.
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