The ionic product of water at $ 25^\circ C $ is $ {10^{ - 14}} $ , the ionic product at $ 90^\circ C $ will be
(a) $ 1 \times {10^{ - 14}} $
(b) $ 1 \times {10^{ - 12}} $
(c) $ 1 \times {10^{ - 20}} $
(d) $ 1 \times {10^{ - 16}} $
Answer
637.2k+ views
Hint: Ionic product of water is the multiplication of the concentrations of the hydrogen ion and the hydroxide ion at a constant temperature and pressure and the ionic product is constant when the temperature and pressure are constant.
Complete step-by-step answer
We start with the definition of the ionic product of the water:
Ionic product of water: The ionic product of water is the product of the concentration of the hydrogen ions and hydroxide ions present in the pure water in aqueous solution at constant temperature is called ionic product of water, which is a constant and equal to $ 1 \times {10^{ - 14}} $ .
As we know, the increase in temperature increases the ionic product of water. So, as the temperature increases from $ 25^\circ C $ to $ 90^\circ C $ the ionic product will increase.
Now, moving towards the option,
In the first option, with increases in temperature the ionic product is constant. Hence, the option is incorrect.
Similarly, in the second option, with increases in temperature the ionic product is increased. Hence, the option is correct.
Similarly, in the third option, with increases in temperature the ionic product is decreased. Hence, the option is incorrect.
Similarly, in the last option, with increases in temperature the ionic product is decreased. Hence, the option is incorrect.
So, from the above discussion,
The ionic product of water at $ 25^\circ C $ is $ {10^{ - 14}} $ , the ionic product at $ 90^\circ C $ will be $ 1 \times {10^{ - 12}} $
Hence, the correct option is (b) $ 1 \times {10^{ - 12}} $ .
Note
With increase in the temperature, the ionic product of the water increases because the dissociation of water increases. According to Le-Chatlier’s Principle, when there is an increase in any temperature, equilibrium shifts towards the reaction which is endothermic in nature and as dissociation of water is endothermic in nature, the reaction will forward and increase the concentration of ions.
Complete step-by-step answer
We start with the definition of the ionic product of the water:
Ionic product of water: The ionic product of water is the product of the concentration of the hydrogen ions and hydroxide ions present in the pure water in aqueous solution at constant temperature is called ionic product of water, which is a constant and equal to $ 1 \times {10^{ - 14}} $ .
As we know, the increase in temperature increases the ionic product of water. So, as the temperature increases from $ 25^\circ C $ to $ 90^\circ C $ the ionic product will increase.
Now, moving towards the option,
In the first option, with increases in temperature the ionic product is constant. Hence, the option is incorrect.
Similarly, in the second option, with increases in temperature the ionic product is increased. Hence, the option is correct.
Similarly, in the third option, with increases in temperature the ionic product is decreased. Hence, the option is incorrect.
Similarly, in the last option, with increases in temperature the ionic product is decreased. Hence, the option is incorrect.
So, from the above discussion,
The ionic product of water at $ 25^\circ C $ is $ {10^{ - 14}} $ , the ionic product at $ 90^\circ C $ will be $ 1 \times {10^{ - 12}} $
Hence, the correct option is (b) $ 1 \times {10^{ - 12}} $ .
Note
With increase in the temperature, the ionic product of the water increases because the dissociation of water increases. According to Le-Chatlier’s Principle, when there is an increase in any temperature, equilibrium shifts towards the reaction which is endothermic in nature and as dissociation of water is endothermic in nature, the reaction will forward and increase the concentration of ions.
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