
Which of the following is a true statement?
A. The ionization constant and ionic product of water are the same.
B. Water is a strong electrolyte.
C. The value of the ionic product of water is less than that of ionization constant.
D. At 298 K the number of H+ ions in a liter of water is $6.023\times {{10}^{16}}$
Answer
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Hint: Ionization constant is the ratio of the concentration of the products to the concentration of the reactants. But in the case of ionic products of water, it is the product of the concentration of the products.
Complete step by step solution:
- In the question it is given that we have to choose which statement is correct among the following options.
- Coming to option A, The ionization constant and ionic product of water are the same. This statement is wrong because they are not the same. The formulas for both are as follows.
Ionization constant of water = ${{K}_{a({{H}_{2}}O)}}=\dfrac{[{{H}^{+}}][O{{H}^{-}}]}{[{{H}_{2}}O]}\to (1)$
Ionic product of water = ${{K}_{w}}=[{{H}^{+}}][O{{H}^{-}}]\to (2)$
- Coming to option B, Water is a strong electrolyte. It is also wrong because pure water won't act as an electrolyte. Pure water is a nonelectrolyte.
- Coming to option C, The value of the ionic product of water is less than that of ionization constant. It is also wrong because from the above formulas (1) and (2) we can say that ionization constant is a ratio, then the ionic product of water is greater than ionization constant.
\[{{K}_{w}}>{{K}_{a}}\]
- Coming to option D, At 298 K the number of H+ ions in a liter of water is $6.023\times {{10}^{16}}$ .
- At 298 K temperature pH of the water is 7, then the concentration of the water is $[{{H}^{+}}]={{10}^{-7}}mol/lit$ .
- Then 1 mole contains $6.023\times {{10}^{23}}ions$ .
- Therefore the concentration of the hydrogen ion is as follows.
\[\begin{align}
& ={{10}^{-7}}\times 6.023\times {{10}^{23}} \\
& =6.023\times {{10}^{16}} \\
\end{align}\]
- Therefore option D is correct.
Note: One mole of all the compounds contains Avogadro number of atoms or ions in it. The Avogadro number is going to be denoted with a symbol ${{N}_{A}}$. The value of Avogadro number is $\text{6}\text{.023 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{23}}}$ atoms or ions.
Complete step by step solution:
- In the question it is given that we have to choose which statement is correct among the following options.
- Coming to option A, The ionization constant and ionic product of water are the same. This statement is wrong because they are not the same. The formulas for both are as follows.
Ionization constant of water = ${{K}_{a({{H}_{2}}O)}}=\dfrac{[{{H}^{+}}][O{{H}^{-}}]}{[{{H}_{2}}O]}\to (1)$
Ionic product of water = ${{K}_{w}}=[{{H}^{+}}][O{{H}^{-}}]\to (2)$
- Coming to option B, Water is a strong electrolyte. It is also wrong because pure water won't act as an electrolyte. Pure water is a nonelectrolyte.
- Coming to option C, The value of the ionic product of water is less than that of ionization constant. It is also wrong because from the above formulas (1) and (2) we can say that ionization constant is a ratio, then the ionic product of water is greater than ionization constant.
\[{{K}_{w}}>{{K}_{a}}\]
- Coming to option D, At 298 K the number of H+ ions in a liter of water is $6.023\times {{10}^{16}}$ .
- At 298 K temperature pH of the water is 7, then the concentration of the water is $[{{H}^{+}}]={{10}^{-7}}mol/lit$ .
- Then 1 mole contains $6.023\times {{10}^{23}}ions$ .
- Therefore the concentration of the hydrogen ion is as follows.
\[\begin{align}
& ={{10}^{-7}}\times 6.023\times {{10}^{23}} \\
& =6.023\times {{10}^{16}} \\
\end{align}\]
- Therefore option D is correct.
Note: One mole of all the compounds contains Avogadro number of atoms or ions in it. The Avogadro number is going to be denoted with a symbol ${{N}_{A}}$. The value of Avogadro number is $\text{6}\text{.023 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{23}}}$ atoms or ions.
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