
Equal volumes of $ 0.1M $ $ AgN{O_3} $ and $ 0.2M $ $ NaCl $ are mixed. The concentration of $ NO_3^ - $ ion in the mixture will be:
(A) $ 0.1M $
(B) $ 0.05M $
(C) $ 0.2M $
(D) $ 0.15M $
Answer
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Hint: Molarity is the concentration term which is calculated in terms of the amount of solute (number of moles of solute) that are present in one litre of the solution. So, when the above two given compounds are mixed together, we will see their dissociation into ions and hence, it will help us to determine the concentration of nitrate ions in the mixture.
Complete answer:
We know that we calculate concentration of any ion by the term molarity. Molarity is calculated in terms of the amount of solute (number of moles of solute) that are present in one litre of the solution.
When a compound or electrolyte dissociates into solution, it produces cations and anions. The minimum concentration of these ions will be equal to the concentration of electrolyte.
So when $ 0.1M $ $ AgN{O_3} $ dissociates, it produces $ 0.1M $ $ NO_3^ - $ ions and $ NO_3^ - $ does not produce any nitrate ions.
Let the volumes of both the solutions be $ 1L. $ Now, $ 0.1mole $ $ AgN{O_3} $ dissociates, it produces $ 0.1mole $ $ NO_3^ - $ ions.
Therefore, moles of nitrate ions in the solution is $ 0.1mole. $ And the total volume of the mixture (solution) is $ 2L. $
Hence, the concentration of nitrate ions in the mixture is calculated as:
$ M = \dfrac{n}{V} $
Where, n is the number of moles of nitrate ions in the solution and V is the total volume of the mixture.
$ M = \dfrac{{0.1}}{2} $
$ Molarity = 0.05M $
Hence, the concentration of $ NO_3^ - $ ions in the mixture is $ 0.05M. $
Therefore, the correct option is (B) $ 0.05M $ .
Note:
If the electrolyte is completely ionised in solution, then we can easily find the concentration of its ions present in that solution. This is because the minimum concentration of these ions will be equal to the concentration of electrolyte. But when the electrolyte is partially ionised, then we cannot directly find out the concentration of its ions in the solution.
Complete answer:
We know that we calculate concentration of any ion by the term molarity. Molarity is calculated in terms of the amount of solute (number of moles of solute) that are present in one litre of the solution.
When a compound or electrolyte dissociates into solution, it produces cations and anions. The minimum concentration of these ions will be equal to the concentration of electrolyte.
So when $ 0.1M $ $ AgN{O_3} $ dissociates, it produces $ 0.1M $ $ NO_3^ - $ ions and $ NO_3^ - $ does not produce any nitrate ions.
Let the volumes of both the solutions be $ 1L. $ Now, $ 0.1mole $ $ AgN{O_3} $ dissociates, it produces $ 0.1mole $ $ NO_3^ - $ ions.
Therefore, moles of nitrate ions in the solution is $ 0.1mole. $ And the total volume of the mixture (solution) is $ 2L. $
Hence, the concentration of nitrate ions in the mixture is calculated as:
$ M = \dfrac{n}{V} $
Where, n is the number of moles of nitrate ions in the solution and V is the total volume of the mixture.
$ M = \dfrac{{0.1}}{2} $
$ Molarity = 0.05M $
Hence, the concentration of $ NO_3^ - $ ions in the mixture is $ 0.05M. $
Therefore, the correct option is (B) $ 0.05M $ .
Note:
If the electrolyte is completely ionised in solution, then we can easily find the concentration of its ions present in that solution. This is because the minimum concentration of these ions will be equal to the concentration of electrolyte. But when the electrolyte is partially ionised, then we cannot directly find out the concentration of its ions in the solution.
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