
What will be the ${{H}^{+}}$ ion concentration, when 4 gram NaOH is dissolved in 1000 ml of water?
(a)- ${{10}^{-1}}$ en
(b)- ${{10}^{-13}}$
(c)- ${{10}^{-4}}$
(d)- ${{10}^{-10}}$
Answer
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Hint: The given compound is sodium hydroxide (NaOH) and we know that it is a strong base because it can release hydroxyl ions very easily and dissociates fully into ions, so the solution will have a high amount of hydroxyl ions and fewer amount of hydrogen ions. First, we have to calculate the molarity of the solution and then the concentration of the hydrogen ions can be calculated.
Complete answer:
The concentration of hydrogen ion and hydroxyl ions are calculated when the compound dissociates into ions in the water. The given compound is sodium hydroxide (NaOH) and we know that it is a strong base because it can release hydroxyl ions very easily and dissociates fully into ions, so the solution will have a high amount of hydroxyl ions and less amount of hydrogen ions.
So, first, we have to calculate the molarity of the solution, this will give the concentration of the hydroxyl ions. We are given that 4 grams of NaOH are dissolved in 1000 ml water, and the molecular mass of NaOH is 40. The molarity can be calculated as:
$Molarity=\dfrac{\text{Given mass x 1000}}{\text{Molecular mass x volume of solution}}$
Putting the values, we get:
$Molarity=\dfrac{\text{4 x 1000}}{\text{40 x 1000}}=0.1M$
So, the molarity is 0.1 or ${{10}^{-1}}$ which is equal to the concentration of hydroxyl ions.
$[O{{H}^{-}}]={{10}^{-1}}M$
As we know that the product of the concentration of hydrogen ions and hydroxyl ions is equal to ${{10}^{-14}}$. So, we can calculate the value of the concentration of hydrogen ions as:
$[{{H}^{+}}][O{{H}^{-}}]={{10}^{-14}}$
$[{{H}^{+}}]=\dfrac{{{10}^{-14}}}{[O{{H}^{-}}]}$
$[{{H}^{+}}]=\dfrac{{{10}^{-14}}}{{{10}^{-1}}}={{10}^{-13}}$
So, the concentration of hydrogen ions is ${{10}^{-13}}M$
So, the correct answer is “Option (b)”.
Note:
The extent of ions tell the pH of the solution, as the above solution has a very high amount of hydroxyl ions and a very less amount of hydrogen ions, then the pH of the solution is very high.
Complete answer:
The concentration of hydrogen ion and hydroxyl ions are calculated when the compound dissociates into ions in the water. The given compound is sodium hydroxide (NaOH) and we know that it is a strong base because it can release hydroxyl ions very easily and dissociates fully into ions, so the solution will have a high amount of hydroxyl ions and less amount of hydrogen ions.
So, first, we have to calculate the molarity of the solution, this will give the concentration of the hydroxyl ions. We are given that 4 grams of NaOH are dissolved in 1000 ml water, and the molecular mass of NaOH is 40. The molarity can be calculated as:
$Molarity=\dfrac{\text{Given mass x 1000}}{\text{Molecular mass x volume of solution}}$
Putting the values, we get:
$Molarity=\dfrac{\text{4 x 1000}}{\text{40 x 1000}}=0.1M$
So, the molarity is 0.1 or ${{10}^{-1}}$ which is equal to the concentration of hydroxyl ions.
$[O{{H}^{-}}]={{10}^{-1}}M$
As we know that the product of the concentration of hydrogen ions and hydroxyl ions is equal to ${{10}^{-14}}$. So, we can calculate the value of the concentration of hydrogen ions as:
$[{{H}^{+}}][O{{H}^{-}}]={{10}^{-14}}$
$[{{H}^{+}}]=\dfrac{{{10}^{-14}}}{[O{{H}^{-}}]}$
$[{{H}^{+}}]=\dfrac{{{10}^{-14}}}{{{10}^{-1}}}={{10}^{-13}}$
So, the concentration of hydrogen ions is ${{10}^{-13}}M$
So, the correct answer is “Option (b)”.
Note:
The extent of ions tell the pH of the solution, as the above solution has a very high amount of hydroxyl ions and a very less amount of hydrogen ions, then the pH of the solution is very high.
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