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Hint: pH is a measure of the acidity and the basicity of the solution. Lower the value of the pH, the more acidic the solution is and higher the value of the pH, the more basic the solution is. Mathematically, it is given as $pH = -log[{H}^{+}]$.
Complete step by step answer: pH of the solution can be increased by either adding a base to the solution or by adding a neutral substance of solution to the given solution.
Now, we know that the pH is related to the concentration of the ${H}^{+}$ ions in the solution and the relation is given as follows:
$pH = -log[{H}^{+}]$
Now, it is given that initially the pH of the HCl solution is 1.
Therefore, the concentration of the ${H}^{+}$ ions in this solution is:
$1 = -log[{H}^{+}]$
$\implies [{H}^{+}] = 0.1M$
Now, when water is added to this solution, the pH becomes 2. Therefore, the concentration of the ${H}^{+}$ ions in this final solution is:
$2 = -log[{H}^{+}]$
$\implies [{H}^{+}] = 0.01M$
Now, from the dilution formula, we know that the no. of moles of HCl in a given solution will be equal always.
We know that, $concentration,\quad M=\cfrac { No.\quad of\quad moles,\quad n }{ Volume,\quad V } $
$\implies n = MV$
Now, as the no. of moles is constant, therefore, ${M}_{1}{V}_{1} = {M}_{2}{V}_{2}$.
where, ${M}_{1} d and {V}_{1}$ are the concentration of the initial solution and ${M}_{2} and {V}_{2}$ are the concentration of the final solution.
$\implies { V }_{ 2 } = \cfrac { { M }_{ 1 }{ V }_{ 1 } }{ { M }_{ 2 } } $
Substituting the values in the above equation, we get
${ V }_{ 2 } = \cfrac { 0.1 \times 1 }{ 0.01 }$
$\implies { V }_{ 2 } =\quad 10L$
Therefore, the total of the solution is 10L. Thus, the amount of water added in the solution is total volume of the solution - the volume of HCl present in the solution
Therefore, Volume of water added = 10 - 1 = 9L.
Hence, the correct answer is option (B).
Note: In these types of questions, we need to keep in mind that the number of moles of any substance in a solution always remains constant unless and until some amount of the solution is taken out or more of that substance is added to the solution.
Complete step by step answer: pH of the solution can be increased by either adding a base to the solution or by adding a neutral substance of solution to the given solution.
Now, we know that the pH is related to the concentration of the ${H}^{+}$ ions in the solution and the relation is given as follows:
$pH = -log[{H}^{+}]$
Now, it is given that initially the pH of the HCl solution is 1.
Therefore, the concentration of the ${H}^{+}$ ions in this solution is:
$1 = -log[{H}^{+}]$
$\implies [{H}^{+}] = 0.1M$
Now, when water is added to this solution, the pH becomes 2. Therefore, the concentration of the ${H}^{+}$ ions in this final solution is:
$2 = -log[{H}^{+}]$
$\implies [{H}^{+}] = 0.01M$
Now, from the dilution formula, we know that the no. of moles of HCl in a given solution will be equal always.
We know that, $concentration,\quad M=\cfrac { No.\quad of\quad moles,\quad n }{ Volume,\quad V } $
$\implies n = MV$
Now, as the no. of moles is constant, therefore, ${M}_{1}{V}_{1} = {M}_{2}{V}_{2}$.
where, ${M}_{1} d and {V}_{1}$ are the concentration of the initial solution and ${M}_{2} and {V}_{2}$ are the concentration of the final solution.
$\implies { V }_{ 2 } = \cfrac { { M }_{ 1 }{ V }_{ 1 } }{ { M }_{ 2 } } $
Substituting the values in the above equation, we get
${ V }_{ 2 } = \cfrac { 0.1 \times 1 }{ 0.01 }$
$\implies { V }_{ 2 } =\quad 10L$
Therefore, the total of the solution is 10L. Thus, the amount of water added in the solution is total volume of the solution - the volume of HCl present in the solution
Therefore, Volume of water added = 10 - 1 = 9L.
Hence, the correct answer is option (B).
Note: In these types of questions, we need to keep in mind that the number of moles of any substance in a solution always remains constant unless and until some amount of the solution is taken out or more of that substance is added to the solution.
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