Answer
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Hint: The pH of a substance is defined as the negative logarithm to the base 10 of the hydrogen ion concentration of a solution. We shall calculate the concentration of hydrogen ions in each of the options and calculate their pH.
Formula Used:
${\text{pH}} = - {\log _{10}}\left[ {{{\text{H}}^ + }} \right]$
Complete step by step answer:
From the formula of pH we have, pH = $ - {\log _{10}}\left[ {{{\text{H}}^ + }} \right]$. So we need to find out the hydrogen ion concentrations of the given solutions.
For the first solution, the molarity of the solution is $0.1$ and as hydrogen chloride dissociates completely in water, so the hydrogen ion concentration is also $0.1$ .
So, the pH of the solution = $ - {\log _{10}}\left[ {0.1} \right]$=$ - {\log _{10}}\left[ {{{10}^{ - 1}}} \right]$= 1.
Coming to the second solution, the molarity of the solution is $0.01$.
So, the pH of the solution = $ - {\log _{10}}\left[ {0.01} \right]$=$ - {\log _{10}}\left[ {{{10}^{ - 2}}} \right]$=2.
The third solution is a basic solution, which instead of releasing hydrogen ions in the medium, releases hydroxyl ions. So for the basic solutions, we calculate the pOH of the solution and the relation between both is that ${\text{pH + pOH}} = 14$.
So the hydroxyl ion concentration of $\dfrac{M}{{10}}$ $NaOH$ solution = $0.1$
pOH of the solution = $ - {\log _{10}}\left[ {0.1} \right]$= $ - {\log _{10}}\left[ {{{10}^{ - 1}}} \right]$= 1.
Hence the pH of the solution = \[14 - 1 = 13\].
For the last solution, which has a concentration of$\dfrac{M}{{100}}$ , the molarity of the solution is $0.01$.
So, the pOH of the solution = $ - {\log _{10}}\left[ {0.01} \right]$=$ - {\log _{10}}\left[ {{{10}^{ - 2}}} \right]$= 2
Thus, the pH of the solution = \[14 - 2 = 12\].
So, the highest pH is that of $\dfrac{M}{{10}}$$NaOH$.
Hence, the correct option is C.
Note:
The pH and pOH solutions are very important factors. These factors are dependent on the temperature of the medium. The range of pH and pOH is from 0-14. Where in the pH scale, 0 is the value for the most acidic solution and 14 is the pH of the most basic solution, whereas this is just the reverse in case of the pOH scale.
Formula Used:
${\text{pH}} = - {\log _{10}}\left[ {{{\text{H}}^ + }} \right]$
Complete step by step answer:
From the formula of pH we have, pH = $ - {\log _{10}}\left[ {{{\text{H}}^ + }} \right]$. So we need to find out the hydrogen ion concentrations of the given solutions.
For the first solution, the molarity of the solution is $0.1$ and as hydrogen chloride dissociates completely in water, so the hydrogen ion concentration is also $0.1$ .
So, the pH of the solution = $ - {\log _{10}}\left[ {0.1} \right]$=$ - {\log _{10}}\left[ {{{10}^{ - 1}}} \right]$= 1.
Coming to the second solution, the molarity of the solution is $0.01$.
So, the pH of the solution = $ - {\log _{10}}\left[ {0.01} \right]$=$ - {\log _{10}}\left[ {{{10}^{ - 2}}} \right]$=2.
The third solution is a basic solution, which instead of releasing hydrogen ions in the medium, releases hydroxyl ions. So for the basic solutions, we calculate the pOH of the solution and the relation between both is that ${\text{pH + pOH}} = 14$.
So the hydroxyl ion concentration of $\dfrac{M}{{10}}$ $NaOH$ solution = $0.1$
pOH of the solution = $ - {\log _{10}}\left[ {0.1} \right]$= $ - {\log _{10}}\left[ {{{10}^{ - 1}}} \right]$= 1.
Hence the pH of the solution = \[14 - 1 = 13\].
For the last solution, which has a concentration of$\dfrac{M}{{100}}$ , the molarity of the solution is $0.01$.
So, the pOH of the solution = $ - {\log _{10}}\left[ {0.01} \right]$=$ - {\log _{10}}\left[ {{{10}^{ - 2}}} \right]$= 2
Thus, the pH of the solution = \[14 - 2 = 12\].
So, the highest pH is that of $\dfrac{M}{{10}}$$NaOH$.
Hence, the correct option is C.
Note:
The pH and pOH solutions are very important factors. These factors are dependent on the temperature of the medium. The range of pH and pOH is from 0-14. Where in the pH scale, 0 is the value for the most acidic solution and 14 is the pH of the most basic solution, whereas this is just the reverse in case of the pOH scale.
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