Answer
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Hint: pH of a solution is generally defined as the negative logarithm (base 10) of the Hydrogen concentration and is generally used to conveniently express relative acidity or alkalinity of a solution. pH units decrease with increasing acidity by a factor of 10.
Complete step by step answer:
The solution given to us is sodium Hydroxide $(NaOH)$ which is a strong base and thus, would release Hydroxide$\left( O{{H}^{-}} \right)$ ions when dissolved in water.
To find the pH of $NaOH$ solution, we first need to the concentration of the solution which here given to us is, Normality$=0.1$
That means there are $0.1mol/l$ of $O{{H}^{-}}$ ions.
To calculate the $pOH$, we take log of $0.1$
$pOH=-\log \left[ O{{H}^{-}} \right]$
$pOH=-\log \left[ 0.1 \right]=1$
We also know that,
$pH+pOH=14$
So, we get $pH=13$
Hence, $pH$ of $NaOH$ solution \[=13\]
So, the correct answer is “Option D”.
Note: The $N/10$, here, refers to the normality of the $NaOH$ solution. Normality is, as already mentioned, is the concentration of the solution and may also equal to the number of moles per liter of the solution, thus, it can also be written as $0.1N$ or $0.1mol/l$. After seeing the above solution we came to conclude that option A, B and C are incorrect.
Complete step by step answer:
The solution given to us is sodium Hydroxide $(NaOH)$ which is a strong base and thus, would release Hydroxide$\left( O{{H}^{-}} \right)$ ions when dissolved in water.
To find the pH of $NaOH$ solution, we first need to the concentration of the solution which here given to us is, Normality$=0.1$
That means there are $0.1mol/l$ of $O{{H}^{-}}$ ions.
To calculate the $pOH$, we take log of $0.1$
$pOH=-\log \left[ O{{H}^{-}} \right]$
$pOH=-\log \left[ 0.1 \right]=1$
We also know that,
$pH+pOH=14$
So, we get $pH=13$
Hence, $pH$ of $NaOH$ solution \[=13\]
So, the correct answer is “Option D”.
Note: The $N/10$, here, refers to the normality of the $NaOH$ solution. Normality is, as already mentioned, is the concentration of the solution and may also equal to the number of moles per liter of the solution, thus, it can also be written as $0.1N$ or $0.1mol/l$. After seeing the above solution we came to conclude that option A, B and C are incorrect.
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