
Find the $pH$ of a $0.001M$ $NaOH$ solution will be?
Answer
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Hint: We can first write the dissociation of sodium hydroxide into its constituent ions. From this, since it is a base, we can get the concentration of the hydroxyl ions. Taking negative logarithms of this will give us the $pOH$ value. As we know, the sum of $pH$ and $pOH$ is $14$. Hence, we can subtract the $pOH$ value from $14$ to get the $pH$ value.
Formulas used: $pOH = - \log [O{H^ - }]$
Where $pOH$ is the concentration of the hydroxyl ions.
$pOH + pH = 14$
Where $pH$ and $pOH$ represents the potential of hydrogen ions and hydroxyl ions respectively.
Complete step by step answer:
Since, the concentration of sodium hydroxide is given, let us see how it dissociates, to find the concentration of hydroxyl ions.
$NaOH \rightleftharpoons N{a^ + } + O{H^ - }$
Since the stoichiometric coefficients (number of moles which react according to the balanced chemical equation) of both $NaOH$ and $O{H^ - }$ are the same, their concentrations will also be the same. Hence:
$[NaOH] = [O{H^ - }] = 0.001M$
As we know, taking the negative logarithm of the concentration of $O{H^ - }$ ion will give us the $pOH$ value of the solution:
$pOH = - \log [O{H^ - }]$
Where $pOH$ is the concentration of the hydroxyl ions
Hence, substituting the concentration value in this equation, we get:
$pOH = - \log (0.001) = - ( - 3)$
$ \Rightarrow pOH = 3$
The sum of values of $pH$ and $pOH$ is $14$. Hence:
$pOH + pH = 14 \Rightarrow pH = 14 - pOH$
Where $pH$ and $pOH$ represents the potential of hydrogen ions and hydroxyl ions respectively.
Substituting the value of $pOH = 3$ in this, we get:
$pH = 14 - 3 = 11$
Hence, the $pH$ of the given solution of sodium hydroxide is $11$
Note: $pH$ of a solution tells us how basic or acidic a particular solution is. A $pH$ of less than seven indicates an acidic solution, while a $pH$ of more than seven indicates a basic solution. Note that the $pH$ of pure water is taken as seven. $pH$ of a solution can be measured using $pH$ papers, which provide a range of colours to compare with or using digital $pH$ meters, which gives a more accurate value.
Formulas used: $pOH = - \log [O{H^ - }]$
Where $pOH$ is the concentration of the hydroxyl ions.
$pOH + pH = 14$
Where $pH$ and $pOH$ represents the potential of hydrogen ions and hydroxyl ions respectively.
Complete step by step answer:
Since, the concentration of sodium hydroxide is given, let us see how it dissociates, to find the concentration of hydroxyl ions.
$NaOH \rightleftharpoons N{a^ + } + O{H^ - }$
Since the stoichiometric coefficients (number of moles which react according to the balanced chemical equation) of both $NaOH$ and $O{H^ - }$ are the same, their concentrations will also be the same. Hence:
$[NaOH] = [O{H^ - }] = 0.001M$
As we know, taking the negative logarithm of the concentration of $O{H^ - }$ ion will give us the $pOH$ value of the solution:
$pOH = - \log [O{H^ - }]$
Where $pOH$ is the concentration of the hydroxyl ions
Hence, substituting the concentration value in this equation, we get:
$pOH = - \log (0.001) = - ( - 3)$
$ \Rightarrow pOH = 3$
The sum of values of $pH$ and $pOH$ is $14$. Hence:
$pOH + pH = 14 \Rightarrow pH = 14 - pOH$
Where $pH$ and $pOH$ represents the potential of hydrogen ions and hydroxyl ions respectively.
Substituting the value of $pOH = 3$ in this, we get:
$pH = 14 - 3 = 11$
Hence, the $pH$ of the given solution of sodium hydroxide is $11$
Note: $pH$ of a solution tells us how basic or acidic a particular solution is. A $pH$ of less than seven indicates an acidic solution, while a $pH$ of more than seven indicates a basic solution. Note that the $pH$ of pure water is taken as seven. $pH$ of a solution can be measured using $pH$ papers, which provide a range of colours to compare with or using digital $pH$ meters, which gives a more accurate value.
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