Question

Calculate the $pH$ of $0.05M\,NaOH\,$solution.
$NaOH \to N{a^ + } + O{H^ - }$

Answer 1

Hint: The $pH$ of a solution is the concentration of ${H^ + }$ ion in the solution.
Since, The above reaction Is of a base
We will need to find out the value of ${H^ + }$ ion concentration using $O{H^ - }$ ion concentration
The Concentration of $O{H^ - }$ ion will the same as the concentration of $NaOH$ which is given as
$0.05M\,NaOH\,$.

Formulae used:
1. $pOH = - \log [O{H^ - }]$
2. $pH = 14 - pOH$

Complete step by step answer:
The $pH$ of a solution refers to the strength of that solution as an acid or a base. The $pH$ scale ranges from $1 - 14$ and hence if a solution has its $pH$ $ < 7$ then the solution is said to be acidic and if the $pH$ is $> 7$ the solution is said to be basic.
Since the reaction given to us in the question is the dissociation of a base, we know that the value $pH$ has to be $> 7$.
The dissociation of Sodium Hydroxide proceeds in such a way:
$NaOH \to N{a^ + } + O{H^ - }$
It is given to us that the concentration of the base is $0.05M\,NaOH\,$ and hence:
$NaOH\,\,\,\, \to \,\,\,N{a^ + }\,\,\,\,\,\,\,\, + \,\,\,\,\,\,\,\,\,\,O{H^ - } \\
 (0.05M)\,\,\,\,\,(0.05M)\,\,(0.05M) \\ $
The concentration of $O{H^ - }$ ions will be the same as the concentration of Sodium Hydroxide.
The formula to calculate $pOH$ is
 $pOH = - \log [O{H^ - }]$
Since, we do not know the value of ${H^ + }$ ion concentration, we will use the value of $O{H^ - }$ concentration since base dissociates to give $O{H^ - }$ and not ${H^ + }$.
we know that $pOH = - \log [O{H^ - }]$
Hence, From the question we know:
$[O{H^ - }] = 0.05M$
Substituting the value in the equation, we get
$pH = - \log [0.05M]$
Solving this equation, we get:
$pOH = 1.3$
Since we know the value of $pOH$, we use it to find the value of $pH$ using the equation:
$pH = 14 - pOH$
Substituting the value of $pOH$ we get,
$pH = 14 - 1.3$
and solving it, we get
$pH = 12.7$

Note: The strength of a base depends on its $O{H^ - }$ ion concentration. If the Solution of sodium Hydroxide had been more stronger than the value of $pH$ would have been greater.
The $pH$ of two solutions of the same molarity will be different depending on their value of acidity.
Acidity refers to the number of replaceable $O{H^ - }$ ions.
For Calcium Hydroxide the acidity is $2$ and for Sodium Hydroxide it is $1$.