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

Calculate the pHpH of 0.05M NaOH 0.05M\,NaOH\,solution.
NaOH→Na++OH−NaOH \to N{a^ + } + O{H^ - }

Explanation

Solution

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

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

Complete step by step answer:
The pHpH of a solution refers to the strength of that solution as an acid or a base. The pHpH scale ranges from 1−141 - 14 and hence if a solution has its pHpH <7 < 7 then the solution is said to be acidic and if the pHpH is >7> 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 pHpH has to be >7> 7.
The dissociation of Sodium Hydroxide proceeds in such a way:
NaOH→Na++OH−NaOH \to N{a^ + } + O{H^ - }
It is given to us that the concentration of the base is 0.05M NaOH 0.05M\,NaOH\, and hence:
NaOH    →   Na+        +          OH− (0.05M)     (0.05M)  (0.05M) NaOH\,\,\,\, \to \,\,\,N{a^ + }\,\,\,\,\,\,\,\, + \,\,\,\,\,\,\,\,\,\,O{H^ - } \\\ (0.05M)\,\,\,\,\,(0.05M)\,\,(0.05M) \\\
The concentration of OH−O{H^ - } ions will be the same as the concentration of Sodium Hydroxide.
The formula to calculate pOHpOH is
pOH=−log⁡[OH−]pOH = - \log [O{H^ - }]
Since, we do not know the value of H+{H^ + } ion concentration, we will use the value of OH−O{H^ - } concentration since base dissociates to give OH−O{H^ - } and not H+{H^ + }.
we know that pOH=−log⁡[OH−]pOH = - \log [O{H^ - }]
Hence, From the question we know:
[OH−]=0.05M[O{H^ - }] = 0.05M
Substituting the value in the equation, we get
pH=−log⁡[0.05M]pH = - \log [0.05M]
Solving this equation, we get:
pOH=1.3pOH = 1.3
Since we know the value of pOHpOH, we use it to find the value of pHpH using the equation:
pH=14−pOHpH = 14 - pOH
Substituting the value of pOHpOH we get,
pH=14−1.3pH = 14 - 1.3
and solving it, we get
pH=12.7pH = 12.7

Note: The strength of a base depends on its OH−O{H^ - } ion concentration. If the Solution of sodium Hydroxide had been more stronger than the value of pHpH would have been greater.
The pHpH of two solutions of the same molarity will be different depending on their value of acidity.
Acidity refers to the number of replaceable OH−O{H^ - } ions.
For Calcium Hydroxide the acidity is 22 and for Sodium Hydroxide it is 11.