What is the pH of a 1.4⋅10−2M NaOH solution?
1 Answer
Explanation:
Even before doing any calculations, you can say that since you're dealing with a ** strong base, the pH of the solution must be higher** than
The higher the concentration of the base, the higher the pH will be.
In your case, you're dealing with a solution of sodium hydroxide,
NaOH(aq]→Na+(aq]+OH−(aq]
)
Notice that the salt dissociates in a
[OH−]=[NaOH]=1.4⋅10−2M
Now, the pH of the solution is determined by the concentration of hydronium ions,
KW=[OH−]⋅[H3O+]
At room temperature, you have
KW=10−14
This means that the concentration of hydronium ions can be determined by using
[H3O+]=KW[OH−]
Plug in your values to get
[H3O+]=10−141.4⋅10−2=7.14⋅10−13M
The pH of the solution is equal to
pH=−log([H3O+])
In your case,
pH=−log(7.14⋅10−13)=12.15
As predicted, the pH is not only higher than
Alternatively, you can use the pOH of the solution to find its pH. As you know,
pOH=−log([OH−])
In your case,
pOH=−log(1.4⋅10−2)=1.85
You know that
pH+pOH=14
and so, once again, you have
pH=14−1.85=12.15
