Question

The $$pH$$of a $$0.001M$$$$NaOH$$will be:
A. 3
B. 2
C. 11
D. 12

Answer 1

Hint: Consider the solution is at ideal room temperature. Use the formula $$pH+pOH=14$$. You can express the $$pH$$of the solution in terms of its $$pOH$$. This simplifies the equation. Since, $$NaOH$$is a strong base, it completely dissociates into its ions in 1:1 ratio. This gives us the relation $$[O{H}^{-}]=[NaOH]=0.001M$$.

Complete step by step answer:
A $$0.001M$$$$NaOH$$at unknown temperature.
Since, the temperature is not given in the question, we will consider the ideal room temperature, that is 298K. At 298K, an aqueous solution has
$$pH+pOH=14$$ (1)
This means that we can express the $$pH$$of the solution in terms of its $$pOH$$.
$$pOH$$-of a solution at 298K can be defined as:
$$pOH=-\log ([O{H}^{-}])$$ (2)
From equations (1) and (2), we get
$pH=14-pOH$
$pH=14--\log [O{H}^{-}]$
$pH=14+\log O{H}^{-}$
So instead of calculating the $$pH$$of the solution by using the concentration of hydronium cations, $$H_3{O}^{+}$$
$$pH=-\log ([H_3{O}^{+}])$$
We can do so indirectly by using the concentration of hydroxide anions.
Now, sodium hydroxide is a strong base, which means that it dissociates completely in aqueous solution to produce sodium cations, which are of no interest to us here, and hydroxide anions.
Both ions are produced in 1:1 mole ratio with the strong base, so we can say that $$[O{H}^{-}]=[NaOH]=0.001M$$
Now that you know the concentration of hydroxide anions in this solution, we can say that its$$pH$$is equal to
$$pH=14+\log (0.001)=11.0$$

Hence the correct answer is C.

Note:
When calculating the $$pH$$of an aqueous solution, always take the temperature into consideration. $$pH$$ of a solution is temperature dependent and can vary with increase or decrease in the temperature. Usually, the temperature is given. If it is not, always considering the reactions at room temperature because it is more ideal and the calculations are straightforward.