Question

The hydroxyl ion concentration of sodium hydroxide having a pH value of 12 at 25 $^{\circ }\text{C}$ is:
(a)0.00001
(b)0.0001
(c)0.001
(d)0.01

Answer 1

Hint: We have been given the pH of NaOH solution, from this we will first the ${{\text{p}}_{OH}}$value of NaOH by the formula: pH+${{\text{p}}_{OH}}$=14 and after by using this value we will find the concentration of the $\text{ }\!\![\!\!\text{ OH}_{aq}^{-}]$in the solution by the formula : ${{\text{p}}_{OH}}$ = -log$\text{ }\!\![\!\!\text{ OH}_{aq}^{-}]$ . Now solve it.

Complete step by step answer:
Before calculation of pH we should know what, it is: pH gives us the measure of acid/base strength of any solution. pH scale ranges from 0-14, 7 is neutral, below 7 it represents acidic and above 7 it represents basic.
Since, the solution is NaOH , which is basic in nature and thus has the value of pH more than 7 and given as12.
Now, solving the numerical:
As we know that:
pH+${{\text{p}}_{OH}}$=14 ----(1)
and pH value of NaOH solution =12(given)
 put this value in equation(1) ,then, ${{\text{p}}_{OH}}$value is as:
12+${{\text{p}}_{OH}}$=14
${{\text{p}}_{OH}}$=14-12
${{\text{p}}_{OH}}$=2
Now, from this ${{\text{p}}_{OH}}$, we can calculate the concentration of the $\text{ }\!\![\!\!\text{ OH}_{aq}^{-}]$ in the NaOH solution as:
${{\text{p}}_{OH}}$ = -log$\text{ }\!\![\!\!\text{ OH}_{aq}^{-}]$
2= -log$\text{ }\!\![\!\!\text{ OH}_{aq}^{-}]$
$\text{ }\!\![\!\!\text{ OH}_{aq}^{-}]$=-log$\dfrac{1}{2}$
$\text{ }\!\![\!\!\text{ OH}_{aq}^{-}]$= 1×${{10}^{-2}}$
Therefore, $\text{ }\!\![\!\!\text{ OH}_{aq}^{-}]$=0.01
Hence, the concentration of $\text{ }\!\![\!\!\text{ OH}_{aq}^{-}]$in NaOH solution having the pH of 12 at $^{\circ }\text{C}$ is 0.01 at mol/dm .

Hence, option(d) is correct.

Note: You can’t directly find the concentration of the $\text{ }\!\![\!\!\text{ OH}_{aq}^{-}]$, we first find the ${{\text{p}}_{OH}}$ value of OH and only ,then we can be able to find the OH concentration and 14 value is taken be the pH scale has total value up to 14 ranging from 0-14.