Question

How do you determine the OH (aq) concentration in 1.0 M aniline ( ${{C}_{6}}{{H}_{5}}N{{H}_{2}}$) solution. (the ${{K}_{b}}$for aniline $4.0\times {{10}^{-10}}$)?

Answer 1

The pOH is the potential of the hydroxyl ion or the power of hydroxyl ion. It helps in determining the strength of the basic solution. It helps in determining the alkalinity of the solution. The solution which has the pOH less than seven are considered to be basic and the pOH whose value is more than 7 are acidic in solution.

Complete step-by-step answer: To approach this question we will use the following formula as the base involved is a weak base that is aniline. The formula is the following:
$pOH=\dfrac{1}{2}[p{{K}_{b}}-\log b]$
Here pOH is the potential of hydroxyl ion
The b is the concentration of the base.
But first we need to calculate the $p{{K}_{b}}$ using the following formula:
$p{K_b}=-log{K_b}$
Here ${{K}_{b}}$is the $4.0\times {{10}^{-10}}$
So on substituting the value in the formula we get,
$p{{K}_{b}}=-\log (4.0\times {{10}^{-10}})$
On simplifying we get,
$p{{K}_{b}}=9.34$
Now substituting the values in the above formula we get,
$pOH=\dfrac{1}{2}[9.34-\log (1)]$
On simplifying we get,
pOH=4.67
then we will use the other formula for calculating the concentration of the hydroxyl ion. the formula is the following:
$pOH=-\log (O{{H}^{-}})$
On substituting the values we get,
$4.67=-\log (O{{H}^{-}})$
Taking antilog we get,
$[O{{H}^{-}}]=2.14\times {{10}^{-5}}mol{{l}^{-1}}$
So the concentration of hydroxyl ion in aniline is $[O{{H}^{-}}]=2.14\times {{10}^{-5}}mol{{l}^{-1}}$ $2.14\times {{10}^{-5}}mol{{l}^{-1}}$.

Note: For determining the hydrogen ion concentration in the solution the pH is used. The pH is defined as the power of hydrogen or the potential of hydrogen. The pH scale ranges from 0 to 14. The pH for acidic solution is less than 7 and basic solution pH is more than 7 and for neutral solution the pH is always equal to 7.