Question

What is the $pH\;$ at the half equivalence point when $50.0\;mL$ of $0.200M\;NaOH$ is titrated with $0.100M\;HCl$ ? The starting $pH\;$ is $\;13.3$ .

Answer 1

The half equivalence point is a point in a titration when exactly half of the given number of moles of base will be neutralized by acid. Hence, from the given data we can find out the half of the given moles of the given base, and as the reaction ratio is $1:1\;$ , the same number of moles of acid will be required. We can find the $pH\;$ of the neutralized solution, by finding the remaining concentration of hydroxyl ion.

Complete answer:
An equivalence point for an acid-base titration can be defined as the point at which the amount of base added to acid or acid added to base is just enough to neutralize the solution i.e. $Moles\;of\;acid=Moles\;of\;base$ .
Now, we can define the half equivalence point as the point at which exactly half of the moles of acid is neutralized by the base, and hence the resulting solution will still have moles of acid left and the solution will be acidic.
Similarly half equivalence point can be defined as the point at which exactly half of the moles of the base will be neutralized by the acid, and hence the resulting solution will still have moles of base left and the solution will be basic.
Here, we are given a solution whose initial $\;pH$ was equal to $\;13.3$ , which shows the solution was basic and thus the concentration of base is greater than acid. Hence, for half equivalence point, we need to find the concentration of acid that will neutralize of exactly half of the total base.
Now, the neutralization reaction of the given acid $HCl\;$ and given base $NaOH\;$ can be expressed as
$HC{{l}_{(aq)}}+NaO{{H}_{(aq)}}\to NaC{{l}_{(aq)}}+{{H}_{2}}{{O}_{(l)}}$
We know that, molar concentration is defined as the ratio of moles of solution to the volume of solution which is expressed as
$M=\dfrac{moles(mol)}{volume(L)}$
Now, from the given data for base, we can find the moles of base present in the solution,
$\therefore 0.200M=\dfrac{moles(mol)}{50\times {{10}^{-3}}}$
$\therefore moles=0.0100mol$
This is the total amount of base present in the solution, but we need to neutralize only half for half equivalence point.
$\therefore \dfrac{1}{2}\times moles=0.0050mol$
Now, from the neutralization reaction, we can understand that the reaction ratio in moles is $1:1\;$ . Hence, moles of acid required to neutralize the half of the moles of the base is same.
$\therefore Moles\;of\;HCl=0.0050mol$
Now, from the given concentration of the acid, we can find the volume of the acid as
$M=\dfrac{moles(mol)}{volume(L)}$
Substituting the values for acid,
$\therefore 0.100M=\dfrac{0.0050mol}{volume(L)}$
$\therefore Volume=0.05L=50mL$
Hence, the total volume of solution will be equal to
$\therefore Total\;Volume=50mL+50mL=100mL$
Now, after the neutralization, half of the total moles of base will be left in the solution.
Hence, after neutralization,
The moles of Hydroxyl ion will be $moles=0.0050mol$
The volume of the solution will be $Volume=100mL=0.1L$
Hence, the molar concentration of the Hydroxyl ion is calculated as
$\left[ O{{H}^{-}} \right]=\dfrac{0.0050moles}{0.1}$
$\therefore \left[ O{{H}^{-}} \right]=0.0500M$
Now, from the concentration of Hydroxyl ion, the $pOH\;$ will be
$pOH=-{{\log }_{10}}\left[ O{{H}^{-}} \right]$
Substituting the value of concentration,
$\therefore pOH=-{{\log }_{10}}(0.0500)$
$\therefore pOH=1.3$
Hence, we can write the $pH\;$ from the relation,
$pH+pOH=14$
Substituting the obtained value,
$\therefore pH=12.70$

Note:
We must remember here that the half equivalence point is not always obtained by taking moles of acid equal to the half of the moles of the base. This depends solely on the $pH\;$ of the given solution. As here $pH>7$ , the solution was basic, and half equivalence is to be taken for base. If $pH<7$ , the solution is acidic, and half equivalence is to be taken for acid. If the half equivalence is to be found for a weak acid then, we can take $p{{K}_{a}}=pH$ at half equivalence point, and for weak base $p{{K}_{b}}=pOH$