Question

0.1 M NaOH is titrated with 0.1 M HA till the end point; ${ K }_{ a }$ for HA is 5.6 x ${ 10 }^{ -6 }$ and degree of hydrolysis is less compared to 1. Calculate pH of the resulting solution at the end point.

Answer 1

Hint:When a base and an acid are titrated such that their acidity and basicity is equal, then if their molarity is equal then their volume will also be equal.

Complete step by step answer:
Here we are titrating solutions of NaOH and HA that are of equal molarities. Therefore the volumes will also be equal since ${ M }_{ 1 }{ V }_{ 1 }={ M }_{ 2 }{ V }_{ 2 }$ will be followed.
Let the volume of NaOH solution used to titrate HA be V. Therefore V volume of 0.1 M NaOH solution will neutralise V volume of 0.1 M solution of HA in order to give a salt ${ Na }^{ + }{ A }^{ - }$ that will undergo partial hydrolysis since it is a salt of a strong base and weak acid. The concentration of the salt formed can be calculated as follows:
$\begin{matrix} NaOH(aq) \\ 0.1V\quad mole \end{matrix}+\begin{matrix} HA(aq) \\ 0.1V\quad mole \end{matrix}\longrightarrow \begin{matrix} { Na }^{ + }{ A }^{ - }(aq) \\ 0.1V\quad mole \end{matrix}+\begin{matrix} { H }_{ 2 }O(l) \\ 0.1V\quad mole \end{matrix}$
Number of moles of NaOH present in V mL of a 0.1 M solution=0.1V mole
Number of moles of HA present in V mL of a 0.1 M solution=0.1V mole
0.1V mole of NaOH will neutralise 0.1V mole of HA to give 0.1V mole of ${ Na }^{ + }{ A }^{ - }$. The concentration of the ${ Na }^{ + }{ A }^{ - }$ salt solution will be=$\cfrac { 0.1V }{ 2V } $=$\cfrac { 0.1 }{ 2 } =0.05\quad M$
(Since 2V is the total volume of the solution and 0.1V is the number of moles of the salt ${ Na }^{ + }{ A }^{ - }$)
This salt will now undergo hydrolysis:
${ Na }^{ + }{ A }^{ - }(aq)+{ H }_{ 2 }O(l)\leftrightharpoons { OH }^{ - }(aq)+{ Na }^{ + }(aq)+HA(aq)$
${ K }_{ H }=\cfrac { \left[ HA \right] \left[ { OH }^{ - } \right] }{ \left[ { A }^{ - } \right] } $
Multiplying and dividing the right hand side by $\left[ { H }^{ + } \right] $:
${ \Rightarrow K }_{ H }=\cfrac { \left[ HA \right] \left[ { OH }^{ - } \right] }{ \left[ { A }^{ - } \right] } \times \cfrac { \left[ { H }^{ + } \right] }{ \left[ { H }^{ + } \right] } $
${ \Rightarrow K }_{ H }=\cfrac { { K }_{ w } }{ { K }_{ a } } =\cfrac { { 10 }^{ -14 } }{ 5.6\times { 10 }^{ -6 } } =1.78\times { 10 }^{ -9 }$
Now let us suppose the degree of hydrolysis of the salt is α. Therefore:
$\begin{matrix} Hydrolysis\quad reaction: \\ Before\quad equilibrium: \\ After\quad equilibrium: \end{matrix}\begin{matrix} { A }^{ - }(aq) \\ c \\ c(1-\alpha ) \end{matrix}+\begin{matrix} { H }_{ 2 }O(l) \end{matrix}\leftrightharpoons \begin{matrix} HA(aq) \\ 0 \\ c\alpha \end{matrix}+\begin{matrix} { OH }^{ - }(aq) \\ 0 \\ c\alpha \end{matrix}$
Therefore, the equation for ${ K }_{ H }$ will become:
${ K }_{ H }=\cfrac { c\alpha \times c\alpha }{ c(1-\alpha ) } $
${ \Rightarrow K }_{ H }=\cfrac { c\times { \alpha }^{ 2 } }{ 1-\alpha } =c\times { \alpha }^{ 2 }$
(Since α is very small when compared to 1)
$\Rightarrow \alpha =\sqrt [ 2 ]{ \cfrac { { K }_{ H } }{ c } } $
$\Rightarrow \alpha =\sqrt [ 2 ]{ \cfrac { (1.78\times { 10 }^{ -9 }) }{ 0.05 } } =1.887\times { 10 }^{ -4 }$
Therefore the concentration of $\left[ { OH }^{ - } \right]$ will be:
$\left[ { OH }^{ - } \right] =c\times \alpha =9.433\times { 10 }^{ -6 }M$
So, $pOH=-\log { (9.433\times { 10 }^{ -6 })=5.025 } $
Since $pH+pOH=14\quad at\quad 25^{ o }{ C }$ therefore,
$pH=14-5.025=8.97\approx 9.00$
Therefore the pH of the solution is 9.0.

Note:
The degree of hydrolysis can only be ignored with respect to 1 when its value is less than 5% i.e. it is less than 0.05. If it is more than 5%, then it cannot be ignored with respect to 1 and therefore the equation will become a quadratic equation and must be solved in order to get the solution.