Question

100 ml of 0.20 M weak acid HA is completely neutralized by 0.20M NaOH ${{K}_{b}}\text{ }for\text{ }{{A}^{-}}$ is ${{10}^{-}}^{5}$, then the pH at the equivalence point is?
a.) 5
b.) 8
c.) 9
d.) 11

Answer 1

Hint: You should know that if you are going to neutralize the weak acid completely, the solution will be left with its conjugate base, which is why you can expect the pH to be greater than 7. Now try to find the pH of the solution.

Complete step by step answer:
In the question it is given that a weak acid HA is completely neutralised by sodium hydroxide which is a strong base.
For such neutralisation i.e. weak acid + strong base, the pH at the equivalence point is slightly towards the basic side.
Here, acid is HA and base is NaOH. We can write the reaction as-
     \[HA+NaOH\rightleftharpoons NaA+{{H}_{3}}{{O}^{+}}\]
Now, pH of a weak acid + strong base solution is given as-
     \[\begin{align}
  & pH=\frac{1}{2}p{{K}_{w}}+\dfrac{1}{2}p{{K}_{a}}+\frac{1}{2}\log C \\
 & or,pH=7+\left\{ \dfrac{1}{2}\times \left( -\log {{K}_{a}} \right) \right\}+\dfrac{1}{2}\log C \\
\end{align}\]

For the above reaction, C is the concentration of NaA.
The concentration of the sodium salt will be $\dfrac{0.2}{2}$ as we have equal concentrations of the acid and the base.
Therefore, C = 0.1M.
Also, in the question ${{K}_{b}}$ is given to us. So we can convert it to ${{K}_{a}}$ using the relation-
     \[{{K}_{b}}=\dfrac{{{K}_{w}}}{{{K}_{a}}}\]

Now, we know that the value of ${{K}_{w}}$ is ${{10}^{-14}}$ as, $p{{K}_{w}}=-\log {{K}_{w}}=-\log {{10}^{-14}}$
Therefore, we can write ${{K}_{a}}$ as-
     \[{{K}_{a}}=\dfrac{{{K}_{w}}}{{{K}_{b}}}\]

Now, putting the values we will get that-
     \[{{K}_{a}}=\dfrac{{{10}^{-14}}}{{{10}^{-5}}}={{10}^{-9}}\]

Now, we can put these values in the equation of pH-
     \[\begin{align}
 & pH=7+\left\{ \dfrac{1}{2}\times \left( -\log {{10}^{-9}} \right) \right\}+\dfrac{1}{2}\log 0.1 \\
 & or, pH = 7 + 4.5 - 0.5 = 11 \\
\end{align}\]

We can see from the above calculations that the pH of the solution is 11 equivalence point.
Therefore, the correct answer is 11.
So, the correct answer is “Option D”.

Note: For a strong acid and a strong base titration, equal moles of both the ions are present at the equivalence point. The equivalence point lies at a neutral point for this titration.
The equivalence point for a strong acid + weak base is at a lower pH and the solution is acidic
You should know that the higher the $[{ H }^{ + }]$ concentration, the lower pH value and the higher the concentration of $[{O H }^{ - }]$, the higher the pH value. So remember this logic so that you can easily solve any given question.