Question

What will be the pH at the equivalence point during the titration of a 1000 ml of 0.2 M solution of $C{{H}_{3}}COONa$ with 0.2 M solution of HCl?
[Given : ${{K}_{a}}=2\times {{10}^{-5}}$ ]
(A)- $3-\log \sqrt{2}$
(B)- $3+\log \sqrt{2}$
(C)- $3-\log 2$
(D)- $3+\log 2$

Answer 1

The equivalence point in a titration is a point at which the amount of titrant added is enough to completely neutralize the analyte solution. At this point, the number of moles of titrant (standard solution) is equal to the moles of the solution of unknown concentration.

Complete answer:
-The point at which moles of acid and moles of the base are equivalent is known as the equivalence point of a chemical reaction.
-We will begin the solution to this question by first writing the titration reaction of $C{{H}_{3}}COONa$and HCl-
$C{{H}_{3}}COONa+HCl\to C{{H}_{3}}COOH+NaCl$
-It is clear from the equation that, one mole of sodium acetate is required to neutralize it using HCl.
Concentration is mathematically the number of moles in a particular volume, that is
$Concentration=\dfrac{\text{Number of moles}}{Volume}$
Number of millimoles of sodium acetate $=Concentration\times Volume=0.2\times 100=20\text{ m moles}$
Number of millimoles of hydrochloric acid $=Concentration\times Volume=0.2\times 100=20\text{ m moles}$
-Then according to Law of Conservation of Mass,
Before reaction, there were 20 millimoles of sodium acetate and 20 millimoles of hydrochloric acid which after reaction produced 20 millimoles of acetic acid and 20 millimoles of sodium chloride.
-The dissociation of acetic acid produces acetate ion and a hydrogen ion.
$C{{H}_{3}}COOH\rightleftharpoons C{{H}_{3}}CO{{O}^{-}}+{{H}^{+}}$
-Finding the concentration of acetic acid,
$[C{{H}_{3}}COOH]=\dfrac{20}{100+100}=0.1M$ (Total volume = Volume of sodium acetate + Volume of HCl)
-pH of the solution is dependent on the products formed in the reaction. More the concentration of acetic acid, more will the hydrogen ions dissociated in the solution, hence more acidic will be the pH.
Since acetic acid is a weak acid, so the concentration of hydrogen ions will be $=[{{H}^{+}}]=\sqrt{{{k}_{a}}.c}$
Multiplying the above equation with -log in both the sides, we will get,
$\begin{aligned} & -\log [{{H}^{+}}]=-\dfrac{1}{2}(\log {{K}_{a}}+\log C) \\\ & \Rightarrow pH=\dfrac{1}{2}(p{{k}_{a}}-\log C) \\\ & =\dfrac{1}{2}(5-\log 2+1) \\\ & =\dfrac{1}{2}(6-\log 2) \\\ & =3-\dfrac{1}{2}\log 2 \\\ & \Rightarrow pH=3-\log \sqrt{2} \\\ \end{aligned}$

Therefore the correct answer is option A.

Note:
Equivalent point and endpoint are two confusing terms, you should not get confused with these terms. The theoretical completion of the reaction is given by equivalence point. Equivalence is the volume of added titrant at which the number of moles of titrant is equal to the number of moles of the analyte. The endpoint is actually the measured physical change in the solution. There is a slight difference between the equivalence point and the endpoint in the titration is referred to as indicator error and is indeterminate.