Question

A weak acid HA after treatment with 12mL of 0.1M strong base has a pH of 5. At the end point, the volume of the same base required is 26.6mL. The value of ${{K}_{a}}$ is:
A) $8.2\times {{10}^{-6}}$
B) $6.4\times {{10}^{=6}}$
C) $5.3\times {{10}^{-5}}$
D) $2.4\times {{10}^{-6}}$

Answer 1

Acid dissociation constant is calculated using the formula for the calculation of pH which is given by, $pH=-\log {{K}_{a}}+\log \dfrac{\left[ salt \right]}{\left[ acid \right]}$ . Substitute the values by making use of the complete and partial neutralisation equations.

Complete answer:
In the lower classes of physical chemistry, we have come across the calculations such as pH calculation and also the related calculations of acid dissociation constant and so on.
We shall see the calculation of the acid dissociation constant in detail for the above given data.
- Let us calculate the milli equivalents of acid and base required for the complete as well as for the half neutralization of the reaction.
- For complete neutralization, we have the milli equivalents of acid and base as the same.
Hence, this will be, total milli equivalents of acid = total milli equivalents of base = $26.6\times 0.1=2.66$
Now, for the partial neutralization, let us see the reaction first and then find the concentrations.
- For partial neutralization,
$HA+BOH\to BA+{{H}_{2}}O$

Before reaction	2.66	1.2	0	C
After reaction	1.46	0	1.2	(C+1.2)

Thus, from the above reaction, we can say that the resultant mixture has HA and BA both which is nothing but this acts as a buffer solution.
Thus, pH of the solution is calculated using the formula,
$pH=-\log {{K}_{a}}+\log \dfrac{\left[ salt \right]}{\left[ acid \right]}$
where, ${{K}_{a}}$is the acid dissociation constant which is to be found.
By substituting the values, we have
$5=-\log {{K}_{a}}+\log \dfrac{\left[ 1.2 \right]}{\left[ 1.46 \right]}=-\log {{K}_{a}}-0.085$
Thus, $\log {{K}_{a}}$ will be equal to $\log {{K}_{a}}=-5.085$
Now, taking the antilog of this value, we get the value of dissociation constant as,
${{K}_{a}}=anti\log (-5.085)=8.21\times {{10}^{-6}}$

Hence, the correct answer is option A) $8.2\times {{10}^{-6}}$

Note:
Note that the buffer of a solution is known by observing the nature of acid and base. That is a buffer solution is the aqueous solution in which there is a mixture of weak acid and conjugate base or vice-versa.