Question

The degree of dissociation of acetic acid in a 0.1 M solution is 1:32 x 10-2
Find out the dissociation constant of the acid and its pKa value.​

Answer 1

Answer:

Dissociation constant is given by the formula:

$\boxed{\bf{K_a = C \alpha ^2}}$

According to the question,

Concentration of the solution (C) = 0.1 M

Degree of Dissociation = 1.32 × 10⁻²

Hence the value of dissociation constant is:

$\implies K_a = 0.1 \times (1.32 \times 10^{-2})^2\\\\\\\implies \boxed{ \bf{K_a = 1.742 \times 10^{-5}}}$

Now we are required to find the value of pKa value.

The formula to find the pKa value from the Ka value is:

$\boxed{ \bf{pK_a = -log(K_a)}}$

Taking log of Ka value, we get:

⇒ pKa = -log ( 1.742 × 10⁻⁵ )

⇒ pKa = - [ log ( 1.742 ) + log ( 10⁻⁵ )]

⇒ pKa = - 0.241 - (-5) log 10

⇒ pKa = (- 0.241 + 5 )

⇒ pKa = 4.759

These are the required answers.

Answer 2

Answer:

Given :-

• The degree of dissociation of acetic acid in a 0.1 M solution is 1:32 × 10-2.

Find Out :-

• Find Out the dissociation constant of the acid and its pka value.

Solution :-

Given that :

$\Rightarrow$ Degree of Dissociation (D.O.D) = 1.32 × 10-²

$\Rightarrow$Concⁿ (C) = 0.1 M

Now, we know that :

$\boxed{\bold{\large{\sf{K_a =\: C\alpha^2}}}}$

$\Rightarrow$ $\tt{ K_a =\: 0.1 \times (1.32 \times 10^-2)^2}$

$\Rightarrow$ $\small{\bf{\underline{\underline{K_a =\: 1.74 \times 10^-5}}}}$

Again, we know that :

$\boxed{\bold{\large{\sf{pK_a =\: logK_a}}}}$

$\Rightarrow$ $\tt{pK_a =\: log(1.74 \times 10^-5)}$

$\Rightarrow$ $\tt{pK_a =\: 5 - log\: 1.74}$

$\Rightarrow$ $\tt{pK_a =\: 5 - 0.26}$

$\Rightarrow$ $\tt{pK_a =\: 5 - \dfrac{26}{100}}$

$\Rightarrow \tt{pK_a =\: \dfrac{500 - 26}{100}}$

$\Rightarrow \tt{pK_a =\: \dfrac{474}{100}}$

$\Rightarrow$$\small{\bf{\underline{\underline{pK_a =\: 4.74}}}}$

∴ The pKa value is 4.74.