The degree of dissociation of acetic acid in a 0.1 M solution is 1.0 × 10^(–2) . The pK_a of acetic acid value.

Correct answer: (C) 5. K_a=(Cα^2)/(1-α)=(0.1×10^(-2))/(21-0.01)=1.01×10^(-5) pK_a=-log⁡1×10^(-5)=5

Q25 / 102 · NTA Abhyas (JEE Main) 2020 · Chemistry · Ionic Equilibrium

The degree of dissociation of acetic acid in a $0 .1 M$ solution is $1.0 \times 1 0^{- 2}$ . The $p K_{a}$ of acetic acid value.

K_{a} = \frac{C α^{2}}{1 - α} = \frac{0.1 \times {(1 0^{- 2})}^{2}}{(1 - 0.01)} = 1.01 \times 1 0^{- 5}

p K_{a} = - \log (1 \times 1 0^{- 5}) = 5

Hence, the correct answer is option: 'C'.