At 25°, Ionization constant of $CH_3COOH$ is $1.8 × 10^-5$ for 2M $CH_3COOH$ solution calculate,

Question:

At 25⁰, Ionization constant of CH­₃COOH is 1.8 x 10^-5 for 2M CH₃COOH solution calculate,

(a) Degree of ionization

(b) Molar Concentration of H⁺ ions in equilibrium

(c) P^H of the solution

Solution:

(a) The dissociation equilibrium is CH₃​COOH⇌CH₃​COO⁻+H⁺.
Let α be the degree of dissociation.

The equilibrium concentrations of CH₃​COOH, CH₃​COO⁻ and H⁺ are c(1−α), c(α) and c(α) respectively.

The equilibrium constant expression is \[\begin{array}{l}{K_c} = \frac{{[C{H_3}CO{O^ - }\;][{H^ + }]}}{{[C{H_3}COOH]}}\\{K_c} = \frac{{(c\alpha )(c\alpha )}}{{c(1 - \alpha )}} \approx c{\alpha ^2}\\\alpha  = \sqrt {\frac{{{K_c}}}{c}}  = \sqrt {\frac{{{\bf{1}}.{\bf{8}}{\rm{ }}{\bf{x}}{\rm{ }}{\bf{1}}{{\bf{0}}^{ - {\bf{5}}}}}}{2}}  = 3 \times {10^{ - 3}}\end{array}\]

Thus, Degree of Ionization is 3 x 10^-3.

(b) [CH₃​CO⁻]=[H⁺]=cα=2×3 x 10^-3=6 × 10⁻⁶M

Thus molar concentration of H⁺ is 6 × 10⁻⁶M

(c) p^H=−log[H⁺]=−log(6 × 10⁻⁶)=5.22

Thus P^H of Solution is 5.22