Question

calculate the pH of 0.01 M solution of CH3COOH Ka for CH3COOH at 298 K is 1.8 × 10-5​

Answer 1

Answer

The dissociation equilibrium of acetic acid :

$\sf CH_3COOH (aq) + H_2O(l) \longleftrightarrow H_3O^+ (aq) + CH_3COO^- (aq)$

Applying law of chemical equilibrium that is,

The ratio of concentration of products to reactants raised to powers of their stoichiometric coefficients at equilibrium in a balanced equation is called law of chemical equilibrium.

we get,

$\sf K_a = \dfrac{[H_3O^+][CH_3COO^-]}{[CH_3COOH]}$

$\sf since, [H_3O^+] = [CH_3COO^-]$

$\sf K_a = \dfrac{[H_3O^+] ^{2} }{[CH_3COOH]} \: \: \: \: ...(1)$

we know that,

[CH₃COOH] = 0.01M = 1 × 10¯²M.

K$\sf _{a}$ = 1.8 × 10¯⁵

Substituting the values in eq (1)

$\sf 1.8 \times {10}^{ - 5} = \dfrac{[H_3O^+] ^{2} }{1 \times {10}^{ - 2} }$

$\sf [H_3O^+] ^{2} = 1.8 \times {10}^{ - 5} \times 10 \times {10}^{ - 2}$

$\sf [H_3O^+] = \sqrt{ 1.8 \times {10}^{ - 7 }}$

$\sf [H_3O^+] = ( 1.8 \times {10}^{ - 7 }) ^{ \frac{1}{2} }$

$\sf [H_3O^+] = 4.242 \times {10}^{ - 4} molL^{-1}$

$\sf pH = -log [H_3O^+]$

$\sf pH = - [log \: 4.242 \times {10}^{ - 4} ]$

$\sf pH = - [log \: 4.242 - 4log10]$

$\sf pH = [4 - log4.242 ]$

$\sf pH = [4 - 0.6276]$

$\underline{ \boxed{ \sf pH = 3.37}}$