Answer to Question #337511 in Chemistry for muhammad

Question

Ka for ethanoic acid is 1.7 × 10^–5 mol dm^–3 at 25 °C.

What is the pH of a 0.125 mol dm^–3 solution of ethanoic acid at this temperature?

Answer 1

Solution:

ethanoic acid ≡ acetic acid ≡ HC₂H₃O₂

K_a for ethanoic acid is 1.7×10^–5 mol dm^–3

The balanced equation for the ionization of HC₂H₃O₂ is

HC₂H₃O₂(aq) + H₂O(l) ⇌ C₂H₃O₂⁻(aq) + H₃O⁺(aq)

Summarize the initial conditions, the changes, and the equilibrium conditions in the following ICE table:

Substitute the equilibrium concentrations into the expression for the acid ionization constant K_a:

K_a = [C₂H₃O₂⁻] × [H₃O⁺] / [HC₂H₃O₂] = (x) × (x) / (0.125 − x)

Make the approximation that 'x' is small (0.125 >> x), substitute the value of the acid ionization constant into the K_a expression and solve for 'x':

1.7×10^–5 = (x²) / (0.125)

x² = 2.125×10^–6

x = 1.458×10^–3

The ratio of 'x' to the number it was subtracted from and added to in the approximation is less than 0.05; therefore, the 'x' is small approximation is valid, and [H₃O⁺] = x = 1.458×10^–3 M.

Finally, calculate the pH as the negative of the base-10 logarithm of [H₃O⁺]:

pH = −log[H₃O⁺] = −log(1.458×10^–3) = 2.836 = 2.84

pH = 2.84

Answer: The pH of ethanoic acid solution is 2.84