Question

Calculate the pH of a 0.10 M solution of acidic acid HC2H3O2(aq) at 25 °C. Ka for HC2H3O2 = 1.8×10−5 at 25 °C g

Answer 1

Answer:

The answer is "3".

Explanation:

acetic acid Dissociation:  

$CH_3COOH \ < - - - - - - - - - - - - - \ > CH_3COO^{-}+H^{+}$

Dissociation constant of the Ka:

$Ka = \frac{[CH3COO^{-}] [H^{+}]}{[CH_3COOH]}$

using the ICE table:

$CH_3COOH \ < - - - - - - - - - - - - - \ > CH_3COO^{-}+H^{+}$ $\to 1.8 \times 10^{-5} =\frac{( x \times x) }{0.1-x} -------(2)$

x is negligible compared to Ka.

$1.8\times 10^{-5} = \frac{x^2}{0.10}\\\\x^2 = 1.8 \times 10^{-6}\\\\x = 1.34 \times 10^{-3}\\\\pH = -\log [H^{+}]$

From the ICE table, $[H^{+}] = x = 1.34 \times 10^{-3}$

$\to pH = -\log(1.34 \times 10^{-3}) = 3$