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

The answer is "3".

Step-by-step explanation:

acetic acid Dissociation:

`CH_3COOH \ < - - - - - - - - - - - - - \ > CH_3COO^(-)+H^(+)\\\\`

Dissociation constant of the Ka:

`Ka = ([CH3COO^(-)] [H^(+)])/([CH_3COOH])\\\\`

using the ICE table:

`CH_3COOH \ < - - - - - - - - - - - - - \ > CH_3COO^(-)+H^(+)\\\\`
`\to 1.8 * 10^(-5) =(( x * x) )/(0.1-x) -------(2)`

x is negligible compared to Ka.

`1.8* 10^(-5) = (x^2)/(0.10)\\\\x^2 = 1.8 * 10^(-6)\\\\x = 1.34 * 10^(-3)\\\\pH = -\log [H^(+)]`

From the ICE table,
`[H^(+)] = x = 1.34 * 10^(-3)`

`\to pH = -\log(1.34 * 10^(-3)) = 3`