Question

What happens to acetic acid when heated up? And what's the chemical equation?

I'm trying to calculate Ka value with the information. I have the pH, and temperature and Concentration.

Answer 1

The degree of dissociation decreases slightly.

Explanation:

You are trying to discover the relation between `K_text(a)` and temperature.

For this, you need the van't Hoff equation.

`color(blue)(bar(ul(|color(white)(a/a)ln(K_2/K_1) = -(ΔH^@)/R(1/T_2 - 1/T_1)color(white)(a/a)|)))" "`

where

`K_1` and `K_2` are the `K_text(a)` values at temperatures `T_1` and `T_2`
`ΔH` is the enthalpy of ionization
`R` is the universal gas constant

For acetic acid, `K_text(a) = 1.75 × 10^"-5"` and `ΔH = "-510 J/mol"`.

Question

At 25°C, `K_text(a) = 4.76 × 10^"-5"` and `ΔH = "-0.51 kJ·mol"^"-1"`. What is the value of `K_text(a)`
at 75 °C?

Answer

`ln(K_2/K_1) = -(ΔH^@)/R(1/T_2 - 1/T_1)`

`= -("-510" color(red)(cancel(color(black)("J·mol"^"-1"))))/(8.314 color(red)(cancel(color(black)("J·K"^"-1""mol"^"-1"))))(1/(348.15 color(red)(cancel(color(black)("K")))) - 1/(298.15 color(red)(cancel(color(black)("K"))))) = 61.3 × ("-4.817 × 10"^"-4") = "-0.0295"`

`K_2/K_1 = 10^"-0.0295" = 0.934`

`K_2 = 0.934K_1 = 0.934 × 1.75 × 10^"-5" = 1.63 × 10^"-5"`

Question

How would the pH of a 0.1 mol/L solution of acetic acid change from 25 °C to 75 °C?

Answer

The calculations are straightforward.

The pH of a 0.1 mol/L acid with `K_text(a) = 1.75 × 10^"-5"` is 2.88.

The pH of a 0.1 mol/L acid with `K_text(a) = 1.63 × 10^"-5"` is 2.90.

The difference of 0.02 pH units is so small that you may have missed it in your measurements.