Question

The K_a for acetic acid drops from 1.76x10^-5 at 25°C to 1.63x10^-5 at 50°C. Between the same two temperatures, K_w increases from 1.00x10^-14 to 5.47x10^-14. At 50°C the density of a .10M solution of acetic acid is 98.81% of its density at 25°C. Will the pH of a .10 M solution of acetic acid in water increase, decrease, or remain the same as the solution is heated from 25°C to 50°C?

Answer 1

We know that

pH = -log[H+]

so the pH increases with decrease in concentration of H+ or vice versa

Now ew need to find out that the H+ decreases on increases with the change in temperature

a) the contribution of H+ by acetic acid is

[H+] = (Ka X concentration of acetic acid) ^1/2

[H+] at 25 C

[H+] = (1.76 X 10^-5 X 0.1) ^1/2 = 1.33 X 10^-3

[H+] at 50 C

The density decreases from 100 to 98.81

V2/ V1 = D1 / D2

V2 = 100 X V1 / 98.81 = 1.012 V1

So concentration will be = Moles / volume = 0.1 X V1 / 1.012 V1 = 0.0988

[H+] = (1.63 X 10^-5 X 0.0988) ^1/2 = 1.269 X 10^-3

b) contribution of H+ due to water

Initally [H+] = 10^-7

At 50 C

[H+] = (5.47 X 10^-14)^1/2 = 2.33 X 10^-7

c) at 25 C

The total [H+] = 10^-7 + 1.33 X 10^-3 = 1.3301 X 10^-3

pH = 2.87

d) at 50 C

The total [H+ ] = 2.33 X 10^-7 + 1.269 X 10^-3  

pH = 2.89

so it will increase slightly