Question

A) The ionization constant for water (Kw) is 5.476 x 10-14 at 50 oC. Calculate [H3O+], [OH-], pH, and

pOH for pure water at 60oC.
b) Look at the following table with temperature, Kw, and pH values for pure water.
Temp. (oC)
Kw
pH
0
0.114 x 10-14
7.47
10
0.293 x 10-14
7.27
20
0.681 x 10-14
7.08
25
1.008 x 10-14
7.00
40
2.916 x 10-14
6.77
60
9.550 x 10-14
6.51
100
51.3 x 10-14
6.14
Notice that the pH falls as the temperature increases.
i) Does this mean that pure water becomes more acidic as the temperature rises? _______
ii) The reason for my answer is in pure water…(choose the correct letter from 1 - 5)._______
1. the pH becomes less than 7 as the temperature rises.
2. there are less H3O+ ions created as the temperature increases.
3. there are more H3O+ ions created as the temperature increases.
4. there are always the same # of hydronium and hydroxide ions even if the pH changes.
5. the OH- ions are evaporated as the temperature rises.

Answer 1

A) To solve this problem, we can use the formula for the ionization constant of water:

Kw = [H3O+][OH-]

At 50 oC, Kw = 5.476 x 10-14. We can assume that [H3O+] and [OH-] are equal since we are dealing with pure water.

Therefore,

[H3O+] = [OH-] = sqrt(Kw) = sqrt(5.476 x 10-14) = 7.40 x 10-8 mol/L

pH = -log[H3O+] = -log(7.40 x 10-8) = 7.13

pOH = -log[OH-] = -log(7.40 x 10-8) = 7.13

To find the values for pure water at 60oC, we can use the new value of Kw at that temperature:

Kw = 9.550 x 10-14

[H3O+] = [OH-] = sqrt(Kw) = sqrt(9.550 x 10-14) = 3.09 x 10-7 mol/L

pH = -log[H3O+] = -log(3.09 x 10-7) = 6.51

pOH = -log[OH-] = -log(3.09 x 10-7) = 6.51

b) i) No, it does not mean that pure water becomes more acidic as the temperature rises.

ii) The correct answer is 2. As the temperature increases, the ionization of water increases and more H3O+ and OH- ions are formed. However, since the concentration of H2O is also decreasing due to the increase in temperature, the increase in ionization does not result in an increase in [H3O+] and pH actually decreases.