Question

As discussed in class, water plays an important biological role and therefore knowing the properties of water plays a role in understanding pH and its impact on biological structures. The ionization constant for water, Kw, is 1.14 x 10–15 at 0 °C and 9.6 x 10–14 at 60 °C. Determine the pH of water at 60 °C.

Answer 1

Answer:

Explanation:

H2O --> H+ + OH-

Ka = [H+][OH-]/[H2O]

Ka × [H2O] = Kw

Kw = [H+] × [OH-]

[H+] = [OH-] = x

Therefore,

9.6 x 10^-14 = x^2

x = 3.098 × 10^-7

[H+] = 3.098 × 10^-7 M

pH = -log[H+]

= -log [3.098 × 10^-7]

= 6.51

Answer 2

Answer:

6.5

Explanation:

Since pure water is neutral, the hydrogen ion concentration equals the hydroxide ion concentration.

[H⁺] = [OH⁻] and pK = pH + pOH. Since [H⁺] = [OH⁻], pH = -log[H⁺] = pOH = -log[OH⁻].

So, pK = 2pH  and K = ionisation constant for water at 60 °C = 9.6 × 10⁻¹⁴

pH = pK/2 = -logK/2 = - log(9.6 × 10⁻¹⁴)/2 = (14 - log9.6)/2 = 6.5

So, pH of water at 60 °C is 6.5